UNIVERSITY OF DEFENCE
BRNO 2011
Jaromír Kuben
Integral Calculus for
Functions of a Single Variable
Prepared under the project of the Education for Competitiveness Operational Programme
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CZ.1.07/2.2.00/07.0256
Innovation of the Military Technology Study Programme/
Extension of Teaching Special Courses in English
This project is co-financed from the European Social Fund
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Kuben Jaromír
Integral Calculus for Functions of a Single Variable
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Preface
The integral is one of the central concepts of mathematics and mathematical analysis. Its development
was motivated by a number of factors, including the following two problems:
1. finding a function whose derivative is known,
2. determining the area of a region defined by the graph of function f on interval ha, bi, the x -axis, and
the lines x = a and x = b.
These two problems led to the concepts of indefinite and definite integrals. The investigation of
the properties, and the evaluation of these forms of interrelated integrals, are the rationale of integral
calculus.
With the development of mathematics and the growing demands of the natural sciences and technology for mathematical solutions, the concept of the integral evolved and was subject to many changes and
generalizations. Gradually new and more general integrals appeared that provided improved solutions to
the aforementioned problems.
Looking into contemporary textbooks of differential and integral calculus of functions of one variable
we find that they usually begin with descriptions of real numbers. This is followed by limits, the use of
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limits to define the derivative and only then is the indefinite and finally the definite integral explored.
These concepts did not arise in this order historically. The concept of the definite integral (evaluating
areas and volumes) was developed initially, followed in the 17th century by the derivative and indefinite
integral, which were based on intuitive understandings of infinitely small and big quantities. It was not
until a century had elapsed that the concept of limit was elaborated and only in the 19th century was the
theory of real numbers posited.
This textbook is addressed to undergraduate students studying mathematics in English, at the Faculty
of Military Technology of the University of Defence. It contains an exposition of integral calculus of
functions of a single variable which, together with differential calculus, underlie math eduction in the
“Ing.” degree programme. A knowledge of integral calculus is a prerequisite for further studies in areas
of mathematics such as differential equations, integral calculus of functions of several variables, vector
analysis, integral transforms and many others. It provides one of the critical building blocks that support
mechanics, physics, and other technical disciplines.
The text is divided into four chapters. The first is devoted to the indefinite integral and the second
to the Riemann definite integral. This ordering facilitates easier explanation. However, it is possible to
begin studies with Sections 1.1 and 1.2 of Chapter 1, then introduce the definite integral, explain its basic
properties and the Fundamental Theorem of Calculus, and return back to the rest of Chapter 1. The various applications found in Chapter 2 can then be addressed. This approach simplifies the organization of
tutorials and enables early practice in the evaluation of definite integrals. Chapter 3 involves the improper
integral. Chapter 4, which is the shortest, provides a brief introduction to the numerical evaluation of the
definite integral.
As this text is designed for students of technical universities numerous statements are not proved. In
Chapter 1, where proofs are not too difficult, almost all results are proved. In Chapter 2, on the other
hand, almost no proofs are provided because they are technically difficult and unnecessary to a basic
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understanding of the subject. In the remaining chapters some simple statements are proved. Extensive
drilling in manual integration is of decreasing importance with the arrival of very helpful modern computer algebra programs. Students really only require a good understanding of how the concepts work in
order to employ these programs properly. However, without this grounding the discovery and explanation of errors, that these programs do make if they are not handled properly, is difficult. For this reason
a great deal of attention is paid to thoroughly defining new concepts, their proper understanding, and
the formulation of mathematical theorems. To improve visualization of the concepts the text contains
numerous illustrative graphs and figures.
Many examples are solved in detail in the text, to help students understand the material. Student
exercises are placed after each topic to facilitate a stronger understanding. Independent work on these
exercises is a critical component of study and the only way in which students will gain the necessary
skill in calculation and fully appreciate the new concepts. To simplify checking results all exercises are
provided solutions. Three self-tests and twelve interactive quizzes are included to help students check
their knowledge. To improve orientation in the text, the ends of proofs are denoted by the symbol and
ends of examples by the symbol N.
There is a long list of textbooks devoted to the integral of functions of a single variable. All proofs
missing in this text can be found in [21] unless a different source is mentioned. [9] is a classical Czech
monography. Likewise it is instructive to read [22]—the first modern Czech textbook of Integral Calculus.
Although the language has become outdated its contents are remarkable. It is interesting to compare
what was emphasized in the exposition of this topic almost hundred years ago and what is emphasized
nowadays. [18] is also worth recommending and for Russian speakers the classical textbook [5].
I thank the reviewers Prof. RNDr. Zuzana Došlá, DSc. and Doc. RNDr. Jaromír Šimša, CSc. of the
Department of Mathematics and Statistics of the Faculty of Science, Masaryk University Brno for their
invaluable comments and contribution to the quality of the text. I owe special thanks to my colleague
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PhDr. Pavlína Račková, Ph.D. for all her help checking the exercises and preparing the source code of
the quizzes. Likewise I thank my colleague Doc. RNDr. Šárka Mayerová, Ph.D., who cooperated on
the preparation of exercises in the Czech edition. Finally, I thank Robert Brukner, MA, for his thorough
language correction of the text.
The text was prepared with the pdfTEX typesetting system using the LATEX 2ε format. Most figures
were created with METAPOST using the TEX macro package mfpic. Two figures were prepared in
Maple.
The translation of the textbook into English was prepared with the support of ESF under the Innovation of the Military Technology Study Programme (reg. num. CZ.1.07/2.2.00/07.0256) of the University
of Defence, Brno. Aside from this electronic, fully hypertexted version, hard copy [13] also exists.
Brno, June 2011
Jaromír Kuben
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Contents
Preface
1
3
Indefinite Integral
1.1 Antiderivatives and the Indefinite Integral
1.2 Basic Integration Methods . . . . . . . .
1.2.1 Basic Integration Formulae . . .
Exercises . . . . . . . . . . . . . . . . .
Answers to Exercises . . . . . . . . . . .
Quiz—Basic Integration Formulae . . . .
1.2.2 Integration by Parts . . . . . . .
Exercises . . . . . . . . . . . . . . . . .
Answers to Exercises . . . . . . . . . . .
Quiz—Integration by Parts . . . . . . . .
1.2.3 Substitution Method . . . . . . .
Exercises . . . . . . . . . . . . . . . . .
Answers to Exercises . . . . . . . . . . .
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Quiz—Substitution Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
1.3 Decomposition into Partial Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . 79
1.4 Integration of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
1.4.1 Integration of Partial Fractions with Real Roots in the Denominator . . . . . . . 86
1.4.2 Integration of Partial Fractions with Complex Roots in the Denominator . . . . 91
1.4.3 Integration of Partial Fractions with real and Complex Roots in the Denominator 99
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
Quiz—Indefinite Integrals of Rational Functions . . . . . . . . . . . . . . . . . . . . 109
1.5 Integration of Some Special Types of Functions . . . . . . . . . . . . . . . . . . . . . 112
1.5.1 Integrals of Rational Expressions of Trigonometric Functions . . . . . . . . . . 112
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
Quiz—Indefinite Integrals of Trigonometric Functions . . . . . . . . . . . . . . . . . 132
1.5.2 Integrals Involving Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148
Quiz—Indefinite Integrals of Functions Involving Roots . . . . . . . . . . . . . . . . . 150
1.6 Final Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
1.6.1 Is the Result of Integrating an Elementary Function Again an Elementary Function?153
1.6.2 Gluing Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
1.7 Final Exercises to Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160
Answers to Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
Quiz—Indefinite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
Self-Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171
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Answers to Self-Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172
2
Definite Integral
2.1 Construction of the Definite Integral . . . . . . . . . .
2.2 Existence of the Definite Integral . . . . . . . . . . .
2.3 Properties of the Definite Integral . . . . . . . . . . .
2.4 Evaluation of the Definite Integral . . . . . . . . . . .
2.4.1 Integration by Parts for the Definite Integral . .
2.4.2 Substitution Method for the Definite Integral .
2.4.3 The Definite Integral as the Function of Limits
Exercises . . . . . . . . . . . . . . . . . . . . . . . .
Answers to Exercises . . . . . . . . . . . . . . . . . .
2.5 Applications of the Definite Integral . . . . . . . . . .
2.5.1 Geometric Applications . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . .
Answers to Exercises . . . . . . . . . . . . . . . . . .
Quiz—Length of a Curve . . . . . . . . . . . . . . .
Quiz—Area of a Plane Figure . . . . . . . . . . . . .
Quiz—Volume of a Solid of Revolution . . . . . . . .
2.5.2 Physical Applications . . . . . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . . .
Answers to Exercises . . . . . . . . . . . . . . . . . .
Quiz—Definite Integral . . . . . . . . . . . . . . . .
Self-Test . . . . . . . . . . . . . . . . . . . . . . . . . . .
Answers to Self-Test . . . . . . . . . . . . . . . . . .
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3
4
Improper Integral
3.1 Improper Integral on Unbounded Intervals . . . . . .
3.2 Improper Integral of Unbounded Functions . . . . .
3.3 Generalization of Improper Integral . . . . . . . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . .
Answers to Exercises . . . . . . . . . . . . . . . . .
Quiz—Improper Integral . . . . . . . . . . . . . . .
3.4 Convergence Tests of Improper Integrals . . . . . . .
3.4.1 Convergence Tests of Nonnegative Functions
3.4.2 Absolute and Conditional Convergence . . .
Exercises . . . . . . . . . . . . . . . . . . . . . . .
Answers to Exercises . . . . . . . . . . . . . . . . .
Self-Test . . . . . . . . . . . . . . . . . . . . . . . . . .
Answers to Self-Test . . . . . . . . . . . . . . . . .
Numerical Methods of Solving Definite Integrals
4.1 Rectangle method . . . . . . . . . . . . . .
4.2 Trapezoidal method . . . . . . . . . . . . .
4.3 Simpson’s rule . . . . . . . . . . . . . . . .
4.4 Exercises to Chapter 4 . . . . . . . . . . . .
Answers to Exercises . . . . . . . . . . . . .
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11
Chapter 1
Indefinite Integral
In previous courses you were introduced to an important concept concerning the derivative of a function.
A new function f 0 was assigned to a function f in some way. The number f 0 (x) can have various
interpretations depending on the value of f . For example, geometrically the value f 0 (x) is defined as
the slope of the tangent line to the graph of function f at a point [x, f (x)], that is, it is the tangent of
an angle contained by the tangent line and the positive part of the x -axis. If we consider a function f
of time which represents the position of a particle moving in a straight line, the number f 0 (t) is the
instantaneous velocity of this particle at time t . If we denote g = f 0 the instantaneous velocity of the
particle, the number g 0 (t) = f 00 (t) is the instantaneous acceleration of this particle at time t , etc. In
general, the value f 0 (x) represents the “rate of change” of the function f depending on the change of
the independent variable x . The greater the value of f 0 (x), the steeper the increase of function f near
point x and vice versa.
The problem we will investigate in this chapter is the converse. For a given function f we will be
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Indefinite Integral
12
looking for a function F , such that F 0 = f will be valid. Thus we will ask which function F we must
differentiate to obtain f . Therefore, with the knowledge of tangent lines to the graph of a function we
will strive to find this function. With the knowledge of the instantaneous velocity of a particle we will
strive to find out the position of this particle, and with the knowledge of the instantaneous acceleration
of a particle we will strive to determine its instantaneous velocity, etc.
In this chapter we will concentrate primarily on the following considerations:
• If function F exists,
• If more functions with this property can exist,
• How to find such a function for a concrete elementary function f .
The answers to the first two questions will have a more theoretical character. For the third question,
to which we will give most attention, we will be interested in practical methods of finding function F .
1.1 Antiderivatives and the Indefinite Integral
Definition 1.1 Let f be a function defined on the interval I . Function F is called the antiderivative,
or primitive function, of f on I if F 0 (x) = f (x) for all x ∈ I .
R
The set of all antiderivatives of function f is called the indefinite integral of f and is denoted f (x) d x .
Thus
Z
f (x) d x = {F : F is an antiderivative of f on I }.
(1.1)
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If interval I in the previous definition is not open, the derivatives at the end points are one-sided
derivatives.
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Indefinite Integral
13
R
The symbol used for the indefinite integral, , arose historically by stretching the letter S, the first
letter of the word sum (the connection will be clarified in Chapter 2). Function f is the integrand. The
term d x is the differential of variable x and at this moment its only meaning is that it tells us which letter
is used for the independent variable. Later on we will see that it simplifies, for example, the mechanism
of evaluating integrals, when a substitution method is used.
Let us try to find an antiderivative to the function cos x , x ∈ R. It is not too difficult to ascertain
that such a function may be, for example, F (x) = sin x as (sin x)0 = cos x . But function sin x + 3
also satisfies (sin x + 3)0 = cos x , thus sin x + 3 is also the antiderivative of cos x . Therefore, all the
functions sin x + c, where c ∈ R is an arbitrary constant, are antiderivatives of cos x .
In general, the next statement is therefore valid: If F is an antiderivative of f on the interval I , then
the functions F + c, where c ∈ R is an arbitrary constant, are also antiderivatives of f on I . Thus if
f has at least one antiderivative, then it has infinitely many antiderivatives. The natural question occurs
whether or not these are all antiderivatives of f . The answer is given by the next statement.
Theorem 1.2 If F is an antiderivative of f on the interval I , then any other antiderivative of f on I
has the form F + c, where c ∈ R.
Proof. Let F and G be two antiderivatives of f . Thus F 0 (x) = G0 (x) = f (x) for any x ∈ I . Since I is
the interval by virtue of the well-known Corollary of Lagrange1 Mean Value Theorem (see [14, p. 233])
these functions differ by a constant, that is there exists c ∈ R such that G(x) = F (x) + c for any x ∈ I ,
which proves the statement.
In other words, the preceding result states that if we know one antiderivative, we know all the antiderivatives. The difference of two antiderivatives is constant on interval I .
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1 Joseph Louis Lagrange (1736–1813) (read
lagraNZ)—notable French mathematician and physicist. He was involved
in many branches of mathematics. Among other contributions, he influenced the development of mathematical analysis and
founded the calculus of variations.
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Indefinite Integral
14
Let us emphasize the fact that I , being an interval, is substantial. If I is not an interval, it is possible for
= G0 but F − G is not constant on I . For example, the function F (x) = sgn x , considered on the set R r {0},
equals −1 on the interval (−∞, 0) and 1 on the interval (0, +∞), therefore its derivative is zero at any point of
the set R r {0}—see Fig. 1.2. Another function with zero derivative on this set is, for example, G(x) = 0. But the
difference F (x) − G(x) = sgn x is not constant on the set R r {0}. Nevertheless, it is constant on each interval
(−∞, 0) or (0, +∞) if they are considered stand-alone, which agrees with Theorem 1.2. (Let us remember that
the concept of antiderivative was introduced only for functions on intervals.)
F0
Keeping in mind the previous result we can modify Formula (1.1). If F is an antiderivative of f ,
then
Z
f (x) d x = F (x) + c,
where c ∈ R.
(1.2)
The number c is called the integration constant. We say that the indefinite integral is determined up to a
constant.
In fact, the expression on the right-hand side of Equality (1.2) should be put in braces, that is {F (x)+
+ c, c ∈ R}, as it is a set, but this notation is not used. So Equality (1.2) means that all antiderivatives
of f are F + c, where F is one fixed (arbitrarily chosen) antiderivative of f and c is any constant. For
example, if f (x) = cos x , we may choose F (x) = sin x as the fixed antiderivative. Then
Z
cos x d x = sin x + c,
where c ∈ R.
The situation is illustrated in Fig. 1.1. Graphs of particular antiderivatives are translated respect to
one another in the direction of the y -axis. For each x the tangent lines to the graphs of functions F + c
at the points [x, F (x) + c] are parallel for any c ∈ R, thus they have the same slope, which corresponds
to the fact that all antiderivatives F + c have the same derivative f (x), at point x . In the figure tangent
lines at two fixed points x0 and x1 are displayed.
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Indefinite Integral
y
y = F (x) + 2.5
y = F (x) + 1.5
y = F (x) = sin x
x
O
x0
x1
y = F (x) − 1.4
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Fig. 1.1: Antiderivatives of cos x
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Indefinite Integral
Now we will tackle the question of whether for a given function f there
exists any antiderivative. In general the answer is negative. For example it
is possible to verify that the function sgn x defined as follows:


−1 for x < 0,
sgn x =
0 for x = 0,


1 for x > 0,
y = sgn x
y
1
O
x
−1
whose graph is in Fig. 1.2, has no antiderivative on the interval (−∞,
+∞). We will not prove this fact. (Function sgn x does not have the
Fig. 1.2
Darboux1 property on R , which is a necessary condition for the existence
of an antiderivative (see, e.g., [4, p. 187]). Simply: if f is the derivative
of function F , it cannot happen that at some point x0 the left-hand (right-hand) limit of f exists and is
different from f (x0 )).
Fortunately, a simple sufficient condition exists that guarantees the existence of an antiderivative:
Theorem 1.3 If a function f is continuous on the interval I , then it has an antiderivative on I .
We will not prove this statement as we do not have necessary tools to do so. In Chapter 2 we discuss
how this antiderivative can be constructed (Corollary 2.29).
The previous statement is a typical example of the existence theorem, which states that something
exists, though it does not say, how to find it. Later on we will discuss this problem, which substantially
complicates the situation of finding antiderivatives (see Section 1.6).
1 Jean Gaston Darboux (1842–1917) (read
darbU)—French mathematician involved in differential geometry and mathe-
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matical analysis.
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Indefinite Integral
17
At the end of this section we formulate a simple but very important statement, which will be used
many times in the following text in the evaluation of indefinite integrals.
R
R
Theorem
1.4
If
the
integrals
f
(x)
d
x
and
g(x) d x exist on an interval I , then the integrals
R
R
(f (x) ± g(x)) d x and αf (x) d x , where α ∈ R is any constant, also exist on I and the following
equalities hold:
Z
Z
Z
f (x) ± g(x) d x = f (x) d x ± g(x) d x,
(1.3)
Z
Z
αf (x) d x = α f (x) d x.
(1.4)
Proof. The statement is the easy consequence of the basic properties of the derivative. If F is an antiderivative of f and G is an antiderivative of g , then (F (x) ± G(x))0 = F 0 (x) ± G0 (x) = f (x) ± g(x)
for any x ∈ I , thus F ± G is an antiderivative of f ± g Similarly (αF (x))0 = αF 0 (x) = αf (x) for
any x ∈ I , thus αF is an antiderivative of αf .
In short, we say that the “indefinite integral of the sum (difference) is the sum (difference) of indefinite
integrals” and that “we might take out a constant by which we multiply (multiplicative constant) from
the indefinite integral”. Evidently, the first statement can be easily extended from two summands, to any
finite number of summands. Formula (1.3) is called the sum rule and Formula (1.4) the constant rule.
Note also, that from the point of view of the existence, the Formulae (1.3) and (1.4) must be read
from right to left—integrals on the right-hand sides must exist; then the integrals on the left-hand sides
also exist, and the corresponding equalities are valid.
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Indefinite Integral
18
Finally, let us remember that from the definition of the indefinite integral we get
Z
0
Z
f (x) d x = f (x)
and
F 0 (x) d x = F (x) + c, c ∈ R,
therefore the operations of differentiation and integration are mutually complementary. The correctness
of the result of integration can be checked each time by differentiating it, from which we must obtain
(often after some rearrangement) the given integrand.
Comment 1.5 Let us take note of Formula (1.3). On the right-hand side we have in fact the sum, or difference of
two
how such a sum is to be understood. We add any element of the set
R infinite sets. Let us explain, for example,
R
f (x) d x with any element of the set g(x) d x . The result will be the set of all such sums. But any element of
the first indefinite integral is F (x) + c1 , c1 ∈ R, and any element of the second indefinite integral is G(x) + c2 ,
c2 ∈ R, where F (x) and G(x) are fixed antiderivatives of f (x) and g(x), respectively. Thus the result is the
set containing functions F (x) + G(x) + c1 + c2 , where c1 and c2 are any independently chosen real numbers.
Therefore, it is the set containing all functions F (x) + G(x) + c, where c is any real number, which is exactly the
set on the left-hand side of Formula (1.3). The difference
R is treated in a similar way.
Analogously, in Formula (1.4) the multiple of set f (x) d x by constant α on the right-hand side will be made
in such a way that any element of this set will be multiplied by the constant α . The elements of the set obtained
are all functions αF (x) + αc, where c is any real constant, which is (for α 6 = 0) the same set as the set on the
left-hand side containing all functions αF (x) + c, where c is any real constant.
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1.2 Basic Integration Methods
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In this section we will learn how to integrate some simple functions that we encounter in common applications. Remember that by elementary functions we mean power functions, exponential and logarithmic
functions, trigonometric and inverse trigonometric functions, hyperbolic and inverse hyperbolic functions, and all the other functions we get from these by applying a finite number of arithmetic operations
(addition, subtraction, multiplication, and division) and compositions of functions.
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Indefinite Integral
Z
0 d x = c,
1.
Z
d x = x + c,
2.
Z
3.
Z
4.
Z
xn dx =
x n+1
+ c,
n+1
where n ∈ R, n 6 = −1,
Z
1
d x = ln |x| + c,
x
x
in general
Z
x
e d x = e + c,
5.
Z
6.
ax d x =
ax
+ c,
ln a
in general
Z
10.
Z
cos x d x = sin x + c,
9.
sin ax d x = −
in general
Z
Z
1 ax
e + c,
a
Z
sin x d x = − cos x + c,
8.
eax d x =
a > 0,
Z
7.
1
d x = ln |x + a| + c,
x+a
1
d x = arctan x + c,
2
x +1
1
cos ax d x =
in general
Z
in general
x2
Z
√
d x = arcsin x + c,
1 − x2
in general
1
cos ax + c,
a
1
sin ax + c,
a
1
1
x
d x = arctan + c,
2
+a
a
a
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1
x
√
d x = arcsin + c,
2
2
a
a −x
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Tab. 1.1: Table of indefinite integrals—Part 1
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20
Indefinite Integral
Z
p
1
√
d x = lnx + x 2 + a + c,
x2 + a
Z
1
d x = tan x + c,
cos2 x
in general
1
d x = − cot x + c,
sin2 x
in general
11.
12.
Z
13.
Z
14.
Z
1
1
d x = tan ax + c,
2
cos ax
a
Z
1
1
d x = − cot ax + c,
2
a
sin ax
f 0 (x)
d x = ln |f (x)| + c.
f (x)
Tab. 1.2: Table of indefinite integrals—Part 2
In the above table a means any non-zero number, i.e. a ∈ R r {0}, apart from Formula 6. c ∈ R is the integration
constant. The formulae are valid on intervals on which both sides are defined.
1.2.1 Basic Integration Formulae
The first group of formulae can be obtained if we reverse basic formulae for the differentiation. After
minor rearrangements we get the Formulae No. 1–10, 12 and 13 in Table 1.1, to which two useful Formulae 11 and 14 are appended. The correctness of all the formulae can be verified by differentiating their
right-hand sides.
Before we demonstrate the use of the above formulae on examples, we begin with a few comments.
R
R
i) Formula 2 is the abbreviation
for 1 d x . In Formula 4 and other similar integrals the notation dxx
R 1
can be used instead of x d x , etc.
ii) Formula 3 enables the integration of power functions, therefore also various roots.
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21
Indefinite Integral
iii) As the derivatives of the inverse tangent and inverse cotangent differ only in the sign, Rand the same
is true for inverse sine and inverse cosine, the Formulae 9, 10 can be expressed as x 21+1 d x =
R
= − arccot x + c, √ 1 2 d x = − arccos x + c, respectively and analogously in their general
1−x
versions.
iv) In all formulae the independent variable is denoted by x . In concrete applications it need not be like
this. From the differential it can be seen how the independent
variable is denoted.
Then the formula
R
R
must
be modified in an adequate way; for example, cos x d x = sin x + c, cos t d t = sin t + c,
R
cos u d u = sin u + c, etc.
Sometimes, this simple modification causes problems for students. For this reason, students should
try to learn the formulae from Table 1.1 without the variable (if that is possible). For example,
— integral of sine equals minus cosine (Formula 7),
— integral of e to the variable equals the same expression (Formula 5),
— integral of one over the variable equals the natural logarithm of the absolute value of the variable
(Formula 4),
— integral of the variable to n equals the variable to n plus one over the same number (Formula 3).
Although it is sometimes rather difficult, the effort will pay off.
v) In the sequel, c added at the end of the evaluation of an indefinite integral, will mean the integration
constant, every time.
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vi) You should learn the formulae from Table 1.1 by heart. Otherwise, even if you have the table at
your disposal, you will be unable to match the right formula, in more complicated cases . If some
manipulation is necessary you will not guess which one to choose, as you will not be able to see in
obtained expressions the particular basic formulae. Do not believe that for successful integration it
is sufficient to have the table in front of you without knowing the formulae by heart.
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Indefinite Integral
Example 1.6 Evaluate the following indefinite integrals:
Z
Z
1
a)
x d x,
b)
d x,
x2
Z
Z
1
√
d)
d
x,
e)
e−x d x,
3
x
Z
Z
1
1
√
√
g)
d x,
h)
d x,
2
2
4−x
x −7
√
x d x,
Z
1
d x,
x2 + 3
Z
3x 2 + 1
d x.
x3 + x + 2
c)
f)
i)
Solution. The first four examples will be evaluated using Formula 3.
Z
x2
a)
x dx =
+ c (we set n = 1),
2
Z
Z
1
1
x −1
−2
b)
+ c = − + c (we set n = −2),
d
x
=
x
d
x
=
2
x
−1
x
Z
Z
√
x 3/2
2√ 3
c)
x d x = x 1/2 d x =
+c =
x + c (we set n = 1/2 ),
3/2
3
Z
Z
1
x 2/3
3√
3
√
d)
d
x
=
x −1/3 d x =
+c =
x 2 + c (we set n = −1/3 ).
3
x
2/3
2
e) In the next example we use Formula 5 for a = −1. We get
Z
e−x
e−x d x =
+ c = −e−x + c.
−1
Z
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Indefinite Integral
f) In this √
example we use Formula 9. As a 2 = 3, we set a =
a = − 3 but why complicate things). Then we have
Z
1
x
1
d x = √ arctan √ + c.
2
x +3
3
3
√
3 (it would be also possible to set
g) In this example we use Formula 10. As a 2 = 4 we choose a = 2. Hence we have
Z
1
x
√
d x = arcsin + c.
2
2
4−x
h) In this example we use Formula 11 for a = −7. We get
Z
p
1
√
d x = lnx + x 2 − 7 + c.
2
x −7
i) In the last example we use Formula 14. It is not difficult to notice that the derivative of the denominator
is (x 3 + x + 2)0 = 3x 2 + 1, which is the numerator. Thus
Z
3x 2 + 1
d x = ln |x 3 + x + 2| + c.
x3 + x + 2
N
In the following examples we apply Theorem 1.4, to help transform more complicated integrals for
the evaluation of several basic integrals.
Contents
Example 1.7 Evaluate the following indefinite integrals:
Z
a)
(2x 5 − x 4 + 3x 3 − 3x 2 + 2) d x,
Z c)
Z b)
2
x
7
4
2
− 3 sin 5x + 2 cos + 3x − x +
−
2
cos x
2
2
3−x
3x + 2
3
2
√
−√
4 − 3x 2
4 + 3x 2
+ 2 e2x/3 d x.
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d x,
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Indefinite Integral
Solution.
a) The integrand is a polynomial so it can be easily integrated using Formula 3 and Relations (1.3) and
(1.4):
Z
(2x 5 − x 4 + 3x 3 − 3x 2 + 2) d x =
Z
Z
Z
Z
Z
5
4
3
2
= 2 x dx − x dx + 3 x dx − 3 x dx + 2 dx =
=2
x6 x5
x4
x3
x 6 x 5 3x 4
−
+3
−3
+ 2x + c =
−
+
− x 3 + 2x + c.
6
5
4
3
3
5
4
b) We separate the integral into two parts and use Formulae 10 and 11.
Z Z
Z
3
2
dx
dx
√
√
−√
dx = 3
−2 √
.
2
2
2
4 − 3x
4 + 3x
4 − 3x
4 + 3x 2
(1.5)
As both integrands must be rewritten before the mentioned formulae can be applied, we evaluate
each integral separately for a better understanding (the integration constant will be added once both
integrals are evaluated).
Z
Z
Z
dx
dx
1
dx
p
p
√
=
=√
=
3
4 − 3x 2
3(4/3 − x 2 )
4/3 − x 2
√
√
√
1
x
1
x 3
3
x 3
= √ arcsin 2 = √ arcsin
=
arcsin
√
2
3
2
3
3
3
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Indefinite Integral
√
(in Formula 10 we have a 2 = 4/3, i.e. a = 2/ 3) and
Z
Z
Z
dx
dx
1
dx
p
p
√
=
=√
=
2
3
4 + 3x
3(4/3 + x 2 )
4/3 + x 2
p
1 = √ lnx + 4/3 + x 2 .
3
Notice that the functions differ only in one sign but their integrals are quite different.
From (1.5) we have
√
Z p
√
3
x 3
2 2
√
−√
d x = 3 arcsin
− √ lnx + 4/3 + x 2 + c.
2
2
2
3
4 − 3x
4 + 3x
c) We separate the integral and apply (eventually after some small modifications) the necessary formulae.
Z 2
x
7
4
2
x
2x/3
− 3 sin 5x + 2 cos + 3 − x +
−
+ 2e
dx =
cos2 x
2
2
3−x
3x + 2
Z
Z
Z
Z
dx
x
=2
− 3 sin 5x d x + 2 cos d x + 3x d x −
2
cos x
2
Z x
Z
Z
Z
1
dx
2
dx
−7
dx − 4
−
+ 2 e2x/3 d x =
2
x−3 3
x + 2/3
1 x
sin x2
− cos 5x
3x
2
= 2 tan x − 3
+2 1 +
−7
− 4 ln |x − 3| −
5
ln 3
ln 12
2
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2x/3
−
2
e
ln |x + 2/3| + 2
3
2
3
+c =
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Indefinite Integral
= 2 tan x +
3
x
3x
7
cos 5x + 4 sin +
+ x
− 4 ln |x − 3| −
5
2 ln 3 2 ln 2
2
ln |x + 2/3| + 3 e2x/3 + c.
3
N
Notice that integration constant must be appended once the last integral is evaluated and then copied
during the following rearrangements.
−
Example 1.8 Evaluate the following indefinite integrals:
Z
Z
a)
tan2 au d u, a 6= 0,
b)
tan bs d s,
Z
dt
.
sin t
Solution. Rewriting all three integrals in an appropriate way we can match them to the basic formulae.
Care must be taken of the notation of the independent variable, this time it is not x .
b 6= 0,
c)
a) Using the identity sin2 α + cos2 α = 1 valid for any α ∈ R and Formula 12 we get
Z
Z
Z
sin2 au
1 − cos2 au
2
tan au d u =
d
u
=
du =
cos2 au
cos2 au
Z Z cos2 au
1
1
−
du =
− 1 du =
=
cos2 au cos2 au
cos2 au
1
= tan au − u + c.
a
sin bs
b) We use Formula 14. As tan bs = cos
and the derivative (with respect to the variable s ) of its
bs
0
denominator is (cos bs) = −b sin bs , we miss −b in the numerator. This can be easily rectified with
the help of (1.4), because −b is a constant. We get
Z
Z
Z
sin bs
1
−b sin bs
1
tan bs d s =
ds = −
d s = − ln | cos bs| + c.
cos bs
b
cos bs
b
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Indefinite Integral
c) This time we will also use Formula 14 (twice), but after applying the well-known trigonometric identities sin2 α + cos2 α = 1 and sin 2α = 2 sin α cos α valid for any α ∈ R. We replace α by t/2.
Z
Z sin2 2t + cos2 2t
sin2 2t
cos2 2t
dt =
+
dt =
2 sin 2t cos 2t
2 sin 2t cos 2t
2 sin 2t cos 2t
Z Z
Z 1
cos 2t
sin 2t
cos 2t
− 12 sin 2t
2
=
+
d
t
=
−
d
t
+
dt =
2 cos 2t
2 sin 2t
cos 2t
sin 2t
sin 2t t t + c = lntan t + c,
= − lncos + lnsin + c = ln
2
2
cos 2t 2
dt
=
sin t
Z
where we added the missing −1 in the numerator as in the previous example.
N
In the previous examples we intentionally neglected the domains so as not to detract attention from
the integration itself. Some domains are easy to determine, some are more complicated. You should
never forget that the results are valid only on intervals on which all functions are defined.
For the sake of brevity, integration constants are not appended to the answers in exercises concerning
indefinite integrals.
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Indefinite Integral
Exercises
1. Find the indefinite integral of the given functions:
√
Z
Z 4
1
x
−1
3
a)
3x d x,
b)
x − +
d x,
x
2
Z
Z
3
7
d x,
d)
5x d x,
e)
4
Z
Z
g)
2.4x −0.16 d x,
h)
4x −3 d x,
Z
j)
1.5
d x,
x
Z k)
3.4
6
+√
3
3
x
x2
2. Find the indefinite integral of the given functions:
Z
Z 5
x + 2x 4 − x 2
3
a)
4x d x,
b)
d x,
x3
Z
Z √
d)
u−5 d u,
e)
z 2 d z,
Z
g)
Z
j)
5
dR,
R6
3
d t,
t
Z
h)
8m
3/5
x 12 d x,
Z
(x + 2)3
d x,
x3
Z
x −a d x, a 6 = 1,
Z
4
d u.
u2
f)
i)
d x,
Z
c)
l)
Z
c)
f)
3z
d z,
4
Z
√
3 ρ dρ,
Z
dm,
i)
x −t d t,
Contents
Z k)
3
1
+√
4
z
z
Z
d z,
l)
√
(3 5 η − 7η) dη.
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Indefinite Integral
3. Find the indefinite integral of the given functions:
Z 3
x − 2x + 1
a)
d x,
x3
Z
5
dy,
c)
2/7
y
Z
e)
(4 x 5 + x 3 − 5) d x,
(R + 1)2
√
dR,
R
Z
50
d t,
(5t)3
Z √
1
1
√
+ K+
dK,
K+
K
K
Z
b)
d)
Z
f)
Z
g)
i)
k)
4. Find the indefinite integral of the given functions:
Z
a)
(x 3 − 3x 2 + 4x − 7) d x,
h)
j)
l)
e)
g)
1
√
d h, g 6 = 0,
2gh
Z 1−x 2
d x,
x
Z √
τ
dτ,
τ2
Z √ 3
14 u
11
4
− 5/3 −
d u.
3
u
3 u2
Z b)
Z
c)
5
√ dM,
M
Z qp
3
4
x 2 d x,
4x
2
√ + (3 − 2x) d x,
3x
x 4 − 10x 2 + 5
d x,
x2
√
Z
4x − 2 x
d x,
x
Z
√ 2
1 + x d x,
Contents
Z
x(2x − 5) d x,
r Z √
2
d x,
2x +
x
Z
√
√
x + 1 x − x + 1 d x,
d)
f)
h)
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Indefinite Integral
1 2
1− √
d x,
3
x
Z √ 4
x + 2 + x −4
d x,
x3
Z i)
k)
c)
Z
e)
Z
g)
Z
i)
k)
1
sin x −
cos2 x
l)
b)
d x,
d)
a
dθ,
b · sin2 θ
f)
5 sin2  + 3 cos2 
d,
2 sin2  cos2 
h)
R · 10x d x,
j)
Z √
x
T d x, T > 0,
6. Find the indefinite integral of the given functions:
Z
3 − 2 cot2 x
a)
d x,
b)
cos2 x
Z
e−u
d)
eu 1 +
d u,
e)
cos2 u
√
x(1 − x 2 ) d x,
Z
2 − x2
√ d x.
x+ 2
j)
5. Find the indefinite integral of the given functions:
Z
a)
(8 cos α − 3 sin α) dα,
Z Z
l)
Z √
(2 σ + 1)2
−2
+
cos
σ
dσ,
σ2
Z
1
d x,
3 cos2 x
Z
cos3 φ − 0.8
dφ,
cos2 φ
Z
3 − 2 tan−2 x
d x,
cos2 x
Z
4λ dλ,
Z
√
0.5 eρ dρ.
Contents
Page 30 from 358
Z
Z
3 · 8τ dτ,
e2t
et
−1
d t,
−1
Z
c)
sin x · cos2 x
Z
f)
dx
2
e3ρ
eρ
+1
dρ,
+1
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Indefinite Integral
Z
g)
Z
j)
4
√
d x,
4 − 4x 2
5
d t,
9 + 9 t2
Z
√
h)
Z
k)
1
3 − 3θ 2
1
d x,
x ln x
dθ,
7. Find the indefinite integral of the given functions:
Z
Z
ex
x
x
a)
2 · 7 d x,
b)
e 1+
d x,
3
3 + e−x sin x
d x,
e−x
Z
−4
√
d x,
16 − 16 x 2
Z 2
x +3
d x,
x2 + 1
Z
d)
g)
j)
e2 x − 1
d x,
ex
2
Z
2x − 3x
d x,
6x
Z
4 (2 u2 + 2)−1 d u,
i)
Z
l)
h)
k)
8. Find the indefinite integral of the given functions:
Z
Z
sin 2υ
x4
a)
d x,
b)
dυ,
2
x +1
sin υ
Z
Z
cos 2β
1
d)
dω,
d
β,
e)
2
1 + cos 2ω
1 − sin β
Z
Z
1
2
g)
tan 9 d9,
h)
dτ,
sin2 2τ
Z
Z
t
3+U
j)
d t,
k)
dU,
t +4
3−U
(2x + 3x )2 d x.
a −x
d x,
a 1+ √
x3
a > 0, a 6 = 1,
Z
B 3 x d x, B > 0,
Z √
1 + x2
√
d x,
1 − x4
Z 2
h −1
d h.
h2 + 1
Z
c)
Z
e)
√
3 − 1 − z2
√
d z,
1 − z2
Z
f)
i)
l)
x
Z
c)
Z
f)
Z
i)
l)
1
w 2 (1 + w2 )
sin2
φ 2
+ 1)2
dw,
Contents
dφ,
(x
d x,
x(x 2 + 1)
Z
η+2
dη.
2η − 1
Page 31 from 358
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Indefinite Integral
Answers to Exercises
3 ln |x|,
b)
x4
2 √
− ln |x| + x 4 x,
4
5
e)
3
x,
4
f)
x + 6 ln |x| −
i)
x 1−a
,
1−a
j)
1.5 ln |x|,
1. a)
2. a)
e)
i)
3. a)
x4,
z1+
x−
b)
x3
+ x 2 − ln |x| ,
3
c)
f)
2ρ 3/2 ,
g)
j)
3 ln |t|,
k)
√
2
√ ,
1+ 2
x −t
,
−
ln |x|
1
2
+ ,
2 x2
x
e)
2 6 1 4
x + x − 5x,
3
4
h)
1
x − 2 ln |x| − ,
x
12
4
− 2,
x
x
c)
x 13
,
13
d)
5 8
x ,
8
g)
20 0.84
x ,
7
h)
−2x −2 ,
k)
−
l)
4
− .
u
3z2
,
8
1
− 5,
R
√
1
− 3 +2 z,
z
−
h)
5m8/5 ,
l)
5 η6/5 − 7η2
.
2
c)
7 y 5/7 ,
f)
2h
,
g
g)
2
i)
1
− 2,
5t
j)
2
−√ ,
τ
l)
28u5/2
33
4
+
+
.
2/3
15
2u
3u
10
s
√
K2
2K K
+ ln |K| +
+ 2 K,
2
3
1
,
4u4
d)
√
M,
b)
√
√
k)
√
1.7
+ 18 3 x,
x2
R+
d)
3 x 4/3 ,
2 R 5/2
4 R 3/2
+
,
5
3
Contents
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33
Indefinite Integral
4. a)
d)
x4
− x 3 + 2x 2 − 7x,
4
x3
5
− 10x − ,
3
x
2 5/2
x + x,
5
2 3/2 2 7/2
j)
x − x ,
3
3
g)
5. a)
d)
g)
j)
8 sin α + 3 cos α,
1
tan x ,
3
5 tan  − 3 cot 
,
2
4λ
,
ln 4
8p 3
4
3x + 9x − 6x 2 + x 3 ,
9
3
√
x
e) 2 2x
+1 ,
3
b)
4 √
x2
x x+ ,
3
2
1
k) ln |x| − 4 ,
4x
e)
l)
3 tan x + 2 cot x,
√ x
T
√ ,
ln T
h)
k)
b)
d)
eu + tan u ,
e)
g)
2 arcsin x,
h)
j)
5
arctan t,
9
k)
3 · 8τ
,
ln 8
et + t,
1√
3 arcsin θ,
3
ln | ln x|,
√
x),
i) x − 3x 2/3 + 3x 1/3 ,
1
8
4 ln |σ | − √ − + tan σ,
σ
σ
a
− cot θ ,
b
b)
2 3 5 2
x − x ,
3
2
f) 4(x −
h) x +
3 tan x + 2 cot x,
6. a)
c)
c)
− cos x − tan x,
f)
sin φ − 0.8 tan φ,
i)
l)
c)
f)
i)
l)
√
x2
2x −
.
2
R · 10x
,
ln 10
√
eρ .
tan x − cot x,
1 2ρ
e − eρ + ρ,
2
−z + 3 arcsin z,
12e2x
9e2x
2e2x
+
+
.
ln 6
2 ln 3
ln 2
Contents
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Indefinite Integral
e2x
,
6
2 · 7x
,
ln 7
b)
ex +
d)
3ex − cos x,
e)
g)
arccos x,
h)
ex + e−x ,
x
2 x
− 32
3
j)
x + 2 arctan x,
k)
2 arctan u,
x3
− x + arctan x,
3
b)
2 sin υ,
d)
2β − tan β,
e)
g)
tan 9 − 9,
h)
j)
t − 4 ln |t + 4|,
k)
7. a)
8. a)
ln 23
− 2x,
1
tan ω,
2
1
− tan 2τ,
2
−U − 6 ln |U − 3|,
c)
f)
c)
ax
2
−√ ,
ln a
x
f)
B 3x
,
3 ln B
i)
arcsin x,
l)
h − 2 arctan h.
1
,
w
φ
φ
φ
− cos sin + ,
2
2
2
− arctan w −
i)
ln |x| + 2 arctan x,
l)
η 5
+ ln |2η − 1|.
2 4
Contents
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Indefinite Integral
Quiz—Basic Integration Formulae
Choose the correct answer (only one is correct).
Z
sin 2x
dx =
cos x
−2 cos x + C
1. (1 pt.)
− ln | cos x| + C
Z
−2x
2. (1 pt.)
dx =
1 − x2
1
+C
1−x
ln |1 − x 2 | + C
√
Z √
x+ 3 x+x
3. (1 pt.)
dx =
x
2√ 3 3√
3
x +
x4 + C
3
4
√
√
2 x + 33 x + x + C
ln | sin 2x| + C
−
2 cos 2x
+C
sin x
x
+C
1−x
1 + x +C
ln
1 − x
6√
6
x 11 + C
11
√
√
2 x + 3 3 x + x ln x + C
Contents
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Indefinite Integral
x2 − 1
dx =
x2 + 1
x + arctan x + C
Z
4. (1 pt.)
x − 2 arctan x + C
Z
8x
dx =
5. (1 pt.)
8x 2 + 2
1
ln |8x 2 + 2| + C
2
8
4x
√ arctan √ + C
2Z
2
1
6. (1 pt.)
dx =
1 + cos 2x
1 − 2 arctan x + C
1
3
x 3 − x ln |x 2 + 1| + C
2 ln |8x 2 + 2| + C
√
4x
2 arctan √ + C
2
ln |1 + cos 2x| + C
1
cot x + C
2
1
tan x + C
2 Z
1
+C
1 − sin 2x
7. (1 pt.)
Contents
(1 − sin2 x) + (1 − cos2 x) d x =
Page 36 from 358
sin3 x − cos3 x + C
1
2x − (sin3 x + cos3 x) + C
3
x+C
1
2x + (− cos 3x + sin 3x) + C
3
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Indefinite Integral
Z
8. (1 pt.)
dx
=
2x − 1
(x − 1)−2 + C
1
(x − 1)−2 + C
2
1
ln |2x − 1| + C
2 Z
ln(2x − 1) + C
9. (1 pt.)
2x d x =
2x
+C
ln 2
2x + C
Z
dx
√
=
10. (1 pt.)
1 − x2
2x+1 + C
2x+1
+C
x+1
ln(1 + x 2 ) + C
1
+C
1 − x2
arcsin x + C
arctan x + C
Contents
Correct Answers:
Points Gained:
Success Rate:
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Indefinite Integral
1.2.2 Integration by Parts
So far we have learned how to integrate functions matching the basic integration formulae and functions
that can be rewritten to these. As noted above, we know that the integral of the sum (difference) is
the sum (difference) of the individual integrals. Unfortunately, nothing so simple as this is true for a
product or a quotient. It is not the case that the integral of a product (quotient) is a product (quotient)
of the individual integrals! We should not be surprised, since the derivative of a product does not equal
the product of the individual derivatives, nor does the derivative of a quotient equal the quotient of the
individual derivatives. Nevertheless, by integrating the derivative product rule formula we gain a very
efficient method for integrating some products.
Theorem 1.9 Let u and v be functions differentiable on the interval I . Then the equality
Z
Z
0
u(x)v (x) d x = u(x)v(x) − u0 (x)v(x) d x
(1.6)
holds if at least one of the integrals involved exists.
Proof. The product rule says that for two differentiable functions u, v the equality (u(x)v(x))0 =
= u0 (x)v(x) + u(x)v 0 (x) holds for x ∈ I . Integrating it we get
Z
Z
0
u(x)v(x) d x = u(x)v(x) + c =
u0 (x)v(x) + u(x)v 0 (x) d x.
Contents
Page 38 from 358
R
0
0
R
0
R
0
uv d x , u v dx exists,R such as
R
RThus0the integral (uv +u v) d x exists. If at least one of the integrals
uv d x , then, given Theorem 1.4, the integral of the difference (uv 0 + u0 v) − uv 0 d x = u0 v d x
exists too, so
Z
Z
0
u(x)v(x) + c = u (x)v(x) d x + u(x)v 0 (x) d x
and this equality implies Formula (1.6).
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39
In the examples below the functions will have continuous derivatives, thus the existence of integrals
will be guaranteed by Theorem 1.3.
The integration method based on Formula (1.6) is called integration by parts. In shorthand, we write
Z
Z
uv 0 d x = uv − u0 v d x.
This method is appropriate for integrals, whose integrands have the form of a product. To be able to
write the right-hand side of (1.6), we need to know how to differentiate one factor of the left-hand side
(in our notation u) to obtain u0 , which is not usually the problem. We also need to know how to integrate
the other factor (in our notation v 0 ) to obtain v , which can be a problem. And finally the integral on the
right-hand side should be easier from the point of view of further integration. We demonstrate the use
with an example.
Z
Example 1.10 Evaluate x sin x d x , x ∈ R.
Solution. The product in the integrand is evident. We can choose either u = x and v 0 = sin x or
u = sin x and v 0 = x .
R
We will start with the first choice. If u = x , then u0 = 1. Further, v 0 = sin x , so v = sin x d x =
= − cos x (the integration constant can be set to zero, because we need one concrete antiderivative).
From (1.6) we get
Z
Z
x sin x d x = x(− cos x) − 1 · (− cos x) d x =
Z
= −x cos x + cos x d x = −x cos x + sin x + c.
Contents
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40
Thus this choice led to the goal. The evaluation is usually arranged in a table and the notation looks as
follows:
Z
Z
u=x
u0 = 1
x sin x d x = 0
= x(− cos x) − 1 · (− cos x) d x =
v = sin x v = − cos x Z
= −x cos x + cos x d x = −x cos x + sin x + c.
As to the notation, we write the table either below or next to the integral. In this text we prefer the
second possibility to save the room and we separate it by vertical lines from the rest of the evaluation.
It is a good idea to write this auxiliary table regarding the position of u, u0 , v , and v 0 in the same way
each time. This habit helps us to avoid unintentional errors. In the left column we have functions u and
v 0 of the given integral, in the leading diagonal of the table we have u and v and in the right column we
have functions u0 and v of the new integral. These pairs are multiplied in Formula (1.6).
Now, let us try the second choice. We get
Z
Z
u = sin x u0 = cos x x 2
x2
=
x sin x d x = 0
sin
x
−
(cos
x)
dx =
2
v =x
v = x2
2
2
Z
x2
1
x 2 cos x d x.
=
sin x −
2
2
The previous equality is correct, but evidently the new integral is more complicated then the original,
therefore this choice did not lead to the goal.
N
Before showing further examples we present the table of typical functions, whose indefinite integrals
can be evaluated using integration by parts. At the same time it shows which function is to be differentiated and which one is to be integrated. Our list in not of course complete. There exist other integrals
Contents
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Indefinite Integral
that can be evaluated successfully using this method. But it is important to know these basic types to be
able to decide correctly and without long hesitation.
Integrals Suitable for Integration by Parts
In the tables below P (x) is a polynomial and a a non-zero constant. The first column contains the
integrand, in the second column it shows which function is to be differentiated, and in the third column
which one is to be integrated. The list is divided into two parts.
In the first group a polynomial is differentiated and the second factor is integrated. The new integral
is the product of the polynomial of degree smaller by one and of the second function, which is of a type
similar to that of the original integral (exponential function eax is kept, sine and cosine are exchanged).
Integrand
u
v0
P (x) eax
P (x)
eax
P (x) sin ax
P (x)
sin ax
P (x) cos ax
P (x)
cos ax
Contents
Tab. 1.3: Integration by Parts—Part One
Page 41 from 358
Z
Example 1.11 Evaluate
(x 2 + 1) e−x d x , x ∈ R.
Solution. The integrand is of the “polynomial times exponential function” type, which can be found in
Table 1.3. Thus we will differentiate the polynomial x 2 + 1 and integrate the exponential function e−x .
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Indefinite Integral
At the same time, in this example we demonstrate a typical feature of integration by parts, namely the
repeated application. As we will see, we get the integral of the same “polynomial times exponential
function” type, but the degree of the polynomial will be smaller by one. Thus we will use integration by
parts once more.
In general, in case of integrands from the first group listed in Table 1.3, we use integration by partsrepeatedly until the polynomial becomes a non-zero constant (if its degree is n, it happens after the nth
derivative). In our example we get
Z
u = x 2 + 1 u0 = 2x =
(x + 1) e d x = 0
v = e−x
v = −e−x Z
2
−x
= (x + 1)(−e ) − 2x(−e−x ) d x =
2
−x
2
= −(x + 1) e
−x
= −(x 2 + 1) e−x
u=x
u0 = 1
+ 2 x e d x = 0
−x
v =e
v = −e−x
Z
−x
−x
+ 2 x(−e ) − 1 · (−e ) d x =
Z
−x
= −(x 2 + 1) e−x − 2x e−x + 2
Z
=
Contents
e−x d x =
Page 42 from 358
J
= −(x 2 + 1) e−x − 2x e−x − 2e−x + c =
= −(x 2 + 2x + 3) e−x + c.
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Indefinite Integral
In the second group a polynomial is integrated and the second factor is differentiated. The reverse
choice would not be possible, because we do not yet know how to integrate a logarithm, inverse sine, etc.
On the contrary, differentiating these “unpleasant” functions we get
√ rid of them. Their derivatives are
much “easier” for the integration ((ln x)0 = 1/x , (arcsin x)0 = 1/ 1 − x 2 , (arctan x)0 = 1/(x 2 + 1)
etc.).
u
Integrand
v0
P (x) ln x
ln x
P (x)
P (x) arcsin ax
arcsin ax
P (x)
P (x) arccos ax
arccos ax
P (x)
P (x) arctan ax
arctan ax
P (x)
P (x) arccot ax
arccot ax
P (x)
Tab. 1.4: Integration by Parts—Part Two
Contents
Z
Example 1.12 Evaluate
(2x − 1) ln x d x , x ∈ (0, +∞).
Page 43 from 358
Solution. The integral is of the “polynomial times logarithmic function” type and can be found in Table 1.4. Thus the polynomial 2x −1 will be integrated and the logarithmic function will be differentiated.
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44
We get
u = ln x
u0 = x1
(2x − 1) ln x d x = 0
=
v = 2x − 1 v = x 2 − x Z
1 2
2
(x − x) d x =
= (ln x)(x − x) −
x
Z
1
= (x 2 − x) ln x − (x − 1) d x = (x 2 − x) ln x − x 2 + x + c.
2
Z
Example 1.13 Evaluate arccot x d x , x ∈ R.
Z
N
Solution. There seems to be no product in this integral. But each time we can take 1 as the second
factor, which is in fact the polynomial of degree zero. Thus it is again the type from Table 1.4. We will
differentiate the function inverse cotangent and integrate 1. We get
Z
u = arccot x u0 = − 21 x
+1
=
arccotx d x = 0
v =1
v=x
Z Z
1
x
= (arccot x)x −
− 2
x d x = x arccot x +
dx =
2
x +1
x +1
Z
1
2x
1
= x arccot x +
d x = x arccot x + ln(x 2 + 1) + c.
2
2
x +1
2
The last integral was evaluated using Formula 14 from Table 1.1.
N
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Indefinite Integral
In the following examples we demonstrate another “trick”, which is commonly used together with
the integration by parts. The idea is that using integration by parts (maybe several times) and appropriate
manipulations, we get an expression that again contains the original unknown integral, so we have an
equation
Z
Z
f (x) d x = h(x) + α
f (x) d x,
α ∈ R, α 6= 0,
from which we can find the unknown integral (if α 6 = 1).
Z
Example 1.14 Evaluate ex sin x d x , x ∈ R.
Solution. This integral matches no type given in Tables 1.3 and 1.4. We apply integration by parts twice,
each time differentiating the exponential function and integrating the second factor (otherwise we would
return back to the original integral). We have
Z
u = ex
u0 = e x
x
=
e sin x d x = 0
v = sin x v = − cos x Z
Z
x
x
x
= e (− cos x) − e (− cos x) d x = −e cos x + ex cos x d x =
Z
u = ex
u0 = ex x
x
= −e cos x + e sin x − ex sin x d x.
= 0
v = cos x v = sin x Contents
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We derived the equation
Z
J
ex sin x d x = −ex cos x + ex sin x −
Z
ex sin x d x,
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Indefinite Integral
from which we easily find
Z
2
Z
ex sin x d x = −ex cos x + ex sin x + c,
ex sin x d x =
1 x
e (sin x − cos x) + c.
2
One might expect the constant c/2 in the result, but if c is any real constant, then c/2 assumes all real
values as well (we denoted c/2 as a new constant and used the same letter for it). We will no longer
comment on origins of similar terms and notation throughout the rest of the text.
N
Example 1.15 Evaluate
Z p
1 − x 2 d x , x ∈ (−1, 1).
Solution. Using integration by parts we will again derive the equation for the integral being evaluated.
We choose one as the second factor. Then we get
√
Z p
u = 1 − x 2 u0 = − √ x 1−x 2 =
1 − x2 dx = v0 = 1
v=x
Z
Z
p
p
−x 2
1 − x2 − 1
√
= x 1 − x2 − √
dx = x 1 − x2 −
dx =
1 − x2
1 − x2
Z p
1 − x2
1
2
√
=x 1−x −
−√
dx =
2
2
1
−
x
1
−
x
Z p
Z
p
dx
= x 1 − x2 −
1 − x2 dx + √
=
1 − x2
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47
Indefinite Integral
Z p
p
2
=x 1−x −
1 − x 2 d x + arcsin x.
We found the equation
Z p
Z p
p
2
2
1 − x dx = x 1 − x −
1 − x 2 d x + arcsin x.
Making the unknown integral the subject of this equation, we get
Z p
xp
1
1 − x2 dx =
1 − x 2 + arcsin x + c.
2
2
The use of integration by parts is not typical for this example and other methods can be applied (see
N
Example 1.30 and the text on page 143).
Z
Example 1.16 Evaluate cos2 x d x , x ∈ R.
Solution. Using integration by parts we find the equation for the integral we want to evaluate. We choose
cosine for both u and v 0 . We have
Z
Z
u = cos x u0 = − sin x 2
2
cos x d x = 0
= cos x sin x + sin x d x =
v = cos x v = sin x
Z
= cos x sin x + (1 − cos2 x) d x =
Z
Z
= cos x sin x + d x − cos2 x d x =
Z
= cos x sin x + x − cos2 x d x,
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48
Indefinite Integral
which gives the equation
Z
cos2 x d x = cos x sin x + x −
Z
cos2 x d x.
Solving it we get
Z
cos2 x d x =
1
x
cos x sin x + + c.
2
2
We used the well-known trigonometric formula cos2 x + sin2 x = 1. This integral is often evaluated
using a different method (see Example 1.47).
N
Let us summarize the features occurring in connection with integration by parts:
• There exists the group of typical integrands having the form of the product of two functions, whose
integrals can be evaluated using integration by parts (at least this method is the starting step)—see
Tables 1.3 and 1.4.
• We choose 1 as the factor to be integrated.
• Integration by parts is often used repeatedly.
• Using integration by parts and appropriate manipulations we derive the equation, from which the
integral which is to be evaluated can be found.
• Recurrence formulae are often derived using integration by parts (see, e.g., Formula (1.15)).
Of course there exist integrals, in addition to those listed in Tables 1.3 and 1.4, that can be successfully
evaluated using integration by parts. One is given in the following example. It is a question of experience
to be able to decide when it is worth using this method.
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Indefinite Integral
Z
Example 1.17 Evaluate
x
d x , x ∈ (−π/2, π/2).
cos2 x
Solution. We will differentiate the polynomial x and integrate the fraction1/ cos2 x . We get
Z
Z
u=x
x
u0 = 1 dx = 0
= x tan x − tan x d x =
v = cos12 x v = tan x cos2 x
Z
− sin x
d x = x tan x + ln | cos x| + c.
= x tan x +
cos x
The absolute value in the logarithmic function can be omitted, because cosine is positive in the considered
interval. Formula 14 from Table 1.1 was used (compare to Example 1.8 b).
N
Note that the integral from the previous example does not match the type given in Table 1.3. That type
has the form “polynomial times cos ax ”, where a is a constant, while in our case we have the expression
“polynomial over cos2 x ”. Thus, instead of the product we have the quotient, and the cosine is raised
to two. Students often only partially remember the types, they know that there is “a polynomial” and “a
cosine” there, they exchange the product and the quotient etc. This can lead to an absolutely unsuitable
2
integration method. The function x ex , for example, does not match the type from Table 1.3. The first
factor is the polynomial, but the second factor should have the form eax , where a is a constant, which is
not true. Integration by parts is worthless in this case. In the next section we will learn that a different
approach must be used to integrate this function.
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Indefinite Integral
Exercises
1. Evaluate the indefinite integral of the following functions:
Z
Z
a)
x arctan x d x,
b)
t e2t d t,
Z
d)
Z
g)
Z
j)
R 3R dR,
Z
θ sin θ dθ,
e)
Z
B 2 sin B dB,
h)
x 3 ex d x,
k)
ε sin
Z
Z
V ln(V − 1) dV ,
Z
g)
Z
j)
Z
√
w ln2 w dw,
ln3 t
d t,
t2
e)
ε
dε,
2
r sin2 r d r,
Z
t 2 sin 2t d t.
l)
Z
(ρ 2 − 3ρ + 2) eρ dρ,
Z
ln R
dR,
R2
c)
m ln m dm,
f)
H ln(H + 1) dH,
i)
Z
x ln x d x,
Z Z
k)
Z
i)
x 2 cos x d x,
2
(3n + 2) cos n dn,
f)
Z
h)
x cos x d x,
c)
Z
2. Evaluate the indefinite integral of the following functions:
Z
Z
2 −2φ
a)
φ e
dφ,
b)
T 2 cos2 T dT ,
d)
Z
5V arctan V dV ,
l)
ln K
K
Contents
2
dK.
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Indefinite Integral
3. Evaluate the indefinite integral of the following functions:
Z
Z
a)
z3 arctan z d z,
b)
4 ln 2 d,
Z
d)
Z
g)
Z
j)
Z
t arcsin t d t,
e)
sin φ2
dφ,
e−φ
h)
ex sin2 x d x,
k)
Z
arctan θ dθ,
c)
arcsin y
p
dy,
1 − y2
Z
eT cos T dT ,
Z
e−2h sin 3h d h,
Z
e3x cos2 3x d x.
f)
Z
ln K
dK,
K
Z
r
e−r/3 sin d r,
3
i)
l)
Answers to Exercises
x
(x 2 + 1)
arctan x − ,
2
2
b)
e2t
(2t − 1),
4
c)
x sin x + cos x,
d)
R 3R
3R
− 2 ,
ln 3
ln 3
e)
sin θ − θ cos θ,
f)
3 cos n + (3n + 2) sin n,
g)
(−B 2 + 2) cos B + 2B sin B,
h)
4 sin
i)
−
r sin 2r
r2
cos2 r
+
−
,
4
4
4
j)
x 3 ex − 3x 2 ex + 6xex − 6ex ,
l)
t 2 cos 2t
cos 2t
t sin 2t
+
+
.
−
2
4
2
1. a)
k)
x 2 sin x − 2 sin x + 2x cos x,
ε
ε
− 2ε cos ,
2
2
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52
Indefinite Integral
2. a)
−
e−2φ
(2φ 2 + 2φ + 1),
4
b)
T3
+
6
T2
1
−
4
8
sin 2T +
T
cos2 T ,
2
c)
eρ (ρ 2 − 5ρ + 7),
d)
1 2
V2
V
(V − 1) ln(V − 1) −
− ,
2
4
2
e)
1 3
m3
m ln m −
,
3
9
f)
−
g)
w3/2
(18 ln2 w − 24 ln w + 16),
27
h)
1
H2
H
(H 2 − 1) ln(H + 1) −
+
,
2
4
2
i)
k)
3. a)
c)
e)
g)
i)
k)
x2
1 2
x ln x −
,
2
4
5 2
(V arctan V − V + arctan V ),
2
j)
l)
arctan z 4
z3 − 3z
(z − 1) −
,
4
12
e−2h
−
(3 cos 3h + 2 sin 3h),
13 3
r
r
− e−r/3 cos + sin
,
2
3
3
1
− (ln3 t + 3 ln2 t + 6 ln t + 6),
t
1
− (ln2 K + 2 ln K + 2).
K
b)
1
θ arctan θ − ln(θ 2 + 1),
2
1
arcsin2 y,
2
2
φ
4
φ
− eφ cos + eφ sin ,
5
2
5
2
d)
f)
h)
j)
l)
ln R
1
− ,
R
R
4 ln 2 − 4,
√
t 2 arcsin t
t 1 − t2
arcsin t
+
−
,
2
4
4
1 T
e (cos T + sin T ),
2
1 2
ln K,
2
(sin x − 2 cos x) ex sin x
2ex
+
,
5
5
e3x
(cos 3x + 2 sin 3x) cos 3x + 2 .
15
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Indefinite Integral
Quiz—Integration by Parts
Choose the correct answer (only one is correct).
Z
1. (1 pt.)
x cos x d x =
1 2
x + sin x + C
2
x sin x + cos x + C
Z
2. (1 pt.) ln x d x =
1
sin x 2 + C
2
1 2
x cos x − x cos x + C
2
x ln2 x
+C
2
1
+C
x
ln2 x
+C
2Z
x ln x − x + C
3. (1 pt.)
xex d x =
Contents
x 2 x+1
e
+C
2
1 x2
e +C
2
ex
ex
+ 2 +C
x
x
xex − ex + C
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Indefinite Integral
Z
4. (1 pt.)
2x+1
2x x d x =
x2
+C
2
2x
2x x
− 2 +C
ln 2Z ln 2
5. (1 pt.)
x2
+C
2
2x x
2x
−
+C
ln 2
ln 2
ex cos x d x =
ex sin x + C
−e2x sin x + e2x cos x
+C
2
Z
√
6. (1 pt.) x 3x + 1 d x =
√
6 x5 x2
+
+C
5
2
p
p
2x (3x + 1)3 4 (3x + 1)5
−
+C
135
Z 9
7. (1 pt.)
2x
ex sin x
+C
2
ex sin x + ex cos x
+C
2
x 2 (3x + 1)
+C
6
√
x 2 3x + 1 x 3 (3x + 1)
−
+C
2
3
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sin2 x d x =
J
sin3 x
+C
3
sin x(1 − cos x)
+C
2
x − sin x cos x
+C
2
2 sin x cos x + C
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Indefinite Integral
Z
8. (1 pt.)
(2x + 1) ln
x
dx =
2
2(x 2 + x)
+C
x
x 2
x
+C
2 ln + (2x + 1) ln
2
Z2
9. (1 pt.) (x 2 + 1) sin x d x =
x3
3
−
+ x sin x − (x 2 + 1) cos x + C
x3
cos x − cos x
3Z
10. (1 pt.)
(x 2 + x) ln
x
x2
−
−x+C
2
2
(2x + 1) ln x2
+C
2
x3
+ x cos x
−
3
−(x 2 + 1) cos x + 2x sin x + 2 cos x + C
2 arctan x d x =
2x arctan x − ln |1 + x 2 | + C
2
+C
1 + x2
arctan2 x + C
2
+ ln |1 + x 2 | + C
1 + x2
Correct Answers:
Points Gained:
Success Rate:
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Indefinite Integral
1.2.3 Substitution Method
In this section we will study another important integrationmethod, which will be obtained by integrating
0
the derivative chain rule. Let us remember that F [ϕ(x)] = F 0 [ϕ(x)] ϕ 0 (x) = f [ϕ(x)] ϕ 0 (x), where
we denoted F 0 (u) = f (u) and u = ϕ(x). The method is formulated in the following statement.
Theorem 1.18 Let function f (u) have an antiderivative F (u) on the open interval J , let function ϕ(x)
be differentiable on the open interval I and ϕ(x) ∈ J for any x ∈ I . Then composite function
f [ϕ(x)] ϕ 0 (x) has an antiderivative on interval I and the equality
Z
f [ϕ(x)] ϕ 0 (x) d x = F [ϕ(x)] + c
(1.7)
holds.
Proof. The statement is the direct consequence of the above mentioned formula for the derivative of the
composite function. The derivative of the right-hand side of Equality (1.7) is the integrand of the integral
on the left-hand side of this equality.
The integration method based on the previous theorem is called the first substitution method. We will
describe its practical application. The assumption concerning the existence of the antiderivative of f (u)
can be written as
Z
f (u) d u = F (u) + c.
Then the statement of Theorem 1.18 can be expressed as
Z
Z
0
f [ϕ(x)] ϕ (x) d x = f (u) d u,
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Indefinite Integral
57
where we substitute ϕ(x) for u in the right-hand side expression. The evaluation is calculated as follows:
• We choose the substitution ϕ(x) = u (the notation of a new variable is not important, the only
condition is that we must use a letter which is different from the letter used for the old variable,
i.e. x ).
• We differentiate the equality ϕ(x) = u. (Let us remember that the differential of a function h(z)
is the product of the derivative of this function, and the increment d z, where z is the independent
variable of h, that is dh(z) = h0 (z) d z.) In our case the independent variable on the left-hand side
is denoted by x , and on the right-hand side by u, thus ϕ 0 (x) = dϕ(x)
and u0 = dd uu = 1. Therefore,
dx
0
0
we get the equality ϕ (x) d x = 1 · d u, that is ϕ (x) d x = d u.
• In the left-hand side integral of Equality (1.8) we replace the function ϕ(x) by the variable u and
the expression ϕ 0 (x) d x by the differential d u. Similarly, as in the case of integration by parts, we
arrange the whole evaluation in a table. Formula (1.8) then looks as follows:
Z
Z
ϕ(x)
=
u
0
= f (u) d u,
(1.9)
f [ϕ(x)] ϕ (x) d x = 0
ϕ (x) d x = d u where the original variable must be put back in the right-hand side, thus we replace u = ϕ(x).
Again, it is reasonable to keep the same arrangement each time to avoid mistakes.
Z
cos x d x
p
Example 1.19 Evaluate
, x ∈ R.
1 + sin2 x
p
Solution. We can easily find the composite function 1
1 + sin2 x in the integrand. Its outer com√
ponent is f (u) = 1 1 + u2 and inner component is ϕ(x) = sin x . Furthermore ϕ 0 (x) = cos x .
Thus
1
1
cos x
f [ϕ(x)] ϕ 0 (x) = p
ϕ 0 (x) = p
cos x = p
2
1 + ϕ 2 (x)
1 + sin x
1 + sin2 x
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Indefinite Integral
is the given integrand. So it is possible to use the substitution method. We choose the substitution
sin x = u and differentiating this equality we easily get the relation cos x d x = d u. The evaluation will
be written as follows:
Z
Z
cos x d x
du
sin x = u p
√
=
=
=
cos x d x = d u
1 + u2
1 + sin2 x
p
p
= lnu + 1 + u2 + c = lnsin x + 1 + sin2 x + c.
We used Formula 11 from Table 1.1.
N
Before evaluating further examples let us think about the required form of the integrand so we can use
the substitution method. It definitely cannot be any expression. On the contrary, the form of the integrand
is quite strictly given. It must be an expression having the form of the product of a composite function
and the derivative of its inner component. As previously, let us denote f (u) the outer component and
ϕ(x) the inner component. Then the integrand must have the form f [ϕ(x)] ϕ 0 (x). Let us show a few
functions of this type in Table 1.5. A composite function is in the first column, its outer component is
in the second column, its inner component is in the third, the derivative of the inner component is in the
fourth and the expected form of the integrand is in the fifth.
Naturally, we cannot expect that the integrand will be “served on a plate” each time, the way we
would like most. For example, the last two integrands from Table 1.5 would be rather written in the form
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(1 + ln x)4
x
ln arctan x
and
.
x2 + 1
√
2
Similarly, the first two integrands would rather look like 2x x 2 − 3 and −2x e−x . We must be able to
“see” a composite function in the given expression and find the derivative of its inner component in it.
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Indefinite Integral
f [ϕ(x)]
f (u)
ϕ(x)
ϕ 0 (x)
f [ϕ(x)] · ϕ 0 (x)
p
x2 − 3
√
u
x2 − 3
2x
p
x 2 − 3 · 2x
eu
−x 2
−2x
e−x · (−2x)
sin6 x
u6
sin x
cos x
sin6 x · cos x
(4 − 7x)10
u10
4 − 7x
(1 + ln x)4
u4
1 + ln x
ln arctan x
ln u
arctan x
−7
1
x
1
2
x +1
(4 − 7x)10 · (−7)
1
(1 + ln x)4 ·
x
1
ln arctan x · 2
x +1
e−x
2
2
Tab. 1.5: Examples of integrands appropriate for substitution method
This fact creates most troubles for students. It is important to know the derivatives of basic elementary
functions by heart. This enables us to find what we are looking for (for example, the derivative of an
inner component), to change the order of factors etc., so that we can decide whether or not the necessary
expression occurs there. This skill is a question of practice. If we have to look for any derivative in a
table, we will hardly “see” something reasonable in the given expression.
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Indefinite Integral
60
Finally, let us point to another√problem. It quite often happens that we miss a multiplicative con2
stant.
√ Consider the expression x x − 3, for example. As we have just explained we would need
2
2x x − 3. This is not a problem, because we can easily complete the constant due to Property (1.4)
from Theorem 1.4. We have
Z p
Z
p
1
2
x x − 3 dx =
2x x 2 − 3 d x,
2
which is what we needed. From a practical point of view we will add another row to the auxiliary table
containing a chosen substitution and its differentials. This row
√will be an appropriate multiple of the
equality between differentials such as the case of the function x x 2 − 3, where the table looks like this:
2
x −3=u
2x d x = d u x dx = 1 du 2
Let us emphasize that we can only complete a multiplicative constant this way. If we really miss a
nonconstant function we cannot proceed like this. We will return to this topic later (see page 66).
Z
(1 + ln x)4
Example 1.20 Evaluate
d x , x ∈ (0, +∞).
x
Solution. The integrand is the last but one expression from Table 1.5. So we choose the substitution
u = 1 + ln x and get:
Z
Z
5
1 + ln x = u (1 + ln x)4
= u4 d u = 1 u5 + c = (1 + ln x) + c.
dx = 1
dx = du x
5
5
x
The correctness of the result can be checked by differentiation.
N
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Indefinite Integral
Z
Example 1.21 Evaluate
sin x cos5 x d x , x ∈ R.
Solution. In this case a good candidate is the composite function cos5 x with the inner component cos x .
Its derivative is − sin x , which is the expression occurring in the integrand up to the multiple −1. Thus
Z
Z
cos
x
=
u
cos6 x
u6
5
+ c.
sin x cos x d x = − sin x d x = d u = u5 (−1) d u = − + c = −
6
6
sin x d x = −d u It is only necessary to calculate that sin x cos5 x d x = cos5 x sin x d x .
Z
2
Example 1.22 Evaluate x e−x d x , x ∈ R.
N
Solution. This is the modification of the second example from Table 1.5. We choose the substitution
u = −x 2 and “complete” the missing constant −2. We get
Z Z
−x 2 = u
−x 2
= eu − 1 d u = − 1 eu + c = − 1 e−x 2 + c.
x e d x = −2x d x = d u
2
2
2
x d x = − 21 d u Again it is sufficient to realize that x e
−x 2
−x 2
dx = e
x dx.
N
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Indefinite Integral
Z
Example 1.23 Evaluate
2
x 3 e−x d x , x ∈ R.
Solution. We choose the substitution s = x 2 and get:
Z
Z
Z
x2 = s
1
3 −x 2
−s 1
s e−s d s =
x e d x = 2x d x = d s = s e · d s =
2
2
x dx = 1 ds 2
(we use the integration by parts method to evaluate this integral. See Table 1.3; the integrand is the
product of the polynomial and the exponential function eas , where a = −1)
u=s
= 0
v = e−s
=
Z
1
u0 = 1
−s
−s
=
−s e − (−e ) d s =
v = −e−s 2
1
1
1
2
−s e−s − e−s + c = − (s + 1) e−s + c = − (x 2 + 1) e−x + c.
2
2
2
N
The integrand in the previous example is similar to that in 1.22. We again “see” the composite
2
function e−x and try the substitution u = −x 2 . But (−x 2 )0 = −2x thus (if we neglect the constant −2)
2
2
the integrand x 3 e−x = x 2 e−x x involves the redundant term x 2 .
That is why it is necessary, this time, to perceive that the integrand involves the composite function
2
f (x 2 ) = x 2 e−x , where f (s) = s e−s , with the inner component x 2 . So we choose the substitution
2
2
s = x 2 and things go smoothly if we imagine that x 3 e−x d x = x 2 e−x x d x .
2
Notice that we can “see” there a different composite function, namely g(−x 2 ) = x 2 e−x , where
g(s) = −s es , with the inner component −x 2 leading to the substitution s = −x 2 . The evaluation
would be analogous and, of course, give the same result. You can try this yourselves.
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Indefinite Integral
In the following two examples we review a simple but very important type of substitution. It is the
linear substitution u = ax + b, where a, b ∈ R, a 6 = 0. As (ax + b)0 = a we have a d x = d u. If the
proper constant a is missing, we can easily complete it using the trick mentioned above.
Z
Example 1.24 Evaluate (4 − 7x)10 d x , x ∈ R.
Solution. This example was also in Table 1.5. Let us use the substitution u = 4 − 7x . We get
Z
4 − 7x = u
Z
Z
1
1
10
10
10
(4 − 7x) d x = −7 d x = d u
= u − 7 du = − 7 u du =
1
dx = − 7 du
=−
Z
Example 1.25 Evaluate
1
1 u11
+ c = − (4 − 7x)11 + c.
7 11
77
N
√
2x − 5 d x , x ∈ h5/2, +∞).
Solution. The linear substitution u = 2x − 5 will be used. We get
2x − 5 = u
Z
Z
Z
√
√ 1
1
2x − 5 d x = 2 d x = d u =
u · du =
u1/2 d u =
2
2
1
dx = du Contents
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2
1u
1√ 3
1p
+c =
u +c =
(2x − 5)3 + c.
2 3/2
3
3
3/2
=
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Indefinite Integral
Comment 1.26 In the case of simple examples, linear substitution can be applied with a little practice
almost by heart, which substantially speeds up the evaluation. If a function f (u) has an antiderivative
F (u), that is
Z
f (u) d u = F (u) + c,
then
Z
f (ax + b) d x =
1
F (ax + b) + c,
a
a, b ∈ R, a 6= 0.
The above formula can be proved either by using the substitution ax + b = u, a d x = d u, that is
d x = a1 d u, or by differentiating the right-hand side, because F 0 (u) = f (u). The formula is valid on
intervals on which function f (ax + b) is defined.
The use will be demonstrated on several examples (integration constants are omitted):
Z
Z
1
u
u
e du = e
⇒
e2x−3 d x = e2x−3 (a = 2, b = −3),
2
Z
Z
du
dx
1
= ln |u|
⇒
= ln |3x + 4| (a = 3, b = 4),
u
3x + 4
3
Z
Z
1
1
u4 d u = u5
⇒
(x + 7)4 d x = (x + 7)5 (a = 1, b = 7),
5
5
Z
Z
1
sin u d u = − cos u
⇒
sin(3 − 5x) d x = − − cos(3 − 5x) =
5
=
1
cos(3 − 5x) (a = −5, b = 3),
5
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Indefinite Integral
Z
Z
cos u d u = sin u
⇒
cos
=
2x − 1
dx =
3
Z
cos
2
1
x−
3
3
dx =
1
2x − 1
3
2x − 1
sin
= sin
2/3
3
2
3
(a = 2/3, b = −1/3).
Note that as a special case for b = 0 we get general versions of Formulae 4, 5, 7, 8, 12, and 13, from
Table 1.1 (right column).
Let us return to the calculation using differentials, which was described on page 60. Students often
proceed mechanically in the following way:
ϕ(x) = u
0
ϕ (x) d x = d u
dx =
du
ϕ 0 (x)
. . . choice of the substitution
. . . differentiation of the previous equality
(1.10)
. . . isolation of the differential of the old variable
Then without thinking they automatically substitute the new variable u for ϕ(x) and replace d x by the
expression d u/ϕ 0 (x).
If the substitution was chosen correctly nothing terrible happens, as the next example demonstrates.
2 + sin x = u Z
Z
Z
cos x d u
du
cos x
d x = cos x d x = d u =
=
=
2
2
(2 + sin x)
u cos x
u2
du d x = cos
x
Z
−1
u
1
1
= u−2 d u =
+c =− +c =−
+ c.
−1
u
2 + sin x
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Indefinite Integral
During the evaluation both the old variable x and the new variable u occurred in the integral involved
at the moment when the differential was already d u. Because the expression containing x (in our example cos x ) was cancelled, everything finished well.
The disaster usually occurs when the choice of the substitution is not correct. We will demonstrate
2
it on the following deterrent “evaluation” of the indefinite integral of the function ex .
Z
x 2 = u Z
du
2
ex d x = 2x d x = d u = eu
.
2x
d x = du 2x
Now students usually consider x to be a constant independent of u, which can be taken out during the
integration with respect to u, and continue as follows:
Z
eu
du
1
=
2x
2x
Z
2
eu d u =
1 u
ex
e +c =
+ c,
2x
2x
!
by which the disaster is ended. The previous evaluation is absolutely false!
Authors of such an evaluation totally ignore the fact that there is a connection between the old and
the new variables given by the equality u = ϕ(x), that is in our case u = x 2 , which implies (for x > 0)
√
that x = u. Thus after the substitution the integral has the form
Z
Z
Z
1 u
eu
1 u
√ e du =
√ d u,
e du =
2x
2 u
2 u
√
which means that in fact the function 1 (2 u ) was taken out! Unfortunately, this mistake happens quite
often.
To avoid something like this you should not use the method indicated in (1.10) if ϕ(x) is a function.
Do not forget the correct form of the integrand, so that the substitution method can be used, that is
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Indefinite Integral
f [ϕ(x)] ϕ 0 (x). Decide what will be considered to be a composite function f [ϕ(x)] and try to find the
derivative of the inner component ϕ 0 (x). If you cannot find it your choice of substitution was probably
not correct. Maybe the example is not appropriate for the substitution method, or at least not for the one
you had chosen.
At the end we note the fact that Formula (1.8) is sometimes used from right to left (less often, but
these cases are important). It is as if we put an inner component into a “simple” function and got an
integral of a composite function, which is seemingly more complicate. But in concrete examples this
integral can be easier for further evaluation. The use is similar but the corresponding statement has
slightly different assumptions and the proof is technically more complicated. Let us remember that ϕ −1
denotes the inverse function to the function ϕ .
Theorem 1.27 Let function f (x) be defined on the open interval J . Let function ϕ(t) have the non-zero
derivative on the open interval I , and map this interval onto interval J . Let us further assume that
function f [ϕ(t)] ϕ 0 (t) has an antiderivative F (t) on interval I .
Then function f (x) has an antiderivative F [ϕ −1 (x)] on interval J . Thus
Z
Z
f (x) d x = f [ϕ(t)] ϕ 0 (t) d t
(1.11)
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if we substitute the function ϕ −1 (x) for t in the antiderivative on the right-hand side.
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The integration method based on the previous theorem is called the second substitution method.
ϕ 0 (t)
ϕ 0 (t)
Proof. As
6 = 0 on I resulting from the Darboux Theorem (see [4, p. 188]) either
> 0 for t ∈ I or
ϕ 0 (t) < 0 for t ∈ I Therefore, function x = ϕ(t) is strictly monotonic on I and has an inverse t = ϕ −1 (x), which
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Indefinite Integral
is differentiable on J , and (see [14])
(ϕ −1 )0 (x) =
1
ϕ 0 (t)
=
1
ϕ 0 [ϕ −1 (x)]
,
x ∈ J.
Because F 0 (t) = f [ϕ(t)] ϕ 0 (t) for t ∈ I using the derivative chain rule and the previous equality we get for
x ∈ J and x = ϕ(t) that
0
F [ϕ −1 (x)] = F 0 [ϕ −1 (x)] · (ϕ −1 )0 (x) = f ϕ[ϕ −1 (x)] ϕ 0 [ϕ −1 (x)] ·
1
= f (x),
ϕ 0 [ϕ −1 (x)]
which was to be proved.
When applying the previous formula we set x = ϕ(t), differentiating this equality we get d x =
= ϕ 0 (t) d t and substitute for x and d x into the left-hand side of (1.11) (this time no manipulations with
the relation d x = ϕ 0 (t) d t are necessary). Then we substitute ϕ −1 (x) for t in the result (sometimes we
also include this relation in the auxiliary table).
Z √
Example 1.28 Evaluate e x d x , x ∈ (0, +∞).
Solution. We use the substitution x = t 2 to remove the unpleasant root in the exponent. As x > 0 we
have J = (0, +∞). Thus the function ϕ(t) = t 2 will be considered on the interval I = (0, +∞) (of
course, it is only a coincidence that we got J = I ). The function ϕ(t) = t 2 is injective on
√the interval I
√
and maps it onto the interval J (its graph is an arc of a parabola). As t > 0 we have x = t 2 = |t| = t .
√
The inverse function of x = ϕ(t) = t 2 is therefore t = ϕ −1 (x) = x . Now we can evaluate the integral
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Indefinite Integral
involved. We get (after applying the integration by parts methods to the integral obtained)
Z
Z √
Z
√
x = t2 x
t2
e dx = = 2t e d t = 2t et d t =
d x = 2t d t Z
u = 2t u0 = 2 t
= 0
= 2t e − 2et d t = 2t et − 2et + c =
v = et v = et √
√
= 2(t − 1) et + c = 2( x − 1) e x + c.
N
It is possible to prove that the result of the previous example holds on the interval h0, +∞), too. If
we wanted to verify it directly from the definition of antiderivative,
that is differentiating the result, we
√
√
x
would have to be very careful because the functions x and e have, at point x = 0, only the right
derivative, which is moreover improper. Thus the standard product rule cannot be used at this point.
As this situation is quite common in connection with the substitution method we will give a simple
statement that easily solves this problem in most cases. The result is formulated for bounded closed
intervals but an analogous statement also holds for half closed intervals (both bounded and unbounded).
Theorem 1.29 Let functions f and F be continuous on the interval hα, βi, α, β ∈ R, and F be the
antiderivative of f on the open interval (α, β), that is, F 0 (x) = f (x) for x ∈ (α, β). Then F is also
the antiderivative of f on the closed interval hα, βi.
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Proof. The statement is the consequence of Exercise 9 in [4, p. 111]. It can also be proved using l’Hospital’s rule.
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Indefinite Integral
Z p
Example 1.30 Evaluate
1 − x 2 d x , x ∈ (−1, 1).
Solution. We have already evaluated this integral using the integration by parts method (see Example 1.15). This time we use the substitution x = sin t . As J = (−1, 1) we choose I = (−π/2, π/2).
Then the function ϕ(t) = sin t maps interval I onto interval J . Function ϕ(t) = sin t is injective on the
interval (−π/2, π/2) and its inverse is ϕ −1 (x) = arcsin x , that is t = arcsin x .
We prepare
p the integrand√after the substitution. As cosine is positive on interval I we obtain that
√
2
1 − x = 1 − sin2 t = cos2 t = | cos t| = cos t . So
Z p
Z p
x = sin t
2
=
1 − x dx = 1 − sin2 t cos t d t =
d x = cos t d t Z
t
1
= cos2 t d t = sin t cos t + + c =
2
2
(we used the result of Example 1.16)
p
1
t
xp
1
sin t 1 − sin2 t + + c =
1 − x 2 + arcsin x + c,
2
2
2
2
√
which is the same result like in Example 1.15. Both the integrand 1 − x 2 and the resulting function
are defined and continuous on the closed interval h−1, 1i. Therefore, due to Theorem 1.29 the result is
also
√ valid on the closed interval h−1, 1i. The direct verification would again be difficult as the functions
1 − x 2 and arcsin x have one-sided improper derivatives at the points x = ±1.
N
=
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Indefinite Integral
Exercises
1. Find the indefinite integral of the given functions:
Z
Z
a)
sin3 ω cos ω dω,
b)
6t sin 3t 2 d t,
Z
cos β
d)
Z
Z
j)
2 sin t
2e
g)
p
sin β dβ,
Z
e)
Z
cos t d t,
2C
dC,
(1 + C 2 )2
h)
2
−4ρ e−2 ρ dρ,
8s 2 d s
k)
2. Find the indefinite integral of the given functions:
Z
Z
(6p − 5) dp
p
a)
,
b)
2 3p 2 − 5p + 6
Z
Z
4 cos t
√
d)
d
t,
e)
3
1 + 2 sin t
Z
Z
6v
√
g)
dv,
h)
4 − 9v 4
Z
Z
1
p
j)
d x,
k)
x 1 − ln2 x
p
3
(8s 3 + 27)2
4 tan3 φ
dφ,
cos2 φ
Z
6r 2 e−2r d r,
Z
3
Z
5W 4 dW
√
.
2 4 + W5
f)
3 ln2 W
dW,
W
Z
Z
c)
i)
,
3 cos φ
dφ,
sin4 φ
sin 2r d r
√
,
2 1 + cos2 r
2 et
√
d t,
2 − 4 e2 t
30k
dk,
3k 4 + 5
l)
3
√
ln y
dy,
y
Z
c)
f)
i)
sin u
√
d u,
2 cos3 u
Z
dx
,
x ln x ln ln x
Z
6 tan 3x d x,
Z
l)
18q d q
.
9 + (3q 2 + 1)2
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Indefinite Integral
3. Find the indefinite integral of the given functions:
Z
Z
−2 dθ
a)
,
b)
4 sin x cos3 x d x,
tan θ sin2 θ
Z
Z
√
2 arctan ρ
d)
d
ρ,
e)
1 + 2x d x,
2
1+ρ
Z p
Z
3x d x
,
h)
x 2x 2 + 7 d x,
g)
(x 2 + 1)2
Z
Z
7 dx
4x d x
√
,
,
k)
j)
3
2
(1
+ 2x)3
8−x
Z
Z
n
dφ
m)
dn,
n)
,
n2 − 1
cos2 (1 − φ)
4. Find the indefinite integral of the given functions:
Z
Z
dx
x − arctan x
√
a)
,
b)
d x,
2
1 + x2
1 − x arcsin x
Z
Z
d)
(4ρ − 3)4 dρ,
e)
(2x + 1)3 d x,
Z
g)
j)
12
d x,
(3x − 7)5
Z
√
3
5 − 6x d x,
Z
h)
p2
Z
k)
dp
,
− 6p + 9
1
√
dm,
4m + 9
Z
c)
f)
i)
2 ln x
d x,
x
Z
dx
√
,
5 − 4x
Z
p
3
9x 2 x 3 + 10 d x,
Z
3 cos4 t sin t d t,
Z
cos y dy
.
3 sin2/3 y
l)
o)
Z
c)
f)
i)
dx
,
x(1 + ln2 x)
Z 1
τ −2
1−
dτ,
6
6
Z
33(8 − 3x)6/5 d x,
Z
l)
√
1
dl.
3 − 2l
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Indefinite Integral
5. Find the indefinite integral of the given functions:
Z
Z
dφ
a)
sin(2ω − 5) dω,
b)
,
2
sin (3φ − 7)
Z
Z
4 dv
d)
,
e)
14e7r−8 d r,
1 − cos 4v
Z q/2
Z 2s
e − e−q/2
e −1
d
s,
h)
d q,
g)
es
2
Z
Z
3
2 dx
j)
d
x,
k)
,
2
x + 3x + 3
1 − 3x + 3x 2 − x 3
Z
Z
√
10
3
m)
d
v,
n)
3e−3θ − 8 5 − 6θ dθ.
2
2v + 8v + 58
6. Find the indefinite integral of the given functions:
Z
Z
dx
50 d x
p
p
a)
,
b)
,
2
1 − (2x + 3)
1 − (25x)2
Z
Z
3 dx
dx
√
√
,
e)
,
d)
2
−2x − x
2x − x 2
Z
Z
2
1
g)
d
x,
h)
dy,
2
2
1 + (x + 1)
y − 2y + 5
Z
1
d t,
cos2 8t
Z
3e−3h+1 d h,
c)
f)
Z
i)
l)
1
dT ,
+ 4T + 5
Z
2
db,
2
b − 2b + 5
T2
Z
c)
f)
i)
2
p
dy,
3 + 2y − y 2
Z
5 dx
p
,
36 − (5x)2
Z
5 dz
.
1 + (2 − 5z)2
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Indefinite Integral
Answers to Exercises
1 4
sin ω ,
4
b)
− cos 3 t 2 ,
d)
2(sin β)3/2
,
3
e)
g)
e2 sin t ,
1. a)
c)
tan4 φ,
e−2 ρ ,
f)
−e−2 r ,
h)
ln3 W,
i)
2 ln3/2 y,
k)
(8 s 3 + 27)1/3 ,
l)
p
2
−1
,
1 + C2
q
3 p 2 − 5 p + 6,
b)
d)
3 (1 + 2 sin t)2/3 ,
e)
1
,
sin3 φ
p
− 1 + cos2 r,
g)
arcsin
3v 2
,
2
h)
arcsin
j)
2. a)
−
√
j)
arcsin ln x,
k)
15 arctan
k2
√
15
,
5
1
,
sin2 θ
b)
− cos4 x,
c)
d)
arctan2 ρ,
e)
(1 + 2x)3/2
,
3
f)
g)
3
,
−
2 (x 2 + 1)
h)
j)
−3 (8 − x 2 )2/3 ,
k)
3. a)
(2x 2 + 7)3/2
,
6
−7
,
4 (1 + 2 x)2
i)
l)
4 + W 5.
f)
1
,
cos u
lnln ln x ,
i)
−2 lncos 3x ,
l)
1
arctan q 2 +
.
3
c)
√ t
2e ,
3
√
ln2 x,
√
5 − 4x
−
,
2
9 (x 3 + 10)4/3
,
4
3
− cos5 t ,
5
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Indefinite Integral
m)
4. a)
d)
g)
1
ln |n2 − 1| ,
2
ln | arcsin x|,
(4ρ − 3)5
,
20
1
,
−
(3x − 7)4
−
5. a)
−
− cot 2v,
g)
es +
1
,
es
√
m)
h)
k)
− arctan2 x + ln(1 + x 2 )
,
2
(2x + 1)4
,
8
1
−
,
p−3
√
4m + 9
,
2
cos(2ω − 5)
,
2
d)
2
e)
(5 − 6x)4/3
,
8
j)
j)
b)
c)
arctan ln x,
f)
6
,
6−τ
i)
−5 (8 − 3x)11/5 ,
l)
√
− 3 − 2l.
c)
1
tan 8 t,
8
2 e7r−8 ,
f)
−e−3h+1 ,
h)
eq/2 + e−q/2 ,
i)
arctan(T + 2),
k)
1
,
(1 − x)2
l)
arctan
n)
−e−3 θ + (5 − 6 θ )4/3 .
b)
−
e)
1
cot(3φ − 7),
3
3 (2x + 3)
,
3
v+2
,
5
1
arcsin(2x + 3) ,
2
b)
d)
arcsin(x + 1),
e)
3 arcsin(x − 1),
g)
arctan(x + 1),
h)
arctan
6. a)
sin1/3 y.
√
3 arctan
arctan
o)
tan(φ − 1),
n)
2 arcsin 25x,
y−1
,
2
b−1
,
2
Contents
c)
f)
i)
y−1
,
2
5x
arcsin
,
6
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2 arcsin
arctan(5z − 2).
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Indefinite Integral
Quiz—Substitution Method
Choose the correct answer (only one is correct).
Z
1. (1 pt.)
(3x − 11)9 d x =
1
(3x − 11)10 + C
10
1
(3x − 11)8 + C
24
3
(3x − 11)10 + C
10 Z
√
2. (1 pt.)
7 − 3x d x =
1
(3x − 11)10 + C
30
2p
3
(7 − 3x)2 + C
9
2p
−
(7 − 3x)3 + C
9Z
1
3. (1 pt.)
dx =
2 − 3x
1
− ln |2 − 3x| + C
3
1p
(7 − 3x)3 + C
2
p
−2 (7 − 3x)3 + C
−
ln |2 − 3x| + C
Contents
−3 ln |2 − 3x| + C
1
ln | − 3x| + C
2
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Indefinite Integral
Z
4. (1 pt.)
2ln x
dx =
x
2ln x
+C
ln x
1
2x
+C
ln 2 √
Z
e x
√ dx =
5. (1 pt.)
x
√ √x
xe +C
2+C
2ln x
+C
ln 2
√
2e
x
+C
√
x
e +C
Z
1
6. (1 pt.)
dx =
(2x + 3)4
1
ln(2x + 3)4 + C
2
1
1
+C
6 (2x + 3)3
Z
7. (1 pt.) x sin(x 2 + 4)d x
−
x2
cos(x 2 + 4) + C
2
−
cos(x 2 + 4) sin(x 2 + 4)
+
+C
2
4x 2
e x
√ +C
x
ln(2x + 3)4 + C
1
+C
(2x + 3)3
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1
− cos(x 2 + 4) + C
2
−x cos(x 2 + 4) + sin(x 2 + 4) + C
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Indefinite Integral
Z
1+x
√
dx
1 − x2
p
arcsin x − 1 − x 2 + C
8. (1 pt.)
p
arctan x − 1 − x 2 + C
Z r
arccos x
dx
9. (1 pt.)
1 − x2
x+C
2
arccos3 x + C
3 Z
1
10. (1 pt.)
dx
2
(1 + x ) arctan x
1
+C
(1 + x 2 )4
ln | arctan x| + C
Correct Answers:
Points Gained:
Success Rate:
2
x + x2
+C
1 − x2
(1 + x) ln |x +
p
1 − x 2| + C
√
arcsin x arccos x
+C
2
2√
−
arccos3 x + C
3
ln |1 + x 2 | +
1
+C
1 + x2
arctan2 x
+C
2
Contents
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Indefinite Integral
1.3 Decomposition into Partial Fractions
When we work with polynomials the decomposition into simpler (linear or quadratic) factors plays an
important role. Likewise in the case of rational functions something similar is important in many applications. While polynomials are decomposed into the product of factors rational functions will be
decomposed into the sum of simple rational functions, also called partial fractions.
Let us briefly review some basic knowledge necessary in the following text:
• Rational function is the quotient of two polynomials.
• Each improper rational function, where the degree of the numerator is greater then or equal to degree
of denominator, can be expressed using the long division algorithm as the sum of a polynomial and
a proper rational function (degree of numerator is less then degree of denominator).
• Degree of polynomial will be denoted by deg(P ).
Partial fractions are special rational functions. We distinguish two types:
A
,
(x − α)k
where k ∈ N, α, A ∈ R,
and
Contents
(x 2
Mx + N
,
+ px + q)k
where k ∈ N, M, N, p, q ∈ R, p 2 − 4q < 0.
Any partial fraction of the first type has a power (the first or higher) of a linear polynomial x − α in the
denominator, and a constant in the numerator. Any partial fraction of the second type has a power (the
first or higher) of a quadratic polynomial x 2 + px + q with complex roots (the discriminant is negative)
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Indefinite Integral
in the denominator, and a linear polynomial (or a constant if M = 0) in the numerator. Partial fractions
are proper rational each time.
P (x)
Theorem 1.31 Let R(x) = Q(x)
be a proper rational function with real coefficients. Let the factorization of the denominator Q(x) into factors irreducible in the real domain have the form
Q(x) = a(x − α1 )k1 · · · (x − αr )kr (x 2 + p1 x + q1 )l1 · · · (x 2 + ps x + qs )ls .
Then R(x) can be written as the sum of partial fractions. To k -tuple real root α of the denominator
there correspond k partial fractions of the form
A1
A2
Ak
,
,...,
2
x − α (x − α)
(x − α)k
and to l -tuple pair of complex conjugate roots of the denominator belonging to the quadratic trinomial
x 2 + px + q there correspond l partial fractions of the form
M2 x + N2
Ml x + Nl
M1 x + N1
, 2
,..., 2
.
2
2
x + px + q (x + px + q)
(x + px + q)l
In this theorem it is important that the function R(x) is a proper rational. If that is not the case, then
first it must be expressed as the sum of a polynomial and of a proper rational function, which can then be
decomposed.
Comment 1.32
i) It is possible to prove that the decomposition from the previous theorem is unique up to the order of
summands, that is undetermined coefficients are unique.
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81
ii) A similar result can be proved in the complex domain (that is coefficients of a rational function can
be complex numbers), but in this case only partial fractions of the first type are necessary. This is due
to the fact that in the complex domain any polynomial can be factored to powers of linear factors. Of
course, coefficients in the decomposition are generally also complex numbers.
Finding of Coefficients of the Decomposition
1. First we check whether the function is a proper rational. Otherwise, using the long division algorithm,
we express it as the sum of a polynomial and a proper rational function, which we then decompose.
2. We decompose the denominator into the product of factors irreducible in the real domain.
3. Based on this decomposition we write the assumed form of the decomposition into the sum of partial
fractions with undetermined coefficients. We equate this expression with the given rational function,
the denominator of which was replaced by its factorization obtained in Item 2.
4. We multiply the equation obtained by the (factored) denominator of the original rational function to
clear the fractions. We get the equality of two polynomials. The coefficients of one are known, the
coefficients of the other are undetermined (they must be calculated).
5. Two polynomials are equal if and only if they have the same degree, and the coefficients at the same
powers are equal. Thus we expand the polynomials on both sides, combine terms with the same
powers, and arrange them in increasing or decreasing powers. Then we equate coefficients at like
powers on the left and right sides of the equation. We obtain a system of linear equations having the
only solution because of the uniqueness of the decomposition.
6. If the denominator has real roots, it is useful to substitute them into the mentioned equation before its
expansion. After the substitution all terms with undetermined coefficients up to one are equal to zero
so we can easily find the undetermined coefficient at the nonzero term. This can be done for each real
root. Then it is sufficient to equate coefficients only at some powers (to obtain the necessary number
of equations for those coefficients we do not yet know).
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Indefinite Integral
7. Another method of finding the coefficients is to substitute an arbitrary set of n + 1 numbers into the
obtained equation; here n is the highest power of the variable which occurs in the equation. Again,
we get a system of linear equations having a unique solution.
Example 1.33 Decompose into partial fractions the rational function
R(x) =
x2
x
.
−1
Solution. The function is a proper rational and no division is necessary. The factorization of the denominator is x 2 − 1 = (x + 1)(x − 1). Thus the denominator has two simple roots, −1 and 1, to which
there correspond two chains, both of the length one of partial fractions of the first type. The assumed
decomposition is
A
B
x
=
+
.
(x + 1)(x − 1)
x+1 x−1
After multiplying by the denominator (x + 1)(x − 1) we get the equation
x = A(x − 1) + B(x + 1).
We successively substitute both real roots and get:
x = −1
=⇒
−1 = −2A
=⇒
x=1
=⇒
1 = 2B
=⇒
1
,
2
1
B= .
2
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A=
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Thus the decomposition is
1
2
I
I
1
2
x
=
+
.
x2 − 1
x+1 x−1
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Indefinite Integral
Example 1.34 Decompose into partial fractions the rational function
2x 3 − x 2 + x − 2
.
x4 + x2
Solution. The function is a proper rational and no division is necessary. The factorization of the denominator is evidently x 4 + x 2 = x 2 (x 2 + 1). Thus the denominator has the double root 0, and the pair of
simple complex conjugate roots i and −i belonging to the binomial x 2 + 1. To root 0 there corresponds
the chain of the length two of partial fractions of the first type, to the polynomial x 2 +1 there corresponds
one chain of the length one of partial fractions of the second type. The assumed decomposition is
R(x) =
2x 3 − x 2 + x − 2
A
B
Cx + D
=
+
.
+
x 2 (x 2 + 1)
x
x2
x2 + 1
After multiplying by the factored denominator x 2 (x 2 + 1) we get the equation
2x 3 − x 2 + x − 2 = Ax(x 2 + 1) + B(x 2 + 1) + (Cx + D)x 2 .
We expand the right-hand side and combine like terms:
2x 3 − x 2 + x − 2 = (A + C)x 3 + (B + D)x 2 + Ax + B.
We equate the coefficients at like powers of x on both sides of the equation.
Contents
x3 :
2 = A + C,
x :
1 = A,
x2 :
−1 = B + D,
x0 :
−2 = B.
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Therefore, A = 1, B = −2, C = 1, and D = 1. The decomposition is
3
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2
1
2
x+1
2x − x + x − 2
= − 2+ 2
.
x4 + x2
x
x
x +1
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Indefinite Integral
Example 1.35 Decompose into partial fractions the rational function
R(x) =
x 3 + 3x 2 + 4
.
x3 + x − 2
Solution. The function is an improper rational therefore we must first divide it. The quotient is 1 and the
remainder is 3x 2 − x + 6 so
3x 2 − x + 6
R(x) = 1 + 3
.
x +x−2
Now we must factorize the denominator, that is find the roots of the equation x 3 + x − 2 = 0. Evidently,
number 1 is the root of this equation. Therefore, x 3 +x −2 = (x −1)(x 2 +x +2). Because the quadratic
trinomial x 2 + x + 2 has complex roots (discriminant is D = −7 < 0) the previous expression is the
factorization into factors irreducible in the real domain. A partial fraction of the first type corresponds
to the simple root 1. A partial fraction of the second type corresponds to the trinomial x 2 + x + 2. The
assumed decomposition is
A
Bx + C
3x 2 − x + 6
=
+ 2
.
2
(x − 1)(x + x + 2)
x−1 x +x+2
After multiplying we get the equation
Contents
3x 2 − x + 6 = A(x 2 + x + 2) + (Bx + C)(x − 1).
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After substituting the real root x = 1 we have 8 = 4A, that is A = 2. Equating coefficients at like
powers of x , on both sides of the equation, we get the equations for the remaining two numbers. After
expanding the right-hand side and combining terms at like powers we have
3x 2 − x + 6 = (A + B)x 2 + (A − B + C)x + 2A − C.
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Indefinite Integral
We choose any two equations containing B and C .
x2 :
3 = A + B,
x0 :
6 = 2A − C.
Thus B = 1 and C = −2. The decomposition is
2
x−2
3x 2 − x + 6
=
+
.
x3 + x − 2
x − 1 x2 + x + 2
For the given improper rational function we have
x 3 + 3x 2 + 4
2
x−2
=
1
+
+
.
x3 + x − 2
x − 1 x2 + x + 2
N
1.4 Integration of Rational Functions
Another important group of functions, that can be integrated (at least theoretically) within the set of
elementary functions, include the rational functions. For their successful integration some results from
algebra, that were summed up in the previous section, are required. From these we know that any rational
function P (x)/Q(x), where P (x) and Q(x) are polynomials, can be written in the form
P (x)
= S(x) + R1 (x) + · · · + Rs (x),
Q(x)
where S(x) is a polynomial and R1 (x), . . . , Rs (x) are partial fractions. These functions are continuous
on any interval which does not include real roots of the denominator Q(x), thus they have antiderivatives
and
Z
Z
Z
Z
P (x)
dx =
S(x) d x + R1 (x) d x + · · · + Rs (x) d x.
Q(x)
As the integration of polynomial S(x) is trivial, it is sufficient to learn how to integrate partial fractions.
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Indefinite Integral
1.4.1 Integration of Partial Fractions with Real Roots in the Denominator
First we tackle partial fractions of the first type, whose denominators have real roots. Their integration
is easy. Using Formula 4 (Table 1.1) for k = 1 we get
Z
A
d x = A ln |x − α| + c.
x−α
For k = 2 we use a substitution and Formula 3 from Table 1.1. We get
Z
Z
Z
x−α =t A
dt
=A
dx = = A t −k d t =
dx = dt (x − α)k
tk
=A
A
A
t −k+1
+c =
+c =
+ c.
k−1
−k + 1
(1 − k)t
(1 − k)(x − α)k−1
We demonstrate the use on a few examples.
Z
3x + 16
Example 1.36 Evaluate
dx.
2
x −x−6
Solution. The integrand is the proper rational function as deg(3x + 16) = 1 < deg(x 2 − x − 6) = 2.
We decompose it into partial fractions. Thus we need the roots of the denominator.
(
√
1 ± 1 + 24
1±5
−2,
2
x −x−6=0
⇒
x1,2 =
=
=
2
2
3.
Therefore, x 2 − x − 6 = (x − 3)(x + 2). Both roots are real and simple. The chain of the length one
of partial fractions corresponds to each root. The expected decomposition is
3x + 16
3x + 16
A
B
=
=
+
,
−x−6
(x − 3)(x + 2)
x−3 x+2
x2
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Indefinite Integral
where A and B are appropriate constants. After multiplying the equality by the factorized denominator
(x − 3)(x + 2) we get the equality of two polynomials:
3x + 16 = A(x + 2) + B(x − 3).
The fastest way to find the constants A and B is to substitute both roots respectively. We have
x = −2 :
10 = −5B
⇒
B = −2,
x=3:
25 = 5A
⇒
A = 5.
Now we can evaluate the integral:
Z
Z 3x + 16
2
5
d
x
=
−
dx =
x2 − x − 6
x−3 x+2
Z
Z
dx
dx
=5
−2
= 5 ln |x − 3| − 2 ln |x + 2| + c.
x−3
x+2
The result is valid on any from the intervals (−∞, −2), (−2, 3), and (3, +∞).
Z
1
Example 1.37 Evaluate
dx.
x5 − x3
N
Contents
Solution. The integrand is the proper rational function because we have deg(1) = 0 < deg(x 5 − x 3 ) =
= 5. The denominator can be easily factored: x 5 − x 3 = x 3 (x 2 − 1) = x 3 (x − 1)(x + 1). All the
roots are real; triple root 0 to which corresponds the chain of three partial fractions, and simple roots 1
and −1 to which corresponds one partial fraction, respectively. The expected decomposition is
x5
1
A
D
E
1
B
C
= 3
= 3+ 2+ +
+
.
3
−x
x (x − 1)(x + 1)
x
x
x
x−1 x+1
(1.12)
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Indefinite Integral
To find the undetermined constants A, B, C, D, E we multiply the previous equality by the denominator
x 3 (x − 1)(x + 1). We get
1 = A(x 2 − 1) + Bx(x 2 − 1) + Cx 2 (x 2 − 1) + Dx 3 (x + 1) + Ex 3 (x − 1).
(1.13)
Substituting the real roots of the denominator we find three constants:
x=0:
1 = −A
⇒
A = −1,
x=1:
1 = 2D
⇒
D = 1/2,
x = −1 :
1 = 2E
⇒
E = 1/2.
The constants B and C remain to be found. We equate coefficients at like powers of the variable x on
both sides of (1.13). After the expansion and the combination of like powers we get
1 = (C + D + E)x 4 + (B + D − E)x 3 + (A − C)x 2 − Bx − A.
It is sufficient to choose two appropriate equations:
x1 :
2
x :
0 = −B
⇒
B = 0,
0=A−C
⇒
C = −1.
Evidently, all constants can be found by equating coefficients, but the combination of this methods and
the substitution of real roots is usually the fastest.
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Indefinite Integral
Now we can come to the evaluation of the integral. From (1.12) we get
Z
Z 1
1
1
1/2
1/2
dx =
− 3− +
+
dx =
x5 − x3
x
x
x−1 x+1
Z
Z
Z
Z
dx
dx
1
dx
1
dx
=−
−
+
+
=
3
x
x
2
x−1 2
x+1
=
1
1
1
−
ln
|x|
+
ln
|x
−
1|
+
ln |x + 1| + c.
2x 2
2
2
(In the first integral you must realize that 1/x 3 = x −3 .) The result is valid on any interval which does
not include real roots of the denominator, that is the numbers 0, 1, and −1.
N
Z 5
x + x 4 − 2x 3 − x 2 + 1
dx.
Example 1.38 Evaluate
x 3 + 3x 2 + 3x + 1
Solution. This time the integrand is the improper rational function (degree of the numerator is 5 and
degree of the denominator is 3) thus first the polynomials must be divided:
(x 5 + x 4 − 2x 3 − x 2 + 1) : (x 3 + 3x 2 + 3x + 1) = x 2 − 2x + 1
−(x 5 + 3x 4 + 3x 3 + x 2 )
−2x 4 − 5x 3 − 2x 2 + 1
−(−2x 4 − 6x 3 − 6x 2 − 2x)
3
2
x + 4x + 2x + 1
−(x 3 + 3x 2 + 3x + 1)
x2 − x
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Indefinite Integral
Therefore,
x 5 + x 4 − 2x 3 − x 2 + 1
x2 − x
2
−
2x
+
1
+
=
x
.
x 3 + 3x 2 + 3x + 1
x 3 + 3x 2 + 3x + 1
The obtained proper rational function must be decomposed into partial fractions. To do so we need the
roots of the denominator. Evidently, x 3 + 3x 2 + 3x + 1 = (x + 1)3 . Thus the denominator has the
unique root −1 of multiplicity three to which there corresponds the chain of three partial fractions. The
expected decomposition is
x2 − x
x2 − x
A
B
C
=
=
+
+
.
3
2
3
3
2
x + 3x + 3x + 1
(x + 1)
(x + 1)
(x + 1)
x+1
After multiplying by the denominator (x + 1)3 and expanding the right-hand side we get:
x 2 − x = A + B(x + 1) + C(x + 1)2 ,
x 2 − x = Cx 2 + (2C + B)x + (A + B + C).
Equating of coefficients at like powers gives:
x2 :
1=C
x1 :
−1 = 2C + B
x0 :
Thus
0=A+B +C
⇒
C = 1,
⇒
B = −3,
⇒
A = 2.
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2
2
3
1
x −x
=
−
+
.
3
3
2
(x + 1)
(x + 1)
(x + 1)
x+1
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Indefinite Integral
All together
x 5 + x 4 − 2x 3 − x 2 + 1
2
3
1
2
=
x
−
2x
+
1
+
−
+
,
(x + 1)3
(x + 1)3 (x + 1)2 x + 1
so
Z
x 5 + x 4 − 2x 3 − x 2 + 1
dx =
x 3 + 3x 2 + 3x + 1
Z
Z
2
= (x − 2x + 1) d x + 2
dx
−3
(x + 1)3
Z
dx
+
(x + 1)2
Z
dx
.
x+1
We evaluate and simplify integrals of the first two partial fractions (the third one is evident).
Z
Z
x+1=u du
1
dx
u−2
1
=
=− 2 =−
=
=
,
dx = du
(x + 1)3
u3
−2
2u
2(x + 1)2
Z
Z
x+1=u du
dx
u−1
1
1
=
=
−
=
−
.
=
=
dx = du
(x + 1)2
u2
−1
u
x+1
The final result, valid on intervals (−∞, −1) and (−1, +∞) is
Z 5
x + x 4 − 2x 3 − x 2 + 1
1 3
1
3
2
d
x
=
x
−
x
+
x
−
+
+ ln |x + 1| + c.
x 3 + 3x 2 + 3x + 1
3
(x + 1)2 x + 1
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N
1.4.2 Integration of Partial Fractions with Complex Roots in the Denominator
Now we tackle partial fractions of the second type. The denominators of these contain the quadratic
trinomial x 2 + px + q with the negative discriminant p 2 − 4q < 0, that is having complex roots.
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Indefinite Integral
First we examine the case p = 0. Then necessarily q > 0 and we can put q = a 2 , where a > 0.
Thus the partial fraction has the form
Mx + N
,
(x 2 + a 2 )n
where M, N, a ∈ R, a > 0, n ∈ N.
Then
Z
Z Mx
N
+
dx =
(x 2 + a 2 )n (x 2 + a 2 )n
Z
Z
x
1
=M
d x.
n dx + N
2
2
2
(x + a )
(x + a 2 )n
Mx + N
dx =
(x 2 + a 2 )n
(1.14)
The first integral can be handled easily. For n = 1 using Formula 14 from Table 1.1 we get
Z
Z
x
2x
1
1
1 d
x
=
d x = lnx 2 + a 2 + c = ln x 2 + a 2 + c,
2
2
2
2
x +a
2
x +a
2
2
because for any x ∈ R there is x 2 + a 2 > 0.
For n = 2 we use the substitution. We have
2
x + a2 = u
Z
Z
Z
x
1 du
1
d x = 2x d x = d u =
·
=
u−n d u =
(x 2 + a 2 )n
un 2
2
1
x dx = 2 du
−n+1
1 u
1
1
·
+c =
· n−1 + c =
2 −n + 1
2(1 − n) u
1
1
=
·
+ c.
2
2(1 − n) (x + a 2 )n−1
=
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Indefinite Integral
The second integral from (1.14) is more difficult to handle. Let us denote
Z
1
Jn (x, a) =
d x,
a > 0.
2
(x + a 2 )n
Using the integration by parts method we derive a recurrence formula
Jn (x, a) =
1
x
2n − 3
·
+
Jn−1 (x, a),
n−1
2
2
2
(2n − 2)a (x + a )
(2n − 2)a 2
(1.15)
which for n = 2 expresses the integral Jn (x, a) using Jn−1 (x, a). This enables us to transform Jn (x, a)
step by step to the known integral J1 (x, a), having Number 9 from Table 1.1:
Z
1
x
1
J1 (x) =
d x = arctan + c.
2
2
x +a
a
a
Let us prove Formula (1.15). Using integration by parts method we get for n = 2 that
Z
u=
1
0 = −(n−1)2x u
1
n n−1
2
2
(x +a ) =
(x 2 +a 2 )
dx = Jn−1 =
n−1
v0 = 1
v=x
(x 2 + a 2 )
Z
x
x2
+
(2n
−
2)
=
n dx =
n−1
(x 2 + a 2 )
(x 2 + a 2 )
Z 2
x
x + a2 − a2
=
+
(2n
−
2)
n dx =
n−1
(x 2 + a 2 )
(x 2 + a 2 )
Z
Z
x
1
1
2
=
+
(2n
−
2)
d
x
−
(2n
−
2)a
dx =
n−1
n−1
2 + a 2 )n
2
2
2
2
(x
(x + a )
(x + a )
x
=
+ (2n − 2)Jn−1 − (2n − 2)a 2 Jn .
n−1
(x 2 + a 2 )
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94
Indefinite Integral
Now from the equation
Jn−1 =
x
(x 2
n−1
+ a2)
+ (2n − 2)Jn−1 − (2n − 2)a 2 Jn ,
we evaluate
(2n − 2)a 2 Jn =
x
(x 2
n−1
+ a2)
+ (2n − 3)Jn−1 .
Dividing by (2n − 2)a 2 we obtain Formula (1.15).
Z
Example 1.39 Evaluate
3x − 8
d x , x ∈ (−∞, +∞).
(x 2 + 2)3
2
Solution.
√ The integrand is a partial fraction of the second type, where p = 0 and q = a = 2, that is
a = 2. We will use the hints provided previously. First
Z
Z
Z
3x − 8
x
1
dx = 3
dx − 8
d x.
2
3
2
3
2
(x + 2)
(x + 2)
(x + 2)3
For the first integral we have
Z
2
Z
x +2=u
Z
x
1 du
1
2x d x = d u =
d
x
=
·
=
u−3 d u =
3
(x 2 + 2)3
u
2
2
x dx = 1 du 2
−2
=
1u
1
1
=− 2 =−
.
2
2 −2
4u
4(x + 1)2
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Indefinite Integral
To evaluate the second integral we use Formula (1.15) twice, first for n = 3 and then for n = 2:
Z
√ √ 1
1
x
3
d x = J3 x, 2 =
· 2
+
J2 x, 2 =
2
3
2
(x + 2)
4 · 2 (x + 2)
4·2
√ x
3
1
x
1
=
=
+
·
+
J1 x, 2
8(x 2 + 2)2 8 2 · 2 x 2 + 2 2 · 2
x
3x
x
3 1
√ arctan √ .
=
+
+
2
2
2
8(x + 2)
32(x + 2) 32 2
2
The complete result is
Z
x
3
3x − 8
3
x
3x
− √ arctan √ + c.
dx = −
− 2
−
2
3
2
2
2
2
(x + 2)
4(x + 1)
(x + 2)
4(x + 2) 4 2
2
N
At the end we notice partial fractions of the second type in case p 6 = 0. The approach will be similar
as when p = 0, that is we separate a partial fraction into two appropriate parts, but the manipulations
will be somewhat more difficult. The denominator of the first fraction will be created in such a way that
it will be a multiple of the derivative of the trinomial x 2 + px + q , that is the binomial 2x + p . Thus
we will find two appropriate numbers r and s such that
Mx + N
r(2x + p) + s
2x + p
1
= 2
=r 2
+s 2
.
(x 2 + px + q)n
(x + px + q)n
(x + px + q)n
(x + px + q)n
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The numbers r and s must satisfy
J
Mx + N = r(2x + p) + s = 2rx + (pr + s)
⇒
M = 2r,
N = pr + s,
from which we get r = M/2, s = N − pM/2. It is useless to remember these formulae by heart, each
time we find them easily equating the coefficients.
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96
Indefinite Integral
Therefore,
Z
Mx + N
dx = r
(x 2 + px + q)n
Z
2x + p
dx + s
(x 2 + px + q)n
Z
1
d x.
(x 2 + px + q)n
The first integral will be evaluated similarly as the first integral in (1.14). For n = 1 we have
Z
2x + p
d x = lnx 2 + px + q + c = ln x 2 + px + q + c,
2
x + px + q
because the denominator is positive each time (graph of function y = x 2 + px + q is a parabola not
intersecting the x -axis and lying above it, since the corresponding quadratic equation has complex roots).
For n = 2 we use the substitution
2
Z
Z
x + px + q = u 2x + p
du
=
d x = =
2
n
(2x + p) d x = d u
(x + px + q)
un
=
1
1
1
1
· n−1 + c =
· 2
+ c.
1−n u
1 − n (x + px + q)n−1
The second integral will be transformed to evaluate the integral Jn using a substitution. First we
complete the square:
p 2 p 2
p 2 4q − p 2
x + px + q = x +
−
+q = x+
+
.
2
4
2
4
2
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As we assume that the discriminant p 2 − 4q is negative we have (4q − p 2 )/4 > 0 so we can set
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Indefinite Integral
(4q − p2 )/4 = a 2 , where a > 0. Now
Z
Z
x+ p =u 1
1
2
=
dx =
d x = dx = du (x 2 + px + q)n
[(x + p/2)2 + a 2 ]n
Z
1
=
d u = Jn (u, a).
2
(u + a 2 )n
We illustrate the whole procedure with the following example.
Z
5x + 1
Example 1.40 Evaluate
d x , x ∈ (−∞, +∞).
x2 + x + 1
Solution. The trinomial x 2 + x + 1 in the denominator has the discriminant 12 − 4 · 1 · 1 = −3, which
is negative, so its roots are complex. Thus it is a partial fraction of the second type. The derivative of the
denominator is 2x + 1. The numerator 5x + 1 will be arranged into the form r(2x + 1) + s . So
5x + 1 = r(2x + 1) + s = 2rx + (r + s)
⇒
5 = 2r, 1 = r + s,
which means that r = 5/2 and s = −3/2. After the separation we get
5x + 1
=
2
x +x+1
therefore
5
2
(2x + 1) −
x2 + x + 1
3
2
=
5 2x + 1
3
1
−
,
2
2
2 x +x+1 2 x +x+1
Z
Z
5x + 1
5
2x + 1
3
1
d
x
=
d
x
−
d x.
x2 + x + 1
2
x2 + x + 1
2
x2 + x + 1
The first integral is easy evaluate (the derivative of the denominator is in the numerator):
Z
2x + 1
d x = ln(x 2 + x + 1).
2
x +x+1
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Z
(1.16)
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98
Before evaluating the second integral we complete the square:
1 2 1
1 2 3
x2 + x + 1 = x +
− +1= x+
+ .
2
4
2
4
√
Now using Formula 9 from Table 1.1, where a = 3/2, we get:
Z
Z
x+ 1 =u 1
1
2
=
dx =
dx = 2
dx = du x2 + x + 1
x + 12 + 34
Z
1
1
u
=
d u = √ arctan √ =
3
2
3
3
u +4
2
2
2 x + 12
2
2x + 1
2
= √ arctan √ .
= √ arctan √
3
3
3
3
Substituting the current results into (1.16) we get:
Z
5x + 1
5
3 2
2x + 1
d x = ln(x 2 + x + 1) − √ arctan √
+c =
x2 + x + 1
2
2 3
3
√
5
2x + 1
= ln(x 2 + x + 1) − 3 arctan √
+ c.
2
3
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Indefinite Integral
1.4.3 Integration of Partial Fractions with Real and Complex Roots in the Denominator
Z 3
x + 3x 2 + 4
dx.
Example 1.41 Evaluate
x3 + x − 2
Solution. The decomposition of the integrand into partial fractions was found in Example 1.35:
x 3 + 3x 2 + 4
2
x−2
=
1
+
+
.
x3 + x − 2
x − 1 x2 + x + 2
The second fraction must be written as follows:
1
5
(2x + 1)
x−2
2
2
=
−
x2 + x + 2
x 2 + x + 2 (x + 12 )2 +
7
4
.
Integrating we get:
Z
1
5
2x + 1
x 3 + 3x 2 + 4
d x = x + 2 ln |x − 1| + ln(x 2 + x + 2) + √ arctan √
+ c.
3
x +x−2
2
7
7
N
Z
Example 1.42 Evaluate
2x 3 + 8x 2 − 8x − 22
dx.
(x + 3)2 (x 2 + 1)
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2
Solution. The denominator has the double real root −3 and the factor x + 1 irreducible in the real
domain. Thus the form of the decomposition is:
3
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2
2x + 8x − 8x − 22
A
B
Cx + D
=
+
+ 2
.
2
2
2
(x + 3) (x + 1)
(x + 3)
x+3
x +1
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100
Indefinite Integral
Multiplying by the factorized denominator we get:
2x 3 + 8x 2 − 8x − 22 = A(x 2 + 1) + B(x + 3)(x 2 + 1) + (Cx + D)(x + 3)2 ,
2x 3 + 8x 2 − 8x − 22 = Ax 2 + Ax + Bx 3 + 3Bx 2 + Bx + 3B + Cx 3 +
+ 6Cx 2 + 9Cx + Dx 2 + 6Dx + 9D,
2x 3 + 8x 2 − 8x − 22 = x 3 (B + C) + x 2 (A + 3B + 6C + D) +
+ x(A + B + 9C + 6D) + 3B + 9D.
Equating the coefficients at like powers we find: A = 2, B = 1, C = 1, and D = −3. Thus
2x 3 + 8x 2 − 8x − 22
2
x−3
1
=
+ 2
,
+
2
2
2
(x + 3) (x + 1)
(x + 3)
x+3 x +1
so
Z
2x 3 + 8x 2 − 8x − 22
dx =
(x + 3)2 (x 2 + 1)
Z
2
dx +
(x + 3)2
Z
1
dx +
x+3
Z
x−3
d x.
x2 + 1
The third integral must be written as follows:
Z
Z
Z
x−3
1
2x
1
dx =
dx − 3
d x.
2
2
2
x +1
2
x +1
x +1
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Evaluating each integral we get:
Z
J
2x 3 + 8x 2 − 8x − 22
−2
1
dx =
+ ln |x + 3| + ln |x 2 + 1| − 3 arctan x + c.
2
2
(x + 3) (x + 1)
(x + 3)
2
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101
Comment 1.43
1) The integration of rational functions consists of two parts:
• decomposition into partial fractions (and perhaps division, if the function is an improper rational),
• integration of particular fractions (and perhaps a polynomial, if the function is an improper rational).
These parts can be of a different difficulty. The first often consumes most of the time necessary to
evaluate the example, but the integration itself is very fast. This is typical if the denominator has only
real roots. In the case of complex roots, the integration can be very long, particularly when they are
multiple.
2) You should realize that the integration can be effectively made only if we are able to decompose
the integrand into partial fractions and the roots of the denominator are needed for this. But we do
not know how to find roots of higher order polynomials in general. We are able to solve linear and
quadratic equations; there exist formulae for equations of the third and fourth orders but they are
too complicated for practical purposes; for equations of order five and more the general solution in
radicals does not exist (see [14]). Students often forget this and consider the integration of rational
functions to be an easy (if time consuming) problem. In the practice we do not work with sample
functions prepared in such a way that the results are nice, but with functions, where the coefficients of
polynomials are obtained for example by measurement. Then we cannot avoid the use of numerical
methods.
3) From the previous text it can be seen that the result of the integration of a rational function can be
obtained as the sum, where only some functions can appear as particular summands (if we do not
combine logarithms, and avoid artificial manipulations like writing eln x instead of x ). The functions
that can occur in the result are
• polynomials,
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Indefinite Integral
102
• rational functions,
• logarithms of linear and quadratic polynomials,
• inverse tangents of linear polynomials.
4) Do not forget to check whether the trinomials x 2 + px + q in the denominators have complex roots,
that is p 2 − 4q < 0 (thus they cannot be factorized into two linear polynomials with real roots). If
you overlook it and try to proceed as described above for partial fractions with complex roots in the
denominator, everything works up to one step. After completing the square and substitution, instead
of binomial u2 + a 2 you get binomial u2 − a 2 , a > 0. The integral
Z
du
2
(u − a 2 )n
cannot be evaluated using recurrence Formula (1.15) and nothing remains than to decompose this
function into partial fractions, which should have been done directly at the beginning. This way you
come to the right result but with a detour requiring some additional work. It would be possible
R d uto
derive an analogy of Formula (1.15) for this case and to evaluate that for n = 1 the integral u2 −a 2
leads to two logarithms. But this approach is not reasonable because the integration of partial fractions
of the first type is much faster.
A
5) Sometimes it is better to write partial fractions of the first type x−α
, where α is a rational number, in
1
a slightly different form. For example, if α = 2 , that is the factor is x − 12 , we look for the partial
fraction
A
A
instead of
.
2x − 1
x − 21
Beware that the constant A will be different each time! In fact, we multiplied the numerator and
the denominator of the second partial fraction by two. The next thing is that when we integrate this
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103
Indefinite Integral
fraction we must use Formula 14 from Table 1.1:
Z
Z
Z
A
1
A
2
A
dx = A
dx =
d x = ln |2x − 1|.
2x − 1
2x − 1
2
2x − 1
2
Exercises
1. Find the indefinite integral of the given functions:
Z
x
d x,
a)
(x + 1)(2x + 1)
Z
5x − 14
c)
d x,
x 3 − x 2 − 4x + 4
Z
dx
e)
,
x(2 + x)
Z
32s
g)
d s,
(2s − 1)(4s 2 − 16s + 15)
Z
dx
i)
,
6 + x − x2
Z
x 2 − 3x + 2
k)
d x,
x 3 + 2x 2 + x
Z
2(x 2 − 4x + 5)
m)
d x,
x+3
Z
b)
d)
f)
h)
j)
l)
n)
u
d u,
2u2 − 3u − 2
Z
33
d h,
6h3 − 7h2 − 3h
Z
12(y − 1)
dy,
(y + 1)(y 2 − 4)
Z
6(x 3 + 1)
d x,
3
x − 5x 2 + 6x
Z
18(3x 2 + 1)
d x,
x 4 − 3x 2 + 2x
Z
4(3x 2 + 1)
d x,
(x 2 − 1)3
Z
x dx
.
x2 − x − 2
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Indefinite Integral
2. Find the indefinite integral of the given functions:
Z
dx
a)
,
(3 + x)(1 + 2x + x 2 )
Z
dx
c)
,
x(16 − 24x + 9x 2 )
Z
2x 2 + 41x − 91
e)
d x,
(x − 1)(x 2 − x − 12)
Z
3x 3 − 5x 2 + 8x
d x,
g)
(x 2 − 2x + 1)(x 2 − 1)
Z 5
x + x 4 + 3x 3 + x 2 − 2
i)
d x,
x4 − 1
Z
x+3
k)
d x,
(x 2 − x + 1)2
Z
5x + 3
m)
d x,
2
(x − 2x + 5)3
3. Find the indefinite integral of the given functions:
Z
x2 dx
,
a)
x 3 + 5x 2 + 8x + 4
Z
4z2
c)
d z,
1 − z4
Z
6r
d r,
e)
r3 + 1
Z
x
d x,
4 − 4x + x 2
Z
x2
d x,
(x + 2)2 (x + 4)2
b)
d)
Z
f)
2x 2
Z
h)
1
d x,
+ 9x − 5
9x − 5
d x,
− 6x + 1
9x 2
x 4 + x 3 − 2x 2 + 2x + 3
d x,
x2 + x − 2
Z
3x + 1
d x,
(x 2 + 2)3
Z
2x + 2
d x.
(x − 1)(x 2 + 1)2
Z
j)
l)
n)
Z
b)
x3
Z
d)
Z
f)
5x − 1
d x,
− 3x − 2
10(7x 2 − 1)
d x,
x 4 + 4x 2 − 5
4t 3
dt
t4 + 1
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Indefinite Integral
Z
x+1 2
x−1
Z
2x
d x,
x
+1
Z
Z
4
8
i)
d
x,
j)
dy,
x(x 4 + 1)
y4 + 1
√
√
Hint: x 4 + 1 = (x 2 + 1)2 − 2x 2 = (x 2 − 2 x + 1)(x 2 + 2 x + 1).
Z
Z
v2
x3 + 1
k)
d x,
d
v,
l)
v 3 + 5v 2 + 8v + 4
x3 − x2
Z
Z
x2
3x 4
m)
d
x,
n)
d x.
2
2
x +2
x +2
g)
d x,
4. Find the indefinite integral of the given functions:
Z
1
a)
d x,
x3 − x2
Z
6m
c)
dm,
3
m −1
Z
4
e)
d z,
(z2 + 1)(z2 + z)
Z
2(p 3 − 6)
g)
dp,
p 4 + 6p 2 + 8
Z
6(3x + 8)
i)
d x,
2
x + 2x + 10
Z
28(−5u + 16)
k)
d u,
2u2 + 7
Z
30
d x,
m)
4x 2 + 4x + 16
h)
x4
Z
b)
2
d x,
x(x 2 + 1)
6x 2 − x + 1
d x,
x3 − x
Z
4
d x,
(x + 1)2 (x 2 + 1)
Z
6b
db,
2
b + 2b + 4
Z
10
d x,
(5x + 4)3
Z
28
dy,
2y 2 + 4y + 6
Z
4p
dp.
(p 2 − 1)(p 2 + 1)
Z
d)
f)
h)
j)
l)
n)
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Indefinite Integral
Answers to Exercises
1. a)
c)
ln |x + 1| −
1
ln |2x + 1|,
2
9 ln |3h + 1| − 11 ln |h| + 2 ln |2h − 3|,
f)
ln |y − 2| + 8 ln |y + 1| − 9 ln |y + 2|,
g)
ln |2s − 1| + 5 ln |2s − 5| − 6 ln |2s − 3|,
h)
6x − 27 ln |x − 2| + ln |x| + 56 ln |x − 3|,
i)
1 2 + x ln
,
5 3 − x k)
2 ln |x| − ln |x + 1| +
m)
x 2 − 14x + 52 ln |x + 3|,
c)
e)
2
1
ln |u − 2| +
ln |2u + 1| ,
5
10
e)
1 x ,
ln
2 2 + x − ln |x − 2| + 3 ln |x − 1| − 2 ln |x + 2|,
d)
2. a)
b)
j)
6
,
x+1
−1
1 3 + x + ln
,
2(1 + x) 4
1+x
x 4
1
,
+ ln
16 4 − 3x
3x − 4 (x − 1)4 (x − 4)5 ,
ln
(x + 3)7
l)
n)
b)
d)
f)
|x|9 · |x − 1|4
−24
+ ln
,
x−1
|x + 2|13
1
1
−
+
,
2
(x − 1)
(x + 1)2
1
2
ln |x + 1| + ln |x − 2|.
3
3
2
+ ln |2 − x|,
2−x
x + 4
−1
− 4 ,
+ 2 ln
x+2
x + 2 x + 4
1 2x − 1 ln
,
11
x+5 Contents
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Indefinite Integral
2
3
−
+ ln |(x − 1)(x + 1)2 |,
2
2(x − 1)
x−1
g)
−
h)
2
+ ln |3x − 1|,
3(3x − 1)
i)
p
x2
+ x + ln |x 2 − 1| · x 2 + 1 + arctan x,
2
j)
p
x3
3
+ ln |x − 1|5 · |x + 2|,
3
k)
14
2x − 1
7x − 5
+ √ arctan √
,
3(x 2 − x + 1) 3 3
3
l)
x−6
3x
3
x
+
+ √ arctan √ ,
8(x 2 + 2)2
32(x 2 + 2) 32 2
2
m)
2x − 7
3
x−1
3x − 3
+
arctan
,
+
4(x 2 − 2x + 5)2
16(x 2 − 2x + 5) 32
2
n)
1 (x − 1)2
1
ln 2
+ 2
− arctan x.
2
x +1
x +1
3. a)
c)
4
ln |x + 1| +
,
x+2
z + 1
− 2 arctan z,
ln
z − 1
b)
d)
√
x − 2
− 2 ,
ln
x + 1 x + 1
√
x − 1
x
,
12 5 arctan √ + 5 ln
x + 1
5
2r − 1
3 arctan √ ,
3
Contents
e)
−2 ln |r + 1| + ln(r 2 − r + 1) + 2
f)
ln(t 4 + 1),
g)
ln |x| −
h)
arctan x 2 ,
i)
4 ln |x| − ln(x 4 + 1),
j)
√
√
√
√
√
√
y2 + y 2 + 1
√
+ 2 2 arctan y 2 + 1 + 2 2 arctan y 2 − 1 ,
2 ln
2
y −y 2+1
4
,
x−1
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Indefinite Integral
k)
m)
4. a)
4
,
v+2
√
x
x − 2 arctan √ ,
2
x − 1
1
,
+ ln
x
x ln |v + 1| +
c)
√
2m + 1
2 ln |m − 1| − ln(m2 + m + 1) + 2 3 arctan √
,
3
d)
4 ln |x + 1| − ln |x| + 3 ln |x − 1| ,
e)
4 ln |z| − 2 ln |z + 1| − ln(z2 + 1) − 2 arctan z,
f)
2 ln |x + 1| −
g)
h)
i)
k)
m)
l)
n)
b)
2
− ln(x 2 + 1),
x+1
√
p
p
3 arctan − ln(p2 + 2) − 3 2 arctan √ + 2 ln(p 2 + 4),
2
2
√
b
+
1
3 ln(b2 + 2b + 4) − 2 3 arctan √ ,
3
x
+
1
,
j)
9 ln(x 2 + 2x + 10) + 10 arctan
3
√
2u
32 14 arctan √ − 35 ln(2u2 + 7),
l)
14
√
2x + 1
15 arctan √
,
n)
15
1
+ 2 ln |x − 1|,
x
√
x
x 3 − 6x + 6 2 arctan √ .
2
x − ln |x| +
2 ln |x| − ln(x 2 + 1),
1
,
(5x + 4)2
√
y+1
7 2 arctan √ ,
2
2
p − 1
.
ln 2
p + 1
−
Contents
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Indefinite Integral
Quiz—Indefinite Integrals of Rational Functions
Choose the correct answer (only one is correct).
Z
1. (1 pt.)
1
dx =
x 2 + 2x
ln |x 2 + 2x| + C
ln |x 2 | +
1
1
ln |x| − ln |x + 2| + C
2 Z
2
1
2. (1 pt.)
dx =
x 2 (x + 1)
1
1
1
− + 2 +C
x+1 x
x
1
− ln |x + 1| + ln |x| −
+C
x−1
Z
x
3. (1 pt.)
dx =
(x + 2)2
1
ln |x| + C
2
arctan(x + 1) + C
ln |x + 1| − ln |x| −
arctan
1
+C
x
x
+C
x+1
Contents
ln |x + 2| − 2 ln(x + 2) + C
2
ln |x + 2| +
+C
x+2
x2
arctan x + C
2
x2
ln(x + 2)2 + C
2
2
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110
Indefinite Integral
Z
4. (1 pt.)
x−1
dx =
x+1
x + 2 ln |x − 1| + C
x − 2 ln |x + 1| + C
Z
1
5. (1 pt.)
dx =
2
x − 6x + 10
arctan(x − 3) + C
2 1
5 1
−
+C
3x−1 3x+1
1
1
+
+C
x−1 x+1
arctan(x + 3) + C
x +C
arctan
3
Z
x4
dx =
6. (1 pt.)
x2 + 1
x3
+ ln |x + 1| + ln |x − 1| + C
3
1
ln |(x − 3)2 + 1| + C
2
x3 x5
+
+C
3
5
Z 2
x + 2x − 1
7. (1 pt.)
dx =
x3 + x
x3
− x + arctan x + C
3
2 arctan x − ln |x| + ln(x 2 + 1) + C
ln |x| + ln |x + 1| + ln |x − 1| + C
x3
+ arctan x + C
3
Contents
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2
+ ln |x| + C
x4
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ln |x| + 2 ln |x + 1| + C
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111
Indefinite Integral
Z
x+5
dx =
+x−2
ln |x 2 + x − 2|
+ 4 arctan(x 2 + x + 2) + C
2
8. (1 pt.)
x2
2 ln |x − 1| − ln |x + 2| + C
Z 2
x + 3x + 3
dx =
9. (1 pt.)
x2 + x + 2
1
+ ln |x + 1| + C
(x + 1)2
x + ln |x 2 + x + 2| + C
Z
1
10. (1 pt.)
dx =
2
x −4
arctan x2
+C
2
ln |x 2 − 4| + C
2 arctan 2x +
1
+C
2
ln |x 2 + x − 2|
+ 2 ln |x − 1| − ln |x + 2| + C
2
x + arctan(x 2 + x + 2) + C
ln |x 2 + x + 2| + C
arctan(x − 2) + C
1
ln |x − 2| − ln |x + 2| + C
4
Contents
Correct Answers:
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Indefinite Integral
1.5 Integration of Some Special Types of Functions
In this section we will study integrals that can be transformed into integrals of rational functions, using
appropriate substitutions. These will be integrals containing trigonometric functions and roots. Such
integrals are common in applications.
To be able to describe exactly the integrals we will deal with we need the concept of rational functions
of two and more variables. This is why we introduce the next notation:
The symbol R(u, v) indicates in the sequel the fraction that has in both the numerator and the denominator only the sums of expressions of the form aum v n , where a is a real constant and m and n are
nonnegative integers, that is a ∈ R, m, n ∈ N ∪ {0}. The mapping (u, v) → R(u, v) is called rational
function of two variables. Rational functions of two variables u and v are for example
uv − 4,
u+v
,
u−v
uv + 2
,
u2 − v 2
u2 v 3 − 2uv + 1
.
uv − u4 v 2 − 3
Analogously we introduce rational functions of three and more variables. Variables can be denoted
by various letters. For example,
R(x, y, z) =
xy 2 z3 − xy − xz + 2x − 3z + 5
x 5 z7 − xyz + y − 4z
is the rational function of three variables x , y , and z.
Contents
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1.5.1 Integrals of Rational Expressions of Trigonometric Functions
In this subsection we will deal with integrals of the form
Z
R(cos x, sin x) d x.
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(1.17)
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Indefinite Integral
Examples of such integrands include:
R(cos x, sin x) = cos2 x sin3 x,
cos2 x
,
sin3 x
1 + 3 cos2 x
R(cos x, sin x) =
.
2 − cos x sin x
R(cos x, sin x) =
Comment 1.44
1) We integrate functions that are obtained from the functions cos x , sin x and real numbers, using a
finite number of arithmetic operations (addition, subtraction, multiplication and division).
2) If we also add tan x and cot x to the initial functions, we do not get anything more. The reason is that
sin x
x
after the substitution tan x = cos
and cot x = cos
into such an expression and a simplification we
x
sin x
again get a rational expression of sines and cosines.
3) More generally, we can consider the functions cos ax and sin ax , where a 6 = 0. As the integration is
analogous we restrict to the case a = 1.
First we notice very frequent integrals
Z
Contents
cosm x sinn x d x,
where m, n ∈ Z.
(1.18)
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The situation is easy if at least one of the numbers m, n is odd. The substitution
sin x = t
if m is odd,
cos x = t
if n is odd,
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Indefinite Integral
transforms the integral (1.18) to the integral of a rational function. If both numbers are odd we can, of
course, choose which substitution to use.
We have already evaluated one integral of this type (see Example 1.21). Let us show further.
Z
Example 1.45 Evaluate cos5 x sin2 x d x , x ∈ R.
Solution. As m = 5 and n = 2, we use the substitution sin x = t . The differential is cos x d x = d t .
We separate one cosine from the integrand and manipulate the remainder, to involve only sines, making
the replacement easy. Using the identity cos2 x + sin2 x = 1 we get
2
2
cos4 x sin2 x = (cos2 x) sin2 x = (1 − sin2 x) sin2 x.
The whole evaluation is as follows:
Z
Z
sin x = t
2
5
2
2
2
cos x sin x d x = (1 − sin x) sin x cos x d x = cos x d x = d t
Z
Z
2
= (1 − t 2 ) t 2 d t = (t 6 − 2t 4 + t 2 ) d t =
=
=
t 7 2t 5 t 3
1
2
1
−
+ + c = sin7 x − sin5 x + sin3 x + c.
7
5
3
7
5
3
Contents
N
The next integral has been already evaluated (see Example 1.8 c). This time we use a different
approach.
Z
dx
Example 1.46 Evaluate
, x ∈ (0, π).
sin x
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Indefinite Integral
Solution. Following (1.18) we have m = 0, n = −1. Thus the substitution will be cos x = t . We obtain
Z
Z
Z
Z
cos x = t
dx
sin x d x
sin x d x
dt
− sin x d x = d t =
=
=
=
.
2
2x
2−1
sin x
1
−
cos
t
sin x
sin x d x = −d t We decompose the rational function into partial fractions. The denominator has two simple roots
t = 1 and t = −1. Therefore, the decomposition is
t2
1
A
B
=
+
.
−1
t −1 t +1
Multiplying by the denominator t 2 − 1 = (t − 1)(t + 1) we get the equality
1 = A(t + 1) + B(t − 1).
Substituting the roots of the denominator gives
t = −1 :
1 = −2B
⇒
t =1:
1 = 2A
⇒
1
B=− ,
2
1
A= .
2
Now we can finish the evaluation of the integral:
Z
Z dx
1/2
1/2
1
1
=
−
d t = ln |t − 1| − ln |t + 1| + c =
sin x
t −1 t +1
2
2
1
1
1 cos x − 1 = ln |cos x − 1| − ln |cos x + 1| + c = ln
+ c.
2
2
2
cos x + 1 Try to imagine what manipulations would be necessary to obtain the same form of the result as in ExamN
ple 1.8 c.
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Indefinite Integral
The remaining case is when both exponents are even in the integral (1.18). The general case will
be described below on page 123. If both numbers m, n are nonnegative, the fastest method relies on the
application of double-angle formulae:
sin2 α =
1 − cos 2α
,
2
cos2 α =
1 + cos 2α
,
2
α ∈ R.
(1.19)
The next two examples illustrate their use.
Z
Example 1.47 Evaluate cos2 x d x , x ∈ R.
Solution. This is the integral that we encounter very often in applications. We have already evaluated it
once (see Example 1.16). Using the previous trigonometric identity we find the result more quickly:
Z
Z
Z
Z
1 + cos 2x
1
1
2
dx =
cos x d x =
dx +
cos 2x d x =
2
2
2
1 sin 2x
x
1
1
= x+ ·
+ c = + sin 2x + c.
2
2
2
2 4
N
Z
Example 1.48 Evaluate sin2 x cos4 x d x , x ∈ R.
Solution. First we rearrange the integrand using Formulae (1.19), in which we set α = x :
1 − cos 2x
1 + cos 2x 2 1
·
= (1 − cos2 2x)(1 + cos 2x) =
2
2
8
1
= (1 + cos 2x − cos2 2x − cos3 2x).
8
sin2 x cos4 x =
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Indefinite Integral
Thus we have:
Z
Z
1
sin x cos x d x =
(1 + cos 2x − cos2 2x − cos3 2x) d x =
8
Z
Z
1
sin 2x
2
3
=
x+
− cos 2x d x − cos 2x d x .
8
2
2
4
We evaluate the two integrals we have obtained. In the first one we again use Formula (1.19), in
which we set this time α = 2x . We get
Z
Z
1 + cos 4x
1
sin 4x
x
1
2
cos 2x d x =
dx =
x+
= + sin 4x.
2
2
4
2 8
The second integral is of the type (1.18), where m = 3, n = 0. We get
Z
Z
sin 2x = t
3
2
cos 2x d x = (1 − sin 2x) cos 2x d x = 2 cos 2x d x = d t =
cos 2x d x = 1 d t 2
Z
3
1
1
t
1
1
=
(1 − t 2 ) d t =
t−
= sin 2x − sin3 2x.
2
2
3
2
6
The final result is:
Z
x
1
x
1
1
sin2 x cos4 x d x = +
sin 2x −
−
sin 4x −
sin 2x +
8 16
16 64
16
1
x
1
1
+
sin3 2x + c =
−
sin 4x +
sin3 2x + c.
48
16 64
48
N
R
Now we will come to the general case of the integral R(cos x, sin x) d x . First we present the
universal substitution. Then we tackle three special types which can be usually evaluated more quickly
using different substitutions. In all cases the integral will be reduced to an integral of a rational function.
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Indefinite Integral
The universal substitution tan x2
We will show that the substitution
tan
x
= t,
2
x = 2 arctan t,
sin x =
2t
,
1 + t2
x ∈ (−π, π),
2
d t,
1 + t2
1 − t2
cos x =
1 + t2
dx =
transforms the integral (1.17) to an integral of a rational function. We will use a
mnemonic device to quickly derive the necessary identities. We use a reference right
triangle with the size of one angle equal to x/2. Setting the length of the adjacent leg
equal to one we get from the definition of tangent (the ratio of the lengths of the opposite 1
and adjacent legs) that the length of the opposite
leg is t . By Pythagoras’ Theorem we
√
2
have that the length of the hypotenuse is 1 + t (see Fig. 1.3). From the definition of
sine and cosine (the ratio of the lengths of the opposite and adjacent leg, respectively,
and the hypotenuse) we get
(1.20)
t
√
1 + t2
x
2
Fig. 1.3
x
t
x
1
=√
,
cos = √
.
2
2
1 + t2
1 + t2
Using the double-angle formulae, in which we replace x by x/2, and Formulae (1.21) we get:
sin
x
x
t
1
2t
· cos = 2 √
·√
=
,
2
2
2
2
1
+
t2
1+t
1+t
2 2
1
t
1 − t2
2 x
2 x
cos x = cos
− sin
= √
− √
=
.
2
2
1 + t2
1 + t2
1 + t2
sin x = 2 sin
(1.21)
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Indefinite Integral
Now applying Theorem 1.27, we can reduce the integral (1.17) involving sine and cosine to an integral
of a rational function:
Z
Z 1 − t2
2t
2
R(cos x, sin x) d x = R
,
·
d t.
2
2
1+t 1+t
1 + t2
We will demonstrate it with an example.
Z
5
Example 1.49 Evaluate
d x on the interval x ∈ (−π, π).
4 + sin x
Solution. We use the substitution tan x2 = t .
tan x
2
Z
x
5
dx = dx
4 + sin x
sin x
5
=
2
Z
=t
= 2 arctan t
2
= 1+t
2 dt
=
2t
1+t 2
Z
5
2
·
dt =
=
2t
4 + 1+t 2 1 + t 2
dt
5
=
t
2
t + 2 +1
2
Z
t+
1
4
4 t+
5
4
= · √ arctan √
2
15
15
dt
1 2
4
+
15
16
=
Contents
10
4t + 1
+ c = √ arctan √
+c =
15
15
4 tan x2 + 1
10
√
= √ arctan
+ c.
15
15
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120
Indefinite Integral
Special substitutions sin x , cos x , and tan x
The universal substitution (1.20) can be applied to each integral (1.17), but the integrals of rational functions we obtain are often complicated (the degree of the denominator is rather high). Sometimes the integrand in (1.17) can be rearranged to a special form and another trigonometric substitution can be used,
leading to an integral of a simpler rational function (the degree of the denominator is lower). Namely,
we have in mind the following types (below, S(w) means a rational function of a single variable):
R(cos x, sin x) = S(sin x) · cos x
substitution
sin x = t,
(1.22)
R(cos x, sin x) = S(cos x) · sin x
substitution
cos x = t,
(1.23)
R(cos x, sin x) = S(tan x)
substitution
tan x = t.
(1.24)
While in Types (1.22) and (1.23), Theorem 1.18 is used and there are no problems with the replacement,
in Type (1.24) Theorem 1.27 must be used. General instructions on how to decide which of the previous
substitutions can be used, are in Comment 1.55. We will illustrate the usage on a few examples.
Z
sin3 x
Example 1.50 Evaluate
d x , x ∈ R.
1 + 4 cos2 x + 3 sin2 x
Solution. The integral is of Type (1.23). Using the identity sin2 x + cos2 x = 1 after the rearrangement
we get
Z
Z
sin3 x
(1 − cos2 x) sin x
d
x
=
dx =
4 + cos2 x
1 + 4 cos2 x + 3 sin2 x
Z 2
Z 2
cos x = t
t −1
t +4−5
= − sin x d x = d t =
dt =
dt =
2
t +4
t2 + 4
sin x d x = −d t Contents
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Indefinite Integral
5
1
t
d t = t − 5 · arctan + c =
t2 + 4
2
2
5
cos x
= cos x − arctan
+ c.
2
2
=
Z 1−
The resulting integrand was an improper rational function, so we reduced it to the sum of the polynomial
(in our case it was the constant 1) and the proper rational function. Formula 9 from Table 1.1 was also
used.
N
The substitution tan x
When the substitution (1.24) is to be applied, it is necessary to express the old variable with the help
of the new one before the differential is evaluated. Furthermore, we need to express sine and cosine as
functions of tangent, as in the following Table.
tan x = t,
x = arctan t,
t
,
sin x = √
1 + t2
x ∈ (−π/2, π/2),
1
dx =
d t,
1 + t2
1
cos x = √
.
1 + t2
(1.25)
Contents
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These identities can be again derived using a mnemonic device—a reference right triangle similar to that
one for the substitution tan x2 = t (see Fig. 1.3). The only difference is that the size of the angle is x.
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Indefinite Integral
Z
Example 1.51 Evaluate
dx
1 + sin2 x
,
x ∈ (−π/2, π/2).
Solution. The integral is of Type (1.24). Manipulating the integrand we get:
1
sin2 x + cos2 x
=
·
1 + sin2 x
2 sin2 x + cos2 x
1
cos2 x
1
cos2 x
=
sin2 x
+1
cos2 x
2
sin x
2 cos
2x + 1
=
tan2 x + 1
.
2 tan2 x + 1
Hence, it is the integral of Type (1.24). We have
Z
tan x = t
dx
tan x + 1
x = arctan t =
dx = =
2
2
2 tan x + 1
1 + sin x
d x = 2d t
t +1
Z 2
Z
Z
t +1
dt
dt
1
dt
=
·
=
=
=
2t 2 + 1 t 2 + 1
2t 2 + 1
2
t 2 + 1/2
√
1 1
t
1
= · 1 arctan 1 + c = √ arctan 2 t + c =
√
2 √
2
2
2
Z
2
√
1
= √ arctan 2 tan x + c.
2
Formula 9 from Table 1.1 was used again. The function ϕ(t) = arctan t fulfils the assumptions of
Theorem 1.27 on the interval (−∞, +∞).
The resulting antiderivative is defined only on the interval (−π/2, π/2). But the integrand is the
function continuous on R . In subsection 1.6.2 we will explain how the antiderivative can be constructed
on the whole real line.
N
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123
Indefinite Integral
Integrals of Type (1.24) also include integral (1.18), in which both m and n are even integers. Even
powers of sine and cosine can be expressed like rational functions of the variable t . As we have already
stated, if the numbers m, n are nonnegative, the use of Formulae (1.19) is faster. If at least one of these
is negative, the substitution tan x = t must be used.
Z
sin4 x
d x , x ∈ (−π/2, π/2).
Example 1.52 Evaluate
cos4 x
Solution. This time it is easy to express the integrand using tangent:
Z
tan x = t
Z
Z
t4
2 +1)2
t4
dt
sin x
(t
x
=
arctan
t
=
d
x
=
=
dt =
·
1
cos4 x
t2 + 1
t2 + 1
d x = 2d t
(t 2 +1)2
t +1
Z 2
Z (t + 1)(t 2 − 1) + 1
1
2
=
dt =
dt =
t −1+ 2
t2 + 1
t +1
4
t3
1
− t + arctan t + c = tan3 x − tan x + arctan(tan x) + c =
3
3
1
= tan3 x − tan x + x + c.
3
=
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The improper rational function was reduced to the sum of a polynomial and a proper rational function.
(The same result could be obtained by dividing t 4 : (t 2 + 1).) Then we used the fact that the functions
tangent and inverse tangent are mutually inverse on the interval (−π/2, π/2), thus arctan(tan x) = x .
If we had noticed that the integrand is in fact tan4 x , then the first manipulation could have been a bit
shorter. This would not have influenced the following evaluation.
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124
Comment 1.53 Sometimes it happens that a concrete integral fits more types. Then it is important to
choose the one leading to the shortest possible solution. The general rule is to choose (if it is possible)
— first Type (1.22) or (1.23), that is the substitution for sine or cosine,
— second Type (1.24), that is the substitution for tangent,
— universal substitution (1.20) for tangent of half-angle, as a last resort.
It is natural that no such rule is valid absolutely and exceptions can occur as the next example demonstrates.
Z
dx
Example 1.54 Evaluate
, x ∈ (0, π).
sin x
Solution. We have already evaluated this integral twice (see Examples1.8 c and 1.46). Now we use the
universal substitution (1.20) and get
tan x = t
Z
Z
Z
2
1
dx
2
dt
x = 2 arctan t =
= ·
=
d
t
=
2t
2
sin x
1
+
t
t
2
dx = 2 2 dt 1+t
1+t
x
= ln |t| + c = lntan + c.
2
Evidently, of the three methods this one was the fastest.
N
Comment 1.55 In case of rational expressions R(cos x, sin x) involving sines and cosines, sometimes it won’t be
clear at first sight that it fits one of special types (1.22)–(1.24), which usually lead to a simpler integration. This
can be found from the properties of rational function R(u, v).
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Indefinite Integral
A rational function R(u, v) is called
odd with respect to the variable u if R(−u, v) = −R(u, v),
odd with respect to the variable v if R(u, −v) = −R(u, v),
even with respect to the variables u, v if R(−u, −v) = R(u, v).
The equality must hold for all values u, v for which the function R(u, v) is defined.
It is possible to prove (see [20, p. 25]) that the integrand R(cos x, sin x) is of type
(1.22) if the function R(u, v) is odd with respect to the variable u (we set sin x = t ),
(1.23) if the function R(u, v) is odd with respect to the variable v (we set cos x = t ),
(1.24) if the function R(u, v) is even with respect to the variables u, v (we set tan x = t ).
For example, in Example 1.50 we had
R(u, v) =
v3
,
1 + 4u2 + 3v 2
thus
(−v)3
−v 3
=
= −R(u, v).
1 + 4u2 + 3(−v)2
1 + 4u2 + 3v 2
Therefore, the function was odd with respect to the variable v and the substitution cos x = t could be used.
Similarly, in Example 1.51 we had
1
R(u, v) =
1 + v2
(the function does not depend on u at all), thus
R(u, −v) =
R(−u, −v) =
1
1
=
= R(u, v).
1 + (−v)2
1 + v2
Therefore, the function was even, with respect to the variables u, v and the substitution tan x = t could be used.
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Indefinite Integral
At the end of this section we mention the integration of expressions that also involve trigonometric
functions, but are not of type (1.17) on page 112. These are the integrals
Z
Z
Z
sin ax cos bx d x,
sin ax sin bx d x,
cos ax cos bx d x,
where a, b ∈ R, a 6 = 0, b 6 = 0. We replace the integrands using formulae
1
sin(α + β) + sin(α − β) ,
2
1
cos(α − β) − cos(α + β) ,
sin α sin β =
2
1
cos α cos β =
cos(α − β) + cos(α + β) .
2
Z
√
Example 1.56 Evaluate sin 2 x cos 3x d x , x ∈ (−∞, +∞).
sin α cos β =
√
√
Solution. Using the corresponding formula (a = 2, b = 3, that is setting α = 2 x , β = 3x ) we get
Z
Z
√
√
√
1
sin 2 x cos 3x d x =
sin 2 x + 3x + sin 2 x − 3x d x =
2
Z
Z
√
√
1
1
=
sin 2 + 3 x d x +
sin 2 − 3 x d x =
2
2
√
√
cos 2 + 3 x
cos 2 − 3 x
√
√
=−
−
+ c.
2 2+3
2 2−3
N
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Indefinite Integral
Exercises
1. Find the indefinite integral of the given functions:
Z
β
a)
cos2 dβ,
2
Z
c)
sin2 x d x,
Z
5
cos x d x,
e)
Z
g)
sin3 ε
dε,
cos2 ε + 1
2. Find the indefinite integral of the given functions:
Z
2
a)
d x,
sin x cos3 x
Z
c)
15 sin2 θ cos3 θ dθ,
Z
e)
Z
g)
Z
i)
cos3 x
d x,
sin2 x
Z
sin3 u d u,
Z
sin5 x d x,
Z
sin6 x d x,
Z
du
.
(2 + cos u) sin u
Z
12 sin3 x cos3 x d x,
Z
cos6 ρ sin5 ρ dρ,
Z
3 sin3 h
d h,
cos4 h
b)
d)
f)
h)
b)
d)
f)
sin3 y + 1
dy,
cos2 y
h)
8 cos4 x d x,
j)
Z
Z
32 sin4 u cos2 u d u,
32 cos6 x d x.
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Indefinite Integral
3. Find the indefinite integral of the given functions:
Z
a)
128 cos4 β sin4 β dβ,
Z
c)
dx
d x,
5 + 4 cos x
g)
i)
k)
1
d x,
1 − cos x
Z
1
d x,
sin x − 1
Z
3
d x,
5 + 4 sin x
Z
2
d x,
5 − 3 cos x
4. Find the indefinite integral of the given functions:
Z
3 dx
,
a)
5 − 4 cos x + 3 sin x
Z
1 − tan z
c)
d z,
1 + tan z
Z
sin 2ω
dω,
e)
cos4 ω
Z
sin 2x
g)
d x,
4
sin x + cos4 x
Z
4
i)
d x,
1 + tan x
60 sin5 α cos5 α dα,
Z
1 + sin u
d u,
1 − sin u
b)
d)
Z
Z
e)
Z
f)
h)
j)
l)
1
dα,
1 + sin α
Z
5 dx
,
3 sin x − 4 cos x
Z
8 dx
,
sin 2x − 2 sin x
Z
2 − sin x
d x.
2 + cos x
Z
b)
d)
f)
h)
j)
5 dx
,
2 sin x − cos x + 5
Z
2(1 + tan u)
d u,
sin 2u
Z
3 cos2 x
d x,
sin4 x
Z
6 cos x
d x,
(1 − cos x)2
Z
1
d x,
(sin x + cos x)2
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Indefinite Integral
Z
k)
3
dα,
cos4 α
5. Find the indefinite integral of the given functions:
Z
x
3x
a)
sin cos
d x,
4
4
Z
c)
cos 3x cos 4x d x,
sin
1
d x.
1 + 3 cos2 x
Z
sin 3x sin x d x,
b)
Z
sin 5x cos 7x d x,
d)
√
Z
e)
Z
l)
Z
3 x cos x d x,
f)
cos
3
1
x cos x d x.
2
2
Answers to Exercises
1. a)
c)
e)
g)
sin β
β
+ ,
2
2
b)
cos3 u
− cos u,
3
x
1
− sin 2x ,
2 4
d)
2
cos5 x
cos3 x −
− cos x,
3
5
sin5 x
2
− sin3 x + sin x,
5
3
cos ε − 2 arctan cos ε ,
f)
h)
5x
sin 2x
3 sin 4x
sin3 2x
−
+
+
,
16
4
64
48
1 u
u
ln tan2 + 3 tan .
3
2
2
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130
Indefinite Integral
tan2 x + 2 ln | tan x|,
b)
3 sin4 x − 2 sin6 x,
c)
5 sin3 θ − 3 sin5 θ,
d)
2
1
1
cos9 ρ −
cos11 ρ − cos7 ρ ,
9
11
7
e)
−
f)
1 − 3 cos2 h
,
cos3 h
h)
2u −
2. a)
g)
i)
3. a)
c)
e)
g)
i)
k)
l)
1 + sin2 x
,
sin x
1 + sin y + cos2 y
,
cos y
sin 4x
2 sin 2x + 3x +
,
4
sin 8β
,
8
1
x
2
arctan
tan
,
3
3
2
1
−
,
tan x2
3β − sin 4β +
2
j)
sin 4u 2 3
− sin 2u,
2
3
3
2
sin 4x − sin3 2x + 10x + 8 sin 2x.
2
3
b)
10 sin6 α − 15 sin8 α + 6 sin10 α,
d)
−
f)
,
h)
−1
x
4
5
tan +
,
j)
2 arctan
3
2 3
x
arctan 2 tan
,
2
√
4
3
x
2 x
2 x
ln tan
+ 3 − ln tan
+ 1 + √ arctan
tan
.
2
2
3
2
3
tan
x
2
4
− u,
tan u2 − 1
2
,
−
tan α2 + 1
x
x
ln2 tan − 1 − lntan + 2,
2
2
1
x
− 2 lntan ,
tan2 x2
2
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Indefinite Integral
4. a)
−
2
3 tan x2 + 1
√
,
c)
ln |1 + tan z| −
e)
b)
1
ln(1 + tan2 z),
2
√ 5
x
5 arctan
3 tan + 1 ,
5
2
d)
tan u + ln | tan u|,
1
,
cos2 ω
f)
−
g)
− arctan(2 cos2 x − 1) or arctan tan2 x,
h)
i)
2 ln |1 + tan x| − ln(1 + tan2 x) + 2x,
j)
k)
tan3 α + 3 tan α,
l)
5. a)
c)
e)
− cos x
x
+ cos ,
2
2
1
(sin 7x + 7 sin x),
14
√
√
cos 3 + 1 x
cos 3 − 1 x
√
√
−
−
,
2 3+1
2 3−1
b)
d)
f)
1
,
tan3 x
3
1
,
x −
tan 2
tan3 x2
−1
,
tan x + 1
1
1
arctan
tan x .
2
2
1
(2 sin 2x − sin 4x),
8
cos 12x
cos 2x
−
+
,
24
4
1
(2 sin x + sin 2x).
4
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Indefinite Integral
Quiz—Indefinite Integrals of Trigonometric Functions
Choose the correct answer (only one is correct).
Z
1. (1 pt.)
sin2 x cos x d x =
sin3 x
+C
3
− cos3 x sin x + C
Z
cos x
2. (1 pt.)
dx =
sin2 x
1
+C
cos x
tan x + C
Z
3. (1 pt.) cos x(5 sin x − 4)7 d x =
1
sin x (−5 cos x − 4x)8 + C
8
1
(5 sin x − 4)8 + C
40
x − sin4 x + C
sin3 x cos x + C
−
1
+C
sin x
1
+C
tan x
1
(5 sin x − 4)8 + C
8
5
7
1
cos2 x
sin2 x − 4x +C
2
2
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Indefinite Integral
Z
4. (1 pt.)
cos3 x d x =
cos4 x
+C
4
sin3 x
sin x −
+C
3
Z
5. (1 pt.) cos2 x tan3 x d x =
sin4 x
+C
4 cos x
cos2 x
− ln | cos x| + C
2Z
cot x
dx =
6. (1 pt.)
sin2 x
2
− 2 +C
sin x
1
+C
2Zsin2 x
sin x
7. (1 pt.)
dx =
1 − sin2 x
− cos x − ln | sin x| + C
−
1
+C
cos x
sin4 x
+C
4
sin x −
cos2 x
+C
2
− sin2 x ln | cos x| + C
sin2 x
− ln | sin x| + C
2
ln | sin3 x| + C
1
+C
2 cos2 x
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− ln | cos2 x| + C
−
cos x
+C
x + cos2 x
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Indefinite Integral
Z
cot(2x + 1) d x =
8. (1 pt.)
−
2
+C
sin (2x + 1)
2
ln | sin(2x + 1)|
+C
Z 2
sin 2x
9. (1 pt.)
dx =
sin2 x + 3
√ √ lnsin x + 3 + lnsin x − 3 + C
ln(sin2 x + 3) + C
tan(2x + 1)
+C
2
2 ln | sin2 (2x + 1)| + C
arctan √x3
√
+C
3
x + ln(sin2 x + 3) + C
Correct Answers:
Points Gained:
Success Rate:
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Indefinite Integral
1.5.2 Integrals Involving Roots
First we will notice integrals having the form
Z
√ R x, s x d x,
where s ∈ N, s = 2.
We will present a few examples of such integrands.
√
√
√
√ x+ x
x4 x
4
√ , R x, x =
√ ,
R x, x =
x− x
1+ 4 x
√ R x, 3 x =
(1.26)
√
3
x
√ .
x+3 x
√
Comment 1.57 Simply said, we consider functions that can be obtained from the functions x and s x
and real numbers with the help of a finite number of arithmetic operations (addition, subtraction, multiplication, and division).
The substitution x = t s transforms the integral (1.26) to the integral of a rational function. We
demonstrate it with an example.
Z 2 √
x + x+1
√
Example 1.58 Evaluate
d x , x ∈ (0, ∞).
x+ x
√
Solution. Using the substitution x = t 2 , that is t = x , we get:
Z 4
Z 2 √
x = t2 x + x+1
t +t +1
=
√
d x = · 2t d t =
d x = 2t d t x+ x
t2 + t
Z 4
t +t +1
=2
d t,
t +1
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Indefinite Integral
which is the integral of an improper rational function. So it is necessary to reduce it to the sum of a
polynomial and a proper rational function. Using the long division algorithm we get:
t4 + t + 1
1
= t3 − t2 + t +
.
t +1
t +1
The complete result is
Z
√
Z 1
x2 + x + 1
3
2
√
dx = 2
t −t +t +
dt =
x+ x
t +1
4
t3 t2
t
− + + ln |t + 1| + c =
=2
4
3
2
√
x2 2 √
=
− x x + x + 2 ln x + 1 + c.
2
3
N
In the previous example only one type of root occurred. But the integrand can involve roots of
various
types.
They all can be expressed as powers with the same base. Thus the integral (1.26), that is
R
√
s
R x, x d x , is sometimes written like
Z
S x,
√
s√
x, . . . , k x d x,
s1
(1.27)
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where k ∈ N, s1 = 2, . . . , sk = 2 are natural numbers and S is a rational function of k + 1 variables.
We evaluate this integral using the next substitution.
If we denote s the least common multiple of numbers s1 , . . . , sk , then each si -th root is a natural
√
√ s/si
power of the s -th root: si x = s x
, where i = 1, . . . , k . Thus the integral of the type (1.27), which
is seemingly more general since it involves more various roots, is in fact equivalent to the integral of the
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137
type (1.26) and can be evaluated using a similar substitution x = t s , where s is the above mentioned
least common multiple of numbers s1 , . . . , sk . This is also true for any greater integer multiple of the
number s . Though in that case we would obtain higher powers of the new variable t and the integration
of the resulting rational function would probably be more complicated.
√
√
Z
1+ x− 3 x
√
Example 1.59 Evaluate
d x , x ∈ (0, +∞).
6
x + x5
√
Solution. The integral is of type (1.27). Using the substitution x = t 6 , that is t = 6 x , satisfying the
assumptions of Theorem 1.27 we get
Z
√
√
Z
x = t6
1 + t3 − t2 5
1+ x− 3 x
=
√
d
x
=
6t d t =
6
d x = 6t 5 d t t6 + t5
x + x5
Z
Z
6t 5 (t 3 − t 2 + 1)
6t 3 − 6t 2 + 6
=
d
t
=
d t,
t 5 (t + 1)
t +1
which is the integral of an improper rational function. So it is necessary to reduce it to the sum of a
polynomial and a proper rational function. Using the long division algorithm we get:
6t 3 − 6t 2 + 6
6
= 6t 2 − 12t + 12 −
.
t +1
t +1
This function can be easily integrated. The complete result is
√
√
Z
Z 1+ x− 3 x
6
2
√
dx =
6t − 12t + 12 −
dt =
6
t +1
x + x5
= 2t 3 − 6t 2 + 12t − 6 ln |t + 1| + c =
√
√
√
√
= 2 x − 6 3 x + 12 6 x − 6 ln 6 x + 1 + c.
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Indefinite Integral
The next type we will deal with has the form
Z
√
s
R x, ax + b d x,
(1.28)
where s ∈ N, s = 2, a, b ∈ R, a 6 = 0. Note that if a = 1 and b = 0, we get an integral of type (1.26).
The substitution determined by the equality t s = ax + b transforms this integral to an integral of
a rational function. Before we evaluate the differential we must first make variable x the object. Thus
t s = ax + b ⇒ x =
ts − b
a
and
dx =
st s−1
d t.
a
We demonstrate the procedure with an example.
Z √
x+1+1
√
Example 1.60 Evaluate
d x , x ∈ (0, ∞).
x+1−1
√
Solution. Using the substitution x + 1 = t 2 , that is t = x + 1, we get
Z
Z √
x + 1 = t2 x+1+1
t +1
=
√
dx = · 2t d t,
d x = 2t d t
t −1
x+1−1
which is the integral of an improper rational function. Dividing it we get
t2 + t
2
=t +2+
.
t −1
t −1
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Indefinite Integral
The result is
Z √
Z x+1+1
2 √
dt =
dx = 2
t +2+
t −1
x+1−1
t2
=2
+ 2t + 2 ln(t − 1) + c =
2
√
√
= x + 1 + 4 x + 1 + 4 ln x + 1 − 1 + c.
N
The next type we will deal with has the form
Z
r
s ax + b
d x,
R x,
cx + d
(1.29)
where s ∈ N, s = 2, a, b, c, d ∈ R, ad − bc 6 = 0.
The condition ad − bc 6 = 0 guarantees that the fraction ax+b
does not cancel to a constant such as
cx+d
4x−2
= 2.
2x−1
Note that for a = d = 1 and b = c = 0 we get the integral of type (1.26), so it is again the
generalization.
The substitution determined by the equality t s =
ax+b
cx+d
transforms this integral to an integral of a
rational function. It is a substitution in the sense of Theorem 1.27, so before evaluating the differential
we must first make the old variable x the object. Thus
ts =
ax + b
cx + d
⇒
cxt s + dt s = ax + b
⇒
x(ct s − a) = b − dt s
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s
⇒
ϕ: x =
b − dt
,
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140
Indefinite Integral
hence
dx =
b − dt s
ct s − a
0
−sdt s−1 (ct s − a) − (b − dt s )sct s−1
dt =
(ct s − a)2
s(ad − bc)t s−1
d t.
=
(ct s − a)2
dt =
It is useless to remember this result, in each concrete case we isolate the old variable x and then evaluate
the differential.
In the resulting integral, we must replace t by the inverse function ϕ −1 (x) the evaluation of which is
easy:
r
ax + b
−1
ϕ :t=s
.
cx + d
r
Z
1 x+1
Example 1.61 Evaluate
d x , x ∈ (1, +∞).
x x−1
Solution. It is the integral of type (1.29). We use the substitution given by the equality
make x the object, that is find the function ϕ(t):
x+1
x−1
t2 + 1
x+1
= t 2 ⇒ xt 2 − t 2 = x + 1 ⇒ x(t 2 − 1) = t 2 + 1 ⇒ ϕ : x = 2
.
x−1
t −1
Now we prepare the differential:
dx =
t2 + 1
t2 − 1
= t 2 and
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0
dt =
2t (t 2 − 1) − (t 2 + 1)2t
−4t
dt = 2
d t.
2
2
(t − 1)
(t − 1)2
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Indefinite Integral
Further we find the inverse function ϕ −1 (x), which will be needed when Theorem 1.27 is applied:
r
x+1
−1
ϕ :t=
.
x−1
A more thorough investigation of the behaviour of the function ϕ(t) shows that using the notation from
Theorem 1.27 we chose J = I = (1, +∞). Another
possibility would be J = (1, +∞), I = (−
p
−1
−∞, −1), but then there would be ϕ (x) = − (x + 1)/(x − 1).
Making the substitution we get:
r
Z
Z
Z
1 x+1
1
−4t
−4t 2
dx =
d
t
=
d t,
·
t
·
2
t +1
x x−1
(t 2 − 1)2
(t 2 − 1)(t 2 + 1)
2
t −1
which is the integral of a proper rational function.
We decompose the integrand into partial fractions. The factorization of the denominator in the real
domain is evidently (t − 1)(t + 1)(t 2 + 1), thus the form of the decomposition is
−4t 2
A
B
Ct + D
=
+
+ 2
,
2
(t − 1)(t + 1)(t + 1)
t −1 t +1
t +1
where A, B , C , and D are appropriate constants. After multiplying by the factored denominator to clear
the fractions we get:
−4t 2 = A(t + 1)(t 2 + 1) + B(t − 1)(t 2 + 1) + (Ct + D)(t 2 − 1).
First we substitute the real roots 1 and −1, which immediately gives two constants:
t =1:
− 4 = 4A
⇒
A = −1,
t = −1 :
− 4 = −4B
⇒
B = 1.
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Indefinite Integral
Further comparing coefficients at appropriate powers we find two equations, from which we determine
the remaining two constants. Even without the expansion it can be seen that:
t3 :
0=A+B +C
⇒
C = 0,
0
0=A−B −D
⇒
D = −2.
t :
We have
Z
1
x
r
Z x+1
1
1
2
dx =
−
+
−
dt =
x−1
t − 1 t + 1 t2 + 1
= − ln |t − 1| + ln |t + 1| − 2 arctan t + c =
r
r
r
x+1
x+1
x+1
= − ln
− 1 + ln
+ 1 − 2 arctan
+ c.
x−1
x−1
x−1
The result can be simplified. As an exercise, try to verify that
r
r
Z
√
√
1 x+1
x+1
d x = 2 ln x + 1 + x − 1 − 2 arctan
+c
x x−1
x−1
holds (the constants c are not the same in the results, they differ by ln 2).
The first result is also valid on the interval (−∞, −1). Try to think of how the modified version of
the result would look on this interval.
N
The integrals of type (1.28) and (1.29) can be also generalized in the case where the integrand involves
more various roots of the same linear term or fraction. In fact, it is not the true generalization, the situation
is the same as with the pair of types (1.26) and (1.27).
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143
Indefinite Integral
Another type of integral with roots that can be reduced to integrals of rational functions is
Z
R x,
p
ax 2 + bx + c d x,
where a, b, c ∈ R, a 6 = 0.
(1.30)
Here, R means a rational function of two variables. Moreover, we assume that the quadratic trinomial
does not have a double root, that is that b2 − 4ac 6 = 0 (otherwise, the root would disappear).
We restrict to the case b = 0 (then, necessarily, c 6 = 0) and present the evaluation using trigonometric
substitutions.
Comment 1.62 Integrals of type (1.30) are usually evaluated using Euler1 substitutions (see, e.g. [9, 20]). Because
of the lack of time we will not study them. From the point of view of applications, the above mentioned special
cases will be sufficient for our purposes. Moreover, we will usually need integrals of roots themselves and in
these cases trigonometric substitutions usually give results quite quickly. Moreover, completing the square we can
transform the general case ax 2 + bx + c, b 6 = 0, to the case b = 0 using an auxiliary simple substitution. Compare
this to the integration of the partial fraction 1/(x 2 + px + q)n on page 96.
To get an idea of how Euler substitutions appear, we will describe them briefly. These are three substitutions
(the first two cover all cases, but the third one is sometimes more appropriate). We denote the new variable by t .
p
p
ax 2 + bx + c = |a| t (x − α),
when b2 − 4ac > 0
(α is a root of ax 2 + bx + c = 0),
√
ax 2 + bx + c = ± a x ± t,
p
√
ax 2 + bx + c = ±xt ± c,
p
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2
when b − 4ac < 0, a > 0,
when c > 0.
1 Leonard Euler (1707–1783) (read oyler)—Swiss mathematician, physicist, and astronomer. He worked mainly in Petersburg. One of the greatest mathematicians of all time. He is the author of about 850 works (including multiple-part monographies). He influenced all basic mathematical disciplines. He went blind in 1766, dictating his ongoing work to his students.
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144
Each time we raise the corresponding relation to two and make x the object (the equation will be linear with respect
to x , the power x 2 will disappear). The relation between x and t is given by a rational function. Then we evaluate
the differential d x . The integral (1.30) is transformed to an integral of a rational function. Some examples are
presented, for example, in [20, 21].
In integral of type (1.30) with b = 0, taking out |a| we obtain three types of roots, depending on
the signs of the coefficients. For each type we give the substitution, reducing the integral involved to an
integral of type S(cos x, sin x), where S(u, v) is a rational function of two variables, which we already
know how to evaluate. In the sequel, k > 0 is a constant.
Z
p
R x, k 2 − x 2 d x
x = k sin t,
t ∈ (−π/2, π/2),
(1.31)
Z
p
R x, x 2 + k 2 d x
x = k tan t,
t ∈ (−π/2, π/2),
(1.32)
Z
p
k
R x, x 2 − k 2 d x
x=
,
t ∈ (−π/2, 0) or
(1.33)
sin t
t ∈ (0, π/2).
We have already evaluated an integral of type (1.31). See Example 1.30 on page 70. Now we will
show another one.
Z
1p 2
Example 1.63 Evaluate
x − 1 d x , x ∈ (1, +∞).
x
Solution. The integral is of type (1.33), where k = 1. After the substitution and some rearrangements
we get
r
Z
Z
x= 1
1p 2
1
1
cos t
sin t
x − 1 dx = ·
− 1 · 2 dt =
=−
1
2
d x = − cos2 t d t x
sin t
sin t
sin t
sin t
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145
Indefinite Integral
s
Z
1 − sin2 t cos t
| cos t| cos t
= − sin t
· 2 d t = − sin t
·
dt =
2
| sin t| sin2 t
sin t
sin t
Z
Z
cos t cos t
cos2 t
= − sin t
· 2 dt = −
dt =
sin t sin t
sin2 t
Z
Z 1 − sin2 t
1
=−
d
t
=
1
−
d t = t + cot t + c.
sin2 t
sin2 t
Z
As x ∈ (1, +∞), we choose t ∈ (0, π/2). On this interval both sin t and cos t are positive, which was
used to remove absolute values.
Next we must put back the original variable x . To do that we need the inverse function. We have
x=
1
sin t
⇒
sin t =
1
x
⇒
t = arcsin
1
.
x
Before substituting into the previous result we modify it. We get in succession
p
Z
1 − sin2 t
1p 2
x − 1 d x = t + cot t + c = t +
+c =
x
sin t
q
r
1 − x12
1
1
x2 − 1
= arcsin +
+
c
=
arcsin
+
x
+c =
1
x
x
x2
x
√
1
x2 − 1
1 p
= arcsin + x
= arcsin + x 2 − 1 + c.
x
x
x
Due to Theorem 1.29 on page 69 the result is also valid on the interval h1, +∞).
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146
Indefinite Integral
The evaluation of seemingly simple integrals is sometimes technically rather complicated and lengthy
as the previous example shows. The situation is the same if Euler substitutions are used. Moreover, it
also must be pointed out that the evaluation of the same integral with an Euler substitution and then with
a trigonometric substitution (or with two different Euler substitutions) can lead to substantially different
results. Often it is not trivial to demonstrate that the results are the same (up to an eventual constant).
Exercises
1. Evaluate the indefinite integrals of the following functions:
Z √
x
a)
d x,
b)
1+x
Z
1
√
d x,
d)
c)
1+ x+1
Z
35x 3
√
e)
d x,
f)
x−1
√
Z
3
x
g)
h)
√
√ d x,
x x+3 x
Z √
k+1+1
√
i)
dk,
j)
k+1−1
Z
√
15x a + x d x,
Z
dp
√
,
(2 + p) p + 1
Z
x+1
√
d x,
3
3x + 1
Z
√
1
√ dv,
v+4 v
√
dx
√
.
x+2+ 3 x+2
Z
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2. Evaluate the indefinite integral of the following functions (use Euler’s substitutions):
Z p
Z p
a)
2 b2 − 6 db,
b)
6 9x 2 − 15 d x,
Z p
Z p
2
c)
4 − 3x + 2x d x,
d)
4 p 2 − 2p − 1 dp,
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Indefinite Integral
e)
Z p
Z
4
g)
Z
2
5q − 6q − 1 d q,
p
√
g)
Z
√
i)
x
√
d x,
x+1+ 3 x+1
1
5 − 2x − x 2
Z
√
k)
Z
m)
√
p
2 + x − x 2 d x,
3 + 2s − s 2 d s.
3. Evaluate the indefinite integral of the following functions:
Z
2x − 3
√
a)
d x,
b)
8 − 2x − x 2
Z
2
p
c)
dy,
d)
2
−4y − 12y − 8
Z
2
√
e)
dw,
f)
−5 + 12w − 4w 2
Z
8
f)
d x,
3
12k − 9k 2 + 4
−4x 2
j)
dk,
8x − 3
− 5 + 12x
h)
l)
Z
4(x + 3)
√
d x,
3 + 4x − 4x 2
Z
35
√
d x,
2 − 49x 2
Z
1
p
dp,
−2p − p 2
q
Z 4 x +3
2−x
q
d x,
x
x 2 2−x
Z
2n
√
dn,
10 − n − n2
Z
u
√
d u,
27 − u2 + 6u
d x.
Hint: In a), b), j), l) and m) you proceed the same as when integrating partial fractions of the second type—
rearrange the numerator to get the derivative of the expression under the radical sign in the denominator and
separate the fraction into two parts.
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148
Indefinite Integral
Answers to Exercises
1. a)
2
√
√
x − 2 arctan x,
b)
6(a + x)5/2 − 10a(a + x)3/2 ,
c)
2
√
√
x + 1 − 2 ln 1 + x + 1 ,
d)
2 arctan
e)
10(x − 1)7/2 + 42(x − 1)5/2 + 70(x − 1)3/2 + 70
h)
(3x + 1)5/3
(3x + 1)2/3
+
,
15
3
√
√
√
2 v − 4 4 v + 4 ln 4 v + 1 ,
j)
2
2. a)
b
f)
p
p + 1,
√
x − 1,
g)
√
√
6 ln 6 x − 6 ln 6 x + 1 ,
i)
4
√
√
k + 1 + k + 1 + 4 ln k + 1 − 1,
√
√
√
√
3
6
6
x + 2 − 3 x + 2 + 6 x + 2 − 6 ln x + 2 + 1 .
p
p
b2 − 6 − 6 lnb + b2 − 6 ,
b)
3x
p
p
9x 2 − 15 − 5 ln9x + 3 (9x 2 − 15) ,
√
c)
d)
e)
f)
4 − 3x 2 + 2x
13 √
3x − 1
+
3 arcsin √
,
6
18
13
q
q
(−2 + 2p) p2 − 2p − 1 − 4 lnp − 1 + p 2 − 2p − 1 ,
−
(−3x + 1)
(5q − 3)
(4x − 2)
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p
q
5q 2 − 6q − 1
7 −
ln5q − 3 + 5(5q 2 − 6q − 1) ,
10
25
p
2+x
− x2
2x − 1
+ 9 arcsin
,
3
g)
(2s − 2)
p
J
3 + 2s
− s2
s−1
+ 8 arcsin
.
2
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Indefinite Integral
3. a)
−2
p
8 − 2x − x 2 − 5 arcsin
x+1
,
3
b)
p
1
− 3 + 4x − 4x 2 + 7 arcsin x −
,
2
c)
arcsin(2y + 3),
d)
7x
5 arcsin √ ,
2
e)
3
arcsin w −
,
2
f)
arcsin(p + 1),
g)
h)
j)
l)
2(x + 1)3/2
3(x + 1)4/3
6(x + 1)7/6
6(x + 1)5/6
3(x + 1)2/3
−
+
−x−1+
−
,
3
4
7
5
2
s
s
x+1
24
2−x 5
2−x 3
−
,
i) arcsin √ ,
−
5
x
x
6
√
p
2 (3k − 2)
2n + 1
,
−2 10 − n − n2 − arcsin √
,
k) arcsin
4
41
−
p
27 − u2 + 6u + 3 arcsin
u−3
,
6
m)
−2
p
−4x 2 − 5 + 12x +
9
3
arcsin x −
.
2
2
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150
Indefinite Integral
Quiz—Indefinite Integrals of Functions Involving Roots
Choose the correct answer (only one is correct).
Z
1
√ dx =
1. (1 pt.)
2+ x
√
√ 2 x − 4 ln 2 + x + C
√
1
x+2 x+C
2 Z
2x
√
2. (1 pt.)
dx =
x2 − 1
x2
arcsin x + C
2
x2
x−
+C
Z2
√ 3. (1 pt.) x 1 + x d x =
x2 2 √ 5
+
x +C
2
5
x2 1 1 x+ √ +C
2
2 x
ln 2 +
arctan
√ x +C
√
x+C
p
2 x2 − 1 + C
2 arcsin
p
x2 − 1 + C
x2 5 √ 5
+
x +C
2
2
√
x2 + 2 x3 + C
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151
Indefinite Integral
Z
2x
√
dx =
1 − x4
p
lnx + 1 − x 4 + C
4. (1 pt.)
arcsin x 2 + C
Z
√
5. (1 pt.) x 2x − 8 d x =
√
2 2 5/2 √ 2
x − 2x + C
5
(2x − 8)2
+ 2(2x − 8) + C
Z8
1
√
6. (1 pt.)
dx =
x x−4
2(x − 4)
+C
x
√
x−4
arctan
+C
2
Z
1
√
7. (1 pt.)
dx =
1 + 3x − 2
√
ln1 + 3x − 2 + C
√
2 √
3x − 2 − ln 1 + 3x − 2 + C
3
arcsin x + C
arctan x 2 + C
xp
1p
(2x − 8)3 +
(2x − 8)5 + C
3
15
p
p
(2x − 8)5 4 (2x − 8)3
+
+C
10
3
x−4
+C
2
√
1
x−4
√
+C
+
2
x−4
arctan
√
2 3x − 2
x+
+C
3
x+
3x − 2
+C
6
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Indefinite Integral
Z
x
√
8. (1 pt.)
dx =
3x 2 + 1
x
√ +x+C
3
√
3x 2 + 1
+C
3√
Z 3
x+1
√
9. (1 pt.)
dx =
x
√
6
√
6 x5
+ ln x + C
5
√ √
6
x5
x
+
+C
6
5
3
Z
2
√
√ dx =
10. (1 pt.)
4
x+ x
√ √
4
x3
x
8
+
+C
3
2
√
√
x √
4
4
8
− x + ln x + 1 + C
2
152
√
3 3x 2 + 1
+C
2
1 p 2
ln 3x + 1 + C
6
√ √
√ 2 x 3 x+1 − 6 x +C
√
33 x
2
+ √ +C
4
x
√
84 x + C
−2 ln
√
√
x + 1 − 2 ln x + C
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Points Gained:
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153
1.6 Final Notes
1.6.1 Is the Result of Integrating an Elementary Function Again an Elementary Function?
On page 18 we reviewed the elementary functions. It is not too difficult to realize that differentiation
rules imply that the derivative of an elementary function is again an elementary function.
Unfortunately, in the case of an indefinite integral, the situation is much more complicated. Because
elementary functions are continuous on intervals on which they are defined, they have an antiderivative
in virtue of Theorem 1.3. But in general it is not true that antiderivatives to elementary functions are
again in the set of elementary functions. We must understand this statement correctly. In no case can we
say that the antiderivative to some elementary function does not exist. We only say that this antiderivative
cannot be expressed the way we would like, that is it cannot be built from some explicitly given basic
functions (polynomials, trigonometric functions etc.) using a finite number of arithmetic operations and
compositions.
Exponential, logarithmic, trigonometric and inverse trigonometric functions are called lower transcendental functions. Antiderivatives to elementary functions, which are not elementary are usually
called higher transcendental functions. Unfortunately, there is no easy test that helps decide whether a
concrete indefinite integral leads to a higher transcendental function. Either we manage to evaluate such
an integral (that is we find its antiderivative in the set of elementary functions) or we don’t. In case of
failure we do not know if it is caused by our insufficient knowledge, ignorance of some methods (thus
worth trying again), or if the antiderivative does not exist in the set of elementary functions (futile to waste
time over). In general, it is very difficult to prove that a concrete integral leads to a higher transcendental
function.
Finally, let us list a few very simple indefinite integrals, which are known to lead to higher transcendental functions. These statements have been transmitted since the 19th century, but the proofs cannot
be found in any standard (even very large) textbook of integral calculus. The proofs are based on the Li-
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Indefinite Integral
ouville1 Theorem, giving the necessary and sufficient condition of integrability in the set of elementary
functions. Relatively accessible proofs, that some of the integrals below lead to higher transcendental
functions can be found in [19, 25, 26]. Due to their frequent occurrence, some of these antiderivatives
(uniquely determined prescribing values at some points) have their own name.
Z
Z
Z
sin x
dx
x
1
dx
ln x
2
e−x d x
Z
(sine integral),
Z
cos x
dx
x
2
Z
(Gauss function),
Z
sin x d x,
(logarithmic integral),
R x,
(cosine integral),
cos x 2 d x
p
P (x) d x
(Fresnel’s2 integrals),
(elliptic integrals).
In the last integral, R(u, v) is a rational function and P (x) is a polynomial of degree three or four not
having multiple roots (the name originates in the fact that the length of the ellipse can be expressed using
this integral). In special cases elliptic integrals can be evaluated using elementary functions (these are
called pseudoelliptic integrals), but in general it is not possible to do (see [10, 22, 25]). The consequence
is that there is no “nice” formula for the length of the ellipse (to the contrary to the circle).
It is impossible to judge if a concrete function has
R depending on
R an elementary antiderivative or not
its complexity. For example, we easily evaluated cos2 x d x (Example 1.16) while cos x 2 d x does
not have an elementary antiderivative. In both cases, the integrand is a composite function withR“square
2
root” and “cosine” components. The difference is only in the order of the components. Likewise, ex d x
1 Joseph Liouville (1809–1882) (read
lIuvIl)—an important French mathematician. He was involved in many domains of
analysis. His papers on integrability of elementary functions date from 1833–1841.
2 Augustin Jean Fresnel (1788–1827) (read frenel)—French mathematician, physicist, and engineer.
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155
(similar to the above integral defining
Gauss function) does not lead to an elementary function while
R √the
x
the seemingly more complicated e d x can be easily evaluated (Example 1.28).
The question of whether or not an elementary function has an elementary antiderivative is really
very complicated. Given today’s requirements for greater exactness and generality, older approaches
from the 19th century do not yield satisfactory results. The high point of interest in this topic arose with
the development of modern mathematical programs (known as Computer Algebra Systems). In such
programs, efficient algorithms must be included, which are able to decide in a finite number of steps if
the integral leads to an elementary function (or another type of a function being in the repertoire of the
program), and if it is the case to then find the result. Information about this topic is available in specialist
journals (e.g. [1, 2, 17, 24, 26]).
Z x
e
Example 1.64 Verify that
d x , x > 0, does not lead to an elementary function.
x
Solution. Applying the substitution x = ln t , t > 1, we get
Z ln t
Z x
Z
Z
x = ln t e
e
1
t
1
1
=
dx = · dt =
· dt =
d t.
1
dx = t dt
x
ln t t
ln t t
ln t
As we mentioned above, the resulting integral is not elementary. Let us denote G(t) an antiderivative
to the function 1/ ln t and assume that the integrand ex /x has an elementary antiderivative F (x). Thus
F (x) = G(ex ) + c, c ∈ R. This means that the function F (ln t) − c = G(t) is also elementary, which
is a contradiction.
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Indefinite Integral
1.6.2 Gluing Technology
Example 1.65 Evaluate
Z
dx
.
2 − cos x
Solution. The integrand is the function defined and continuous on the whole real line R (the denominator
2−cos x can never be equal to zero). Using the general substitution tan x2 = t (see (1.20)) we can evaluate
this integral on each interval (−π + 2kπ, π + 2kπ), k ∈ Z. For k = 0 we have
tan x = t
Z
Z
2
1
dx
2
x = 2 arctan t =
= dt =
2 ·
1−t
2 − cos x
1
+
t2
2 − 1+t 2
dx = 2 2 dt 1+t
Z
Z
2
2
dt
t
2 1
=
dt =
= · 1 arctan 1 + c =
1
2
2
√
√
3t + 1
3
3
t +3
3
3
√
√
2
x
2
= √ arctan 3 t + c = √ arctan 3 tan
+ c = G(x) + c.
2
3
3
1
Due to periodicity, the obtained function G(x) is an antiderivative to the function f (x) = 2−cos
on any
x
open interval (−π + 2kπ, π + 2kπ), k ∈ Z. If we want to find an antiderivative on the whole set R,
we must use gluing technology, which we present below.
Let us think about how antiderivatives look on the open intervals of the form (2k − 1)π, (2k + 1)π ,
k ∈ Z, on which function G is continuous.
• For each x ∈ (−π, π) we have G0 (x) = f (x). Thus all antiderivatives on this interval have the
form G(x) + c0 , c0 ∈ R.
Functions f and G are periodic with the period 2π. So it is sufficient to deal with these functions on
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Indefinite Integral
the interval of length 2π, for example (−π, π). Graphs of these functions on further intervals are copies
of the part of the graph on the interval (−π, π). Therefore:
• For each x ∈ (−3π, −π) we have G0 (x) = f (x). Thus all antiderivatives on this interval have the
form G(x) + c1 , c1 ∈ R.
• For each x ∈ (π, 3π) we have G0 (x) = f (x). Thus all antiderivatives on this interval have the
form G(x) + c2 , c2 ∈ R.
And so on. Now, from these antiderivatives on particular intervals, we will create the function F ,
which will be the antiderivative of f on the whole set R. We want function F to be continuous on R.
Hence let us choose the constants c0 , c1 , c2 , . . . in such a way that the antiderivatives connect each other
on contiguous intervals.
• Let us start from the interval (−π, π) and put F (x) = G(x) for any
x ∈ (−π, π). (We set c0 = 0).
• Now, let us evaluate the limit on the left of function F at the right end point of this interval. This
is the point at which we have to properly define the function value to guarantee the continuity. This
limit helps us to determine the constant c2 so as to know, which of the antiderivatives on the interval
(π, 3π) to choose to join continuously to function F on the interval (−π, π).
√
x
π
2
=√ ,
lim− F (x) = lim− √ arctan 3 tan
x→π
x→π
2
3
3
√
2
x
π
lim+ G(x) = lim− √ arctan 3 tan
= −√ .
x→π
x→π
2
3
3
As we can see it is necessary to set c2 =
(π, 3π) up by
2π
√
.
3
2π
√
.
3
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Thus we move the graph of function G on the interval
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158
Indefinite Integral
• We still have a continuous antiderivative on the interval (−π, 3π):

√
x
√2 arctan

3
tan
for x ∈ (−π, π),

2

 3
F (x) = √π3
for x = π,


√

 √2 arctan 3 tan x + √
2π
for x ∈ (π, 3π).
2
3
3
(1.34)
• Analogously we evaluate the limit on the right at the point x = −π.
√
2
x
π
lim + F (x) = lim + √ arctan 3 tan
= −√ ,
x→−π
x→−π
2
3
3
√
π
x
2
=√ .
lim G(x) = lim + √ arctan 3 tan
x→−π−
x→−π
2
3
3
2π
As we can see it is necessary to set c1 = − √
. Thus we move the graph of function G on the
3
interval (−3π, −π) down by
2π
√
.
3
• We have a function continuous on the interval (−3π, 3π):

√
2π
x

√2 arctan
3
tan
− √3

2
 3




− √π


 3
√
F (x) = √23 arctan 3 tan x2




π

√


3



2π
 √2 arctan √3 tan x + √
3
2
3
for x ∈ (−3π, −π),
Contents
for x = −π,
for x ∈ (−π, π),
for x = π,
for x ∈ (π, 3π).
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Indefinite Integral
y
√
3π/ 3
y = F (x)
√
2π/ 3
√
π/ 3
x
−3π
3π
−π
π
O
√
−π/ 3
π
2π
3π
y = G(x)
√
−3π/ 3
Fig. 1.4: Graph of an antiderivative to function
1
2−cos x
on R
In the same way we construct function F on all the intervals (2k − 1)π, (2k + 1)π , k ∈ Z.
Theorem 1.29 implies that the function constructed this way has the derivative also at points (2k + 1)π,
k ∈ Z (that is at the points where we “glued” the function) and that F 0 (x) = f (x) holds at these points.
(The derivative at these points cannot be evaluated in the standard way using chain rule, because the
inner component tan x2 is not defined there; the direct evaluation from the definition of derivative would
1
be rather difficult.) Therefore, the constructed function is the antiderivative to the function 2−cos
. N
x
The result is displayed in Fig. 1.4. The graph of function F is plotted with the solid line, and the graph
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Indefinite Integral
of function G (not defined at odd multiples of π) is plotted with the dashed line. Notice that the graphs of
F and G coincide on the interval (−π, π). We encounter a similar situation (“gluing” of graphs) quite
often when integrating rational expressions of trigonometric functions. If we need an antiderivative on
larger intervals, we must be very careful. Otherwise we can easily get absolutely wrong results. (For
example, when evaluating the definite integral using the indefinite one. See Example 2.18).
Thinking about why we were forced to glue the graphs we can see that it was due to substitution
x
tan 2 = t .The advantage of the substitution tan x2 = t is its universality, each time it transforms the
considered integral into the integral of a rational function. But it has two big disadvantages. The first
is that the evaluations are often rather laborious and the second is we must often use “gluing” to get the
antiderivative on maximal intervals on which the integrand is continuous (see the previous example).
Hence, if we can avoid this universal substitution, then we should avoid it.
1.7 Final Exercises to Chapter 1
1. Evaluate the indefinite integrals of the following functions:
Z
Z
a)
arcsin x d x,
b)
Z p
Z
c)
4 −x 2 + 2x + 3 d x,
d)
Z p
Z
2
e)
2 1 − x − 2x d x,
f)
Z
Z
4x − 1
d x,
h)
g)
x 2 + 5x + 7
Z
Z
u
√
i)
d u,
j)
4 − 9u4
cos η
dη,
1 − sin η
B
dB,
8 − 3B 2
sin x
d x,
1 + cos x
1
d x,
4−x
cot x d x,
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Indefinite Integral
Z
k)
m)
o)
1
√
d x,
3 − 9x 2
Z
√
arctan x d x,
Z
ln 5x d x,
Z
q)
1
d x,
√
√
3
3
x
x−1
Z
l)
n)
p)
r)
1
dY,
1 + cos 4Y
Z
1
dB,
5 + 3B 2
Z
x+2
d x,
x4 + x3
r
Z
1+x
1
d x.
2
x
x
2. Evaluate the indefinite integrals of the following functions:
Z
Z
a)
cos ln x d x,
b)
sin ln x d x,
Z p
Z p
x 2 + 4 d x,
d)
2 9 − x 2 d x,
c)
4x 2
√
d x,
x2 + 9
Z
W +2
dW,
2W − 1
Z
1
d u,
1 + cos u2
Z
144
√
d x,
144x 2 − 52
Z
Z
e)
g)
i)
k)
f)
9(3δ + 5)−1 dδ,
Z
(4 − cos 2α) dα,
h)
Z
j)
l)
21
dr,
9 + 7r 2
Z
3
√
d t,
3t 2 − 2
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Indefinite Integral
Z
m)
Z
o)
2
Z
2(sin ω + cos 2ω) dω,
n)
x 3 ln2 x 4 d x,
p)
cos 2ψsin ψ cos ψ dψ,
Z
3. Evaluate the indefinite integrals of the following functions:
Z
Z
4
a)
8 sin x d x,
b)
Z
Z
√
x ln x d x,
d)
c)
Z
Z
1 + cos 2θ
e)
dθ,
f)
1 − cos 2θ
Z
Z
9(2p − 1)
p
g)
dp,
h)
9p2 − 4
Z
Z
1
√
i)
d z,
j)
z2 − 8
Z
Z
x
√
k)
d x,
l)
8 − x2
Z
Z
m)
2(3w2 − w + 7)(6w − 1) dw,
n)
Z
o)
5e2x + 4ex
d x,
e2x + ex + 4
Z
p)
3x 2 + 2x − 3
d x.
x3 − x
2 sin2
t
d t,
2
1
d x,
1 − cos 2x
1
√
dy ,
y(1 − y)
2(3x − 1)
d x,
x2 + 9
x
√
d x,
x 2 − 32
x tan2 x d x,
Contents
4y − 8
dy,
2
2y − 8y + 7
ln x
d x.
x2
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Indefinite Integral
4. Evaluate the indefinite integrals of the following functions:
Z
Z
a)
(8 cos 2x − 3 sin 3x) d x,
b)
Z
c)
cos3 x − 1
d x,
cos2 x
3 − 2 cot2 z
d z,
cos2 z
Z
3 sin2 x − 2 cos2 x + 5
d x,
4 cos2 x
Z
1
d z,
(sin z cos z)2
Z
4 cos 2α
dα,
sin2 2α
Z
10
dλ,
tan 5λ
Z
arctan 3x d x,
Z
d)
Z
e)
g)
i)
k)
m)
o)
Z
q)
√
x
√ d x,
1+ x
Z
f)
h)
tan2 ε dε,
sin
x 2
x
− cos
d x,
2
2
Z
cos 2t
d t,
cos t − sin t
Z
1 + cos2 y
dy,
1 + cos 2y
Z
ex
d x,
ex + 1
l)
n)
Z
r)
5 sin2 β + 3 cos2 β
dβ,
2(cos β sin β)2
Z j)
p)
1
dα,
sin2 α
1
√ d x.
(x − 1) 3 x
√
Z
3
x
√ d x.
x+ x
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Indefinite Integral
5. Prove that the following integrals are higher transcendental functions:
Z
Z
Z
sin x
x
a)
ee d x,
b)
ln ln x d x,
c)
d x,
x2
Z x
Z
Z
e
sin x
cos x
√
√ d x,
e)
d
x,
f)
d
x,
g)
2
x
x
x
Z
Z
Z
ln x
sin x
i)
d x,
j)
ex ln x d x,
k)
d x,
ln x + 1
x3
Z
cos x
d x,
x2
Z
ex
√ d x,
x
d)
h)
Z
l)
e1/x d x.
Hint: Using the appropriate manipulation transform the given integral, to an integral which is not
elementary, or to an expression which is the sum of an elementary function and an integral which is
not elementary (see Chapter 1.6 and Example 1.64).
Answers to Exercises
1. a) x arcsin x +
p
1 − x2,
p
x−1
c) (2x − 2) −x 2 + 2x + 3 + 8 arcsin
,
2
p
x+1
e) (x + 1) 1 − x 2 − 2x + 2 arcsin √ ,
2
22
2x + 5
2
g) 2 ln(x + 5x + 7) − √ arctan √ ,
3
3
2
1
3u
i)
arcsin
,
6
2
b) − ln |1 − sin η|,
1
d) − ln |8 − 3B 2 |,
6
f) − ln |1 + cos x|,
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h) − ln |4 − x|,
j) ln | sin x|,
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Indefinite Integral
k)
√
1
arcsin 3 x,
3
√
√
m) x arctan
x−
√
x + arctan
x,
o) x ln 5x − x,
q) 3
√
√
3
x + ln 3 x − 1
x
x
cos ln x + sin ln x,
2
2
√
p
2
x x +4
c)
+ 2 ln x + x 2 + 4 ,
2
p
p
e) 2x x 2 + 9 − 18 ln x + x 2 + 9 ,
2. a)
g)
5
W
+ ln |2W − 1|,
2
4
u
,
4
p
k) 12 ln6x + 36x 2 − 13 ,
i) 2 tan
m) ω +
o) x 4
1
tan 2Y,
4
√
√
15
B 15
n)
arctan
,
15
5
x − 1 + 1 ,
p) ln x + 1 x2
x
s
3
2
1+x
r) −
.
3
x
l)
1
sin 2ω,
2
1
2
4 ln |x| − 2 ln |x| +
,
2
x
x
cos ln x + sin ln x,
2
2
p
x
d) x 9 − x 2 + 9 arcsin ,
3
b) −
f) 3 ln |3δ + 5|,
sin 2α
,
2
√
√
7r
j)
7 arctan
,
3
p
√ √
l)
3 ln 3t + 3t 2 − 2 ,
h) 4α −
cos4 ψ + sin4 ψ
,
4
3
x (x − 1) .
p) ln
x+1 n) −
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Indefinite Integral
3. a)
c)
1
sin 4x,
4
2p 3
2
x ln x −
,
3
3
3x − 2 sin 2x +
e)
− cot θ − θ,
g)
2
b)
− sin t + t,
d)
1
− cot x,
2
f)
arcsin(2y − 1),
2
x
3 ln(x 2 + 9) − arctan ,
3
3
p
j)
x 2 − 32,
q
i)
q
9p 2 − 4 − 3 ln3p + 9p 2 − 4 ,
p
lnz + z2 − 8 ,
k)
p
− 8 − x2,
l)
x tan x + ln | cos x| −
m)
(3w2 − w + 7)2 ,
n)
ln |2y 2 − 8y + 7|,
p)
−1
(ln x + 1).
x
h)
x2
,
2
√
o)
4. a)
d)
g)
ln
p
(e2x
+ ex
+ 4)5
4 sin 2x + cos 3x,
3
5
tan β − cot β,
2
2
5x
2 tan x −
,
4
+
15
2ex + 1
arctan √
,
5
15
b)
− cot α,
c)
sin x − tan x,
e)
3 tan z + 2 cot z,
f)
tan ε − ε,
h)
x + cos x,
i)
tan z − cot z,
j)
sin t − cos t,
k)
−
2
,
sin 2α
l)
1
y
tan y + ,
2
2
m)
2 ln | sin 5λ|,
n)
ln(ex + 1),
o)
x arctan 3x −
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ln(1 + 9x 2 ),
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Indefinite Integral
p)
r)
5. a)
3 √
√
1 3 x − 1 √
23 x+1
√
ln
+ 3 arctan
,
2 x−1 3
2
√
√
6
√
√
x+1
26 x − 1
3
√
√
3 x + ln √
.
−
2
3
arctan
3
x −6 x + 1
3
sub.
x = ln t,
u = cos x,
d)
by p.
g)
sub.
j)
sub. x = ln t,
x = t 2,
u = ln ln x,
b)
by p.
e)
by p. u = ex ,
h)
sub.
k)
by p.
x = t 2,
u = sin x,
q)
√
√
x − 2 x + 2 ln x + 1 ,
c)
by p. u = sin x,
f)
sub. x = t 2 ,
i)
sub. x = et−1 ,
l)
sub. x = 1/t.
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Indefinite Integral
Quiz—Indefinite Integral
Answer Yes or No to the following questions.
1. (2 pts.) If f (x) =
1
1
and
F
(x)
=
, then
x2
x
Z
f (x) d x = F (x) + C holds.
Yes.
Z
1
2. (2 pts.) x sin x 2 d x = − cos x 2 + C
2
Yes.
Z
√
1
√ dx = 2 x + C.
3. (2 pts.)
x
No.
No.
Yes.
4. (2 pts.) If F 0 (x) = f (x), then
No.
Z
f 0 (x) d x = F (x) + C .
Yes.
Z
5. (2 pts.) sin x d x = cos x + C
No.
Contents
Yes.
6. (2 pts.) If F 0 (x) = f (x) and a, b ∈ R, then
Yes.
No.
Page 168 from 358
Z
f (ax + b) d x = aF (ax + b) + C .
No.
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Indefinite Integral
7. (2 pts.) Converting the chain rule (derivative of composite function formula) we obtain the substitution
method for the indefinite integral.
Yes.
No.
Z
8. (2 pts.) Is the formula
Z
u d v = uv −
v d u only a different form of the integration by parts
formula?
Yes.
No.
9. (2 pts.) The integration by parts method is based on the derivative product rule.
Yes.
No.
Z1
2
10. (2 pts.) The substitution t = 3 + x reduces
2x 3
d x to
(3 + x 2 )5/2
0
Yes.
Z1
t −3
dt.
t 5/2
0
No.
Z1 p
Zπ/2
11. (2 pts.) The substitution x = sin t reduces x 1 − x 2 d x to
sin t cos2 t d t .
0
0
Yes.
No.
dx
1
12. (2 pts.) If a, b ∈ R, a 6 = 0, then
= ln |ax + b| + C .
ax + b
a
Yes.
No.
A
A
13. (2 pts.) Integrating the partial fraction
, A ∈ R, we obtain
.
2
(x + 1)
x+1
Yes.
No.
Contents
Z
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Indefinite Integral
14. (2 pts.)
Z Suppose S(u0 , u1 , . . . , uk ) is a rational function in k+1 variables, k ∈ N, and s1 , . . . , sk ∈ N.
√
√
Can S(x, s1 x , . . . , sk x) d x each time be reduced to the integral of a rational function using the
substitution x = t s , where s = max{s1 , . . . , sk }?
Yes.
No.
Correct Answers:
Points Gained:
Success Rate:
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171
Self-Test
1. Evaluate the indefinite integrals of the following functions:
Z
Z
Z √
√ 2
3x − 5
1 3
a)
x + 3 x d x,
b)
d
x,
c)
x
+
d x.
x2 + 1
x
2. Evaluate the indefinite integrals of the following functions:
Z
Z
√
5ex
a)
3x + 1 d x,
b)
d x,
ex + 1
Z
Z
p
5
2
c)
sin x cos x d x,
d)
x 2 1 + x 3 d x,
Z
Z
x ln x
d x,
f)
e3x sin x d x,
e)
4
Z
Z
x
g)
arctan d x,
h)
x tan2 x d x.
4
3. Evaluate the following integrals and simplify the results:
Z
Z
3x + 1
dx
a)
d
x,
b)
,
2
2
x − 3x + 2
x (x − 1)
Z
Z
(x + 1) d x
7 − 3x
c)
,
d)
d x.
2
3
(x − 2)(x + 3)
x + x 2 + 9x + 9
4. Evaluate the following integrals and simplify the results:
Z
Z
2 + sin x
sin x
a)
d
x,
b)
d x,
(1 + cos x)2
sin x(1 + cos x)
√
Z √
Z
1−x
x
√
c)
d x,
d)
d x.
3
x
x+1
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Indefinite Integral
Answers to Self-Test
1. a)
c)
2. a)
c)
e)
g)
3. a)
c)
√
√
6
3
x2
12x x 5
3x x 2
+
+
,
2
11
5
3
ln (x 2 + 1) − 5 arctan x,
2
x 4 + 6x 2
1
+ 3 ln |x| − 2 .
4
2x
√
2
(3x + 1) 3x + 1 ,
9
b)
5 ln (ex + 1),
sin3 x
,
3
d)
p
5
5
(1 + x 3 ) 1 + x 3 ,
18
f)
e3x (3 sin x − cos x)
,
10
h)
x tan x + ln | cos x| −
x2
(2 ln x − 1) ,
16
x
x arctan − 2 ln(16 + x 2 ) ,
4
7 ln |x − 2| − 4 ln |x − 1| ,
b)
3
3
1
ln |x − 2| −
ln(x 2 + 3) + √ ,
7
14
7 3
d)
c)
1
(1 + cos x)2 ,
2
√
√
√
2 1 − x + ln1 − 1 − x − ln1 + x − 1,
d)
6
4. a)
b)
1√
√
√ 1p
1p
6
6
6
x7 −
x5 +
x 3 − 6 x + arctan 6 x .
7
5
3
x
.
2
1
+ ln |x − 1| − ln |x| ,
x
2
x
1
ln |x + 1| − ln(x 2 + 9) + arctan .
2
3
3
1
x
x
x b)
tan2 + tan + lntan ,
2
2
2
2
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Chapter 2
Definite Integral
In the previous chapter we became acquainted with the concept of the indefinite integral, which assigned
a function to another function (or more exactly, the whole set of functions). The definite integral we will
deal with in this chapter will, to the contrary, assign a number to a function. Depending on the meaning
of the function the resulting number can express, for example:
• the area of the plane figure,
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• the length of the curve,
• the area of the lateral surface of the solid of revolution,
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• the volume of the solid of revolution or more generally of any solid,
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• the mass of the plane figure,
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• the system moments of the plane figure, serving to evaluate the centre of mass,
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174
• the moment of inertia of the plane figure,
• the total electric charge spread on the plane figure
and values of dozens of other geometric and physical quantities.
2.1 Construction of the Definite Integral
Before we formally describe the construction of the definite integral we will use two examples to explain
the concept which leads to this somewhat complicated construction. One example comes from geometry,
the other from physics. It would be possible to present a number of similar examples originating in
geometry, physics and other technical disciplines.
Geometric Motivation
Consider the nonnegative bounded function f (x) defined on an interval ha, bi, which for the sake of
simplicity is continuous. The graph of this function together with two vertical straight lines x = a and
x = b and the x -axis bounds a plane figure P (see Fig. 2.1 a)). Our task is to evaluate its area.
Let us leave aside for a moment that the concept area of a plane set has not yet been defined exactly.
From secondary school you know the formulae for the area of a triangle, rectangle, disc and a few further
simple figures. In case of more complicated sets, it should be intuitively evident what this number means.
(Generally, this topic is studied within what is known as measure theory, see, e.g., [10].) If we denote
the area of a set A ⊂ R2 by the symbol m2 (A) (m as measure, subscript 2, because units are length units
raised to two, for example cm2 ), then area surely should have the following properties:
• It is a nonnegative number, that is m2 (A) = 0.
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Definite Integral
x=a
x=b
x=a
y = f (x)
x=b
y = f (x)
P1
P
P3
P2
x
a
b
x
a = x0
ξ1
x1
ξ2
a)
x2
ξ3
x3 = b
b)
Fig. 2.1: Evaluation of the area of a plane figure
• If we divide set A into two disjoint parts B and C , that is A = B ∪ C , B ∩ C = ∅, the area of A
equals the sum of the areas of B and C , that is m2 (A) = m2 (B) + m2 (C).
• The area of rectangle O , with sides having lengths a and b, equals the number ab, that is m2 (O) =
= ab.
We will propose how to determine the area of set P (see Fig. 2.1 b)).
1. We divide set P into strips with straight lines parallel to the y -axis (there are three such, denoted P1 ,
P2 , and P3 in Fig. 2.1 b)). We have
m2 (P ) = m2 (P1 ) + m2 (P2 ) + m2 (P3 ).
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Definite Integral
2. We evaluate the areas of individual “strips”. In general, we are not able to do so, because while
bounded by line segments from three sides, the fourth is defined by the graph of the function f (x).
Therefore, it will be only an approximation. Inside the base of each “strip” we choose a point (in our
picture these points are denoted in succession ξ1 , ξ2 , and ξ3 ), evaluate the function value at this point,
and at this height we square the “strip” up to a rectangle by a straight line parallel to the x -axis. Of
course, we commit an error—in some places the rectangle overlaps the “strip”, in others it does not.
Using the notation from Fig. 2.1 b) we get an approximate value of the area of set P :
.
m2 (P ) = (x1 − x0 )f (ξ1 ) + (x2 − x1 )f (ξ2 ) + (x3 − x2 )f (ξ3 ).
(2.1)
(Note that x1 − x0 is the length of the base of the first rectangle, f (ξ1 ) is its height etc.)
3. In the case of “reasonable” functions we expect that the more “strips” we make the smaller error we
commit in replacing “strips” with rectangles. Thus using some kind of a limiting process, that is
increasing unlimitedly the number of “strips” and making them at the same time narrower we assume
that the approximate value (given by the sum of areas of rectangles) approaches more and more the
exact value of the area m2 (P ). Therefore, solving this problem it seems useful to study sums having
the form of the right-hand side of Equation (2.1) as the number of summands increases unlimitedly.
Physical Motivation
Let us consider a nonhomogeneous rod T of negligible thickness and width, lying on the x -axis so that
it covers the interval ha, bi. Let ρ(x) be its length density at point x . Our task is to determine the mass
M(T ) of the rod. The situation is displayed in Fig. 2.2.
Mass has the following properties (note an analogy with the area of the plane set):
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• It is nonnegative, that is M(T ) = 0.
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Definite Integral
y = ρ(x)
x
a = x0
ξ1
T1
x1
ξ2
T2
x2
ξ3
x3 = b
T3
Fig. 2.2: Finding of the mass of a rod
• If we divide the rod into two disjoint parts T1 and T2 , that is T1 ∪ T2 = T , T1 ∩ T2 = ∅, the mass of
the whole rod equals the sum of the masses of the individual parts, that is M(T ) = M(T1 )+M(T2 ).
• If the rod is homogeneous, that is the density is constant, the mass of the rod equals product of its
length and density, that is M(T ) = (b − a)ρ .
Again, we will propose a method to determine the mass of rod T (see Fig. 2.2).
1. We divide the rod into a few disjoint smaller parts (there are three, denoted T1 , T2 , and T3 in Fig. 2.2).
We have
M(T ) = M(T1 ) + M(T2 ) + M(T3 ).
2. We evaluate the masses of the individual parts. We cannot do it exactly, because the density is not
constant, so we approximate. Inside each part we choose a point (in our Figure these points are denoted
ξ1 , ξ2 , and ξ3 ) and we assume that density is constant across the whole and equals to the value at the
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Definite Integral
chosen point. So we get an approximate value of the mass of the rod T :
.
M(T ) = (x1 − x0 )ρ(ξ1 ) + (x2 − x1 )ρ(ξ2 ) + (x3 − x2 )ρ(ξ3 ).
178
(2.2)
(Note that x1 − x0 is the length of the first “homogenized” part, ρ(ξ1 ) is its constant density etc.)
3. It is possible to expect that with more and shorter parts the more truthful will be our assumption that
the density is almost constant on such a small part. Thus, by increasing unlimitedly the number of parts
into which we divide the rod, and making them shorter and shorter we can expect that the approximate
value will approach the exact mass of the rod.
Similarly, we could evaluate, for example, the total electric charge spread on a nonconducting rod if
we knew its density ρ(x) at the point x . In this case, the function ρ(x) could be even negative.
Notice that up to the notation of functions (f versus ρ ) the sums in the right-hand sides of (2.1)
and (2.2) are identical. Both problems, where absolutely different quantities are to be evaluated, thus
lead to the investigation of the same sums. We could give many similar examples. They all would have
analogous properties as the above mentioned properties of area or mass. Namely, the second property is
essential (the total quantity equals the sum of quantities corresponding to disjoint parts). All this provides
the rationale for the following general construction and ensuing definitions.
First, let us introduce a few necessary concepts and notations to be able to define the definite integral.
Let us consider a function f defined on the bounded closed interval ha, bi, which is bounded on this
interval. So two constants m and M must exist such that m 5 f (x) 5 M holds for any x ∈ ha, bi.
Hence, the graph of f is closed in the rectangle, the sides of which lie on the straight lines x = a , x = b,
y = m, and y = M (see Fig. 2.3). Students often forget about the assumption of boundedness of f ,
which is substantial for the construction.
1. A sequence x0 < x1 < · · · < xn−1 < xn , n ∈ N, where x0 = a and xn = b, is called the partition of
the interval ha, bi. We denote the partition by the letter D . Thus, the interval ha, bi will be partitioned
to n intervals hx0 , x1 i, hx1 , x2 i, . . . , hxn−1 , xn i called intervals of partition D .
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Definite Integral
y
y=M
y = f (x)
x
O
y=m
x=a
x=b
Fig. 2.3
2. Norm of the partition D is the number
max{x1 − x0 , x2 − x1 , . . . , xn − xn−1 },
which we denote ν(D). This number gives the length of the greatest interval of the partition. (Of
course, it is possible that there are more intervals with this maximal length; namely, all the intervals
can have the same length—such a partition is called equidistant.) Thus the norm characterizes how
fine is partition D .
3. In each interval of partition D we choose a point. If we denote the point chosen in the i -th interval
hxi−1 , xi i, i = 1, . . . , n, n ∈ N, by the letter ξi , then
x0 5 ξ1 5 x1 5 ξ2 5 x2 5 · · · 5 xn−1 5 ξn 5 xn .
The set Ξ = {ξ1 , ξ2 , . . . , ξn } of these points is called the choice of representatives of partition D .
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Definite Integral
y = f (x)
. . . xi−1
a = x0 ξ1
x1
ξ2 x2
ξi xi . . .
xn−1 ξn
x
xn = b
Fig. 2.4: Visualization of an integral sum
4. If D is a partition of the interval ha, bi and Ξ is a choice of representatives of this partition, then we
define integral sum S (f, D, Ξ ) corresponding to function f , partition D and choice of representatives Ξ by the equality
n
X
S (f, D, Ξ ) =
f (ξi )(xi − xi−1 ),
i=1
or, without summation notation, by
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S (f, D, Ξ ) = f (ξ1 )(x1 − x0 ) + f (ξ2 )(x2 − x1 ) + · · · + f (ξn )(xn − xn−1 ).
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The geometrical meaning of the integral sum is visualized in Fig. 2.4. It is the sum of areas of rectangles with base lengths xi − xi−1 and heights f (ξi ), where i = 1, . . . , n, n ∈ N. Naturally, if
f (ξi ) < 0, the contribution of such a rectangle is negative. Aside from function f , the integral sum
also depends on a concrete partition and its choice of representatives.
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Definite Integral
Now we can formulate the definition of a definite integral.
Definition 2.1 Let f be a function defined and bounded on the bounded closed interval ha, bi, a < b.
We say that function f is integrable or has the definite integral on interval ha, bi if a number I ∈ R
with the following property exists:
To any number ε > 0 there exists a number δ > 0 such that for any partition D of the
interval ha,
ν(D) < δ and for any choice of representatives Ξ of this
bi with the norm
partition S (f, D, Ξ ) − I < ε holds.
The number I is called the value of the definite integral and is denoted
Z
b
f (x) d x = I.
(2.3)
a
The number a is lower limit, the number b is upper limit, the interval ha, bi is integration domain, and
the function f is an integrand. The common name for lower and upper limits is limits of integration.
The meaning of the previous definition is as follows: If we evaluate integral sums for finer and
finer partitions of the interval ha, bi, then (for any choices of representatives) the values S (f, D, Ξ )
stabilize around the number I . If it is not the case (integral sums “oscillate” even for very fine partitions),
the function is not integrable on the interval ha, bi. Below, in Example 2.4, we will show that such a
situation can occur.
Comment 2.2
1) It is easy to verify that number
R b I , with the property given in the previous definition, is unique. Therefore, the definite integral a f (x) d x is defined uniquely.
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182
2) The integral from definition 2.1 is called the Riemann1 integral. A deeper study shows that the properties of this integral are not ideal. For some more theoretical considerations other, more general
integrals, but with more complicated constructions are appropriate. There exists a whole series of
such constructions. The most important and extended is surely the Lebesgue2 integral—see [10]. The
most general in this regard is perhaps the Henstock-Kurzweil integral (see [15, 16, 27]). But for the
common needs of engineers, the Riemann integral is fully sufficient.
3) The Riemann integral is often defined in a different way. Instead of integral sums, what are known as
upper and lower sums are used (see, e.g., [9, 20, 21]). It is possible to prove that both definitions are
equivalent (see, e.g., [6], [20, p. 45]).
4) The differential d x in the notation of the definite integral in (2.3), indicates what letter is used for the
independent variable of f . From the construction of the definite integral
R b it is evident that
R b the notation
of the independent variable by the letter x is not important. Thus, a f (x) d x = a f (y) dy =
Rb
= a f (t) d t etc.
5) The symbols of definite and indefinite integrals are quite similar. The only distinction is that the
definite integral has limits of integration. This similarity often causes students to perceive that these
two concepts are in fact the same. This is absolutely wrong. It is necessary to realize that definite and
indefinite integrals differ fundamentally. Although the two integrals are evaluated for a function,
the result is different:
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1 Georg Friedrich Bernhard Riemann (1826–1866) (read ri:man)—an excellent German mathematician. He was involved
in function theory, geometry, mathematical and theoretical physics and differential equations. One of the greatest mathematicians of all time. His Riemann hypothesis on the distribution of zeros of ζ -function remains unsolved and is considered to be
one of the most difficult mathematical problems.
2 Henri Léon Lebesgue (1875–1941) (read lebeg)—important French mathematician. He was involved in function theory
and integral. The measure he introduced had the substantial impact on modern mathematics.
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183
• In the case of the indefinite integral it is a function (more exactly, the whole set of functions).
• In the case of the definite integral it is a number.
Below, we will see that there is in important relation between these two concepts, defined in a fully
different way by Theorem 2.14. Nevertheless, this does not change the fact that these concepts are
different.
R
6) As we have already mention the symbol has its origins in the letter S, denoting sum. Now, it is clear,
what sum (integral sum) we have in mind.
2.3 Let f be constant on an interval ha, bi, that is f (x) = c for any x ∈ ha, bi. Evaluate
RExample
b
c
d
x
.
a
Solution. Let us choose any partition D : a = x0 < x1 < · · · < xn = b and an arbitrary choice of
representatives Ξ of this partition. Then
S = f (ξ1 )(x1 − x0 ) + f (ξ2 )(x2 − x1 ) + · · · + f (ξn )(xn − xn−1 ) =
= c(x1 − x0 ) + c(x2 − x1 ) + · · · + c(xn − xn−1 ) = c(xn − x0 ) = c(b − a).
So all integral sums of this function
R b are the same, having the value c(b − a). This evidently implies that
the function is integrable and a c d x = c(b − a).
Notice that if c > 0, the result is the area of a rectangle of height c, built above the interval ha, bi. N
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In general, it is very difficult to verify the existence of, and evaluate, the integral directly from the
definition as we did in the previous example. Later we will describe much more efficient and simpler
tools.
In the following paragraphs we will concentrate on three domains of problems concerning the Riemann definite integral:
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Definite Integral
• the existence of the definite integral,
• properties of the definite integral,
• evaluating the definite integral.
2.2 Existence of the Definite Integral
We will start with an example showing that not any function, bounded on a bounded closed interval, is
Riemann integrable.
Example 2.4 Show that the Dirichlet1 function
(
1 for rational x ∈ h0, 1i,
0 for irrational x ∈ h0, 1i
R1
is not Riemann integrable on the interval h0, 1i, that is 0 χ (x) d x does not exist.
χ (x) =
Solution. Between any pair of different real numbers there are infinitely many rational and irrational numbers, so
the graph of the Dirichlet function is totally “disconnected” and cannot be drawn.
We will show that for any number r , 0 5 r 5 1, there exists a partition with an arbitrarily small norm and
its appropriate choice of representatives such that the corresponding integral sum equals r . This means that if we
refine the partitions, integral sums do not approach any fixed value I , on the contrary, they oscillate between 0 and
R1
1, so 0 χ (x) d x does not exist.
First, let us set r = 0. We choose any partition D : 0 = x0 < x1 < · · · < xn = 1 and take irrational numbers
for all representatives, that is χ (ξi ) = 0 for i = 1, . . . , n. Then
S (χ , D, Ξ ) = f (ξ1 )(x1 − x0 ) + f (ξ2 )(x2 − x1 ) + · · · + f (ξn )(xn − xn−1 ) =
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= 0 · (x1 − x0 ) + 0 · (x2 − x1 ) + · · · + 0 · (xn − xn−1 ) = 0.
1 Johann Peter Gustav Lejeune Dirichlet (1805–1859) (read dIrIkle)—notable German mathematician. He was involved
in number theory, mathematical analysis and equations of mathematical physics.
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Definite Integral
Now, let us set r = 1. Again, we choose any partition D : 0 = x0 < x1 < · · · < xn = 1 and take rational
numbers for all representatives, that is χ (ξi ) = 1 for i = 1, . . . , n. Then
S (χ , D, Ξ ) = f (ξ1 )(x1 − x0 ) + f (ξ2 )(x2 − x1 ) + · · · + f (ξn )(xn − xn−1 ) =
= 1 · (x1 − x0 ) + 1 · (x2 − x1 ) + · · · + 1 · (xn − xn−1 ) = xn − x0 = 1.
Finally, let 0 < r < 1 be any number. First, using any points, we partition the interval h0, ri, that is x0 <
< x1 < · · · < xk = r . Then we partition arbitrarily the interval hr, 1i, that is r = xk < xk+1 < · · · < xn = 1.
In the first k intervals of partition we take rational representatives, in the remaining irrational. We get
S (χ , D, Ξ ) = f (ξ1 )(x1 − x0 ) + f (ξ2 )(x2 − x1 ) + · · · + f (ξk )(xk − xk−1 ) +
+ f (ξk+1 )(xk+1 − xk ) + · · · + f (ξn )(xn − xn−1 ) =
= 1 · (x1 − x0 ) + 1 · (x2 − x1 ) + · · · + 1 · (xk − xk−1 ) +
+ 0 · (xk+1 − xk ) + · · · + 0 · (xn − xn−1 ) = xk − x0 = r.
Evidently, in all cases the partitions can have an arbitrarily small norm, which proves that the integral involved
does not exist.
N
We need some simple conditions, which can be easily verified, guaranteeing that the Riemann integral
exists for a sufficiently large set of functions that we encounter in common applications. Such conditions
are given in the following theorem.
Theorem 2.5 Let function f be defined on the bounded closed interval ha, bi. Let any of the following
conditions be satisfied:
(1) f is monotonic.
(2) f is continuous.
(3) f is bounded and has at most a finite number of points of discontinuity.
Rb
Then a f (x) d x exists.
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Definite Integral
R2
Rπ
Condition (2) of the previous theorem implies the existence of definite integrals 0 sin x d x , 1 x1 d x ,
R 1 x2
Re
−1 e d x , 1 ln x d x and so on. In the case of the second and fourth examples, their existence is also
guaranteed by Condition (1), because their integrands are monotonic.
Rπ
From condition (3) of the previous theorem we can conclude that the integrals 0 f (x) d x and
R 4.5
0 g(x) d x , where
(
f (x) =
sin x
x
for x 6 = 0,
0
for x = 0,
(
sin 12
x
g(x) =
0
for x 6 = 0,
(2.4)
for x = 0,
exist. Both functions are bounded and only have the point of discontinuity at zero (see Fig. 2.5).
y
y
y = f (x)
1
y = g(x)
1
x
O
π
x
O
4.5
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−11
a) f (x) = sinx x for x > 0
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b) g(x) = sin 12
x for x > 0
Fig. 2.5: Graphs of discontinuous integrable functions
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Definite Integral
Indeed, using l’Hospital’s rule for f (x) we get
lim+
x→0
0
sin x
cos x
1
LH
= lim+
=
= = 1 6= 0 = f (0).
x→0
x
0
1
1
Thus the function is discontinuous at zero, but has the finite limit there, and is continuous at any other
point, which together with Weierstrass’s Theorem (see [14]) guarantees the boundedness.
does not exist (the values oscillate between
In the case of the function g(x) the limit lim sin 12
x
x→0+
±1), so it is discontinuous
at zero. It is continuous at other points. The boundedness results from the
5 1 valid for x 6= 0.
inequality sin 12
x
R3
Similarly, we can verify the existence of −2 sgn x d x , where the integrand is the function sign, the
graph of which is in Fig. 1.2. This function is evidently bounded and discontinuous only at zero.
The following very useful result is also connected with the existence of the definite integral.
Theorem 2.6 Let functions f and g be defined on the interval ha, bi and let these functions differ at
finitely many points.
If function f is integrable on ha, bi, then function g is also integrable there and
Z
b
Z
f (x) d x =
a
b
g(x) d x.
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a
The previous theorem implies that the change of a function at finitely many points does not change
its definite integral. More exactly:
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Definite Integral
Let function g be obtained from function f by changing it at finitely many points.
• If f is integrable
R b on the interval
R b ha, bi, then function g is integrable on this interval as well, and
the equality a f (x) d x = a g(x) d x holds.
• If f is not integrable on the interval ha, bi, then function g is not integrable on this interval, as
well.
The previous result enables us to neglect the fact that the integrand is not defined at finitely many
points. We can set function values at these points arbitrarily. Neither the property “to be integrable” nor
(if the definite integral exists) the value of the integral depend on it. This feature is very useful in the
evaluation of integrals. E.g, as to the functions from (2.4), we can write
Z
0
π
sin x
dx
x
Z
4.5
sin
or
0
12
dx
x
and needn’t bother about the fact that none of the integrands is defined at zero.
Theorem 2.5 gives simple sufficient conditions of the existence of the definite Riemann integral. The
necessary and sufficient condition of the existence can be also found—see, for example, [10]. It is possible
to prove that the definite Riemann integral exists if and only if the set of points at which the integrand is
discontinuous is “small”. The exact meaning of the word “small” is that it has a zero Lebesgue measure
on the straight line (this measure is the generalization of the length of the interval for much more complex
sets on the straight line).
For example, the Dirichlet function from Example 2.4 is discontinuous at any point of the domain of
integration h0, 1i. The length of this interval is 1, thus it is not a “small” set, which confirms our previous
result that the Riemann integral of this function does not exist.
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Definite Integral
2.3 Properties of the Definite Integral
In this section we describe the basic properties of the definite integral that will be used later in the practical
evaluation of integrals.
Theorem 2.7 Let f and g be functions integrable on the interval ha, bi. Then the functions f ± g and
cf , where c is any constant, are integrable on this interval and:
Z b
Z b
f (x) ± g(x) d x =
f (x) d x ±
g(x) d x,
a
a
a
Z b
Z b
cf (x) d x = c
f (x) d x.
Z
a
b
(2.5)
(2.6)
a
The first property of the previous theorem is called additivity with respect to the integrand or sum rule,
and the second homogeneity with respect to the integrand or constant rule.
Notice that these properties are similar to those of the indefinite integral (see Theorem 1.4). The first
equality can easily be extended to any finite number of summands. As to existence, again the formulae
have to be read from right to left.
It can be proved that integrability of functions f and g on the interval ha, bi implies integrability of their
product f g on this interval.
The situation with the quotient is more complicated. First, it is necessary to assume that g(x) 6 = 0 for
x ∈ ha, bi with the eventual exception of finitely many points (see Theorem 2.6). Due to this it is possible to
redefine function g at these points without changing its integrability and the value of the integral. Moreover, function f/g needn’t be bounded (for example, f (x) = 1 for x ∈ h0, 1i and g(x) = x for x ∈ (0, 1i, g(0) = 1; then
f (0)/g(0) = 1 and f (x)/g(x) = 1/x for x ∈ (0, 1i, which is an upper unbounded function with a graph on the
interval (0, 1i being an arc of a rectangular hyperbola). But if function f/g is bounded, integrability of f and g
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Definite Integral
implies integrability of their quotient f/g (the proof can be done using the necessary and sufficient condition of
the Riemann integrability—see p. 188).
Unfortunately, unlike sums, differences and constant multiples, there is no simple way to express the definite
integral of f g or f/g with the help of integrals of f and g .
The next group of properties concerns the change of domain of integration.
Theorem 2.8 Let function f be integrable on the interval ha, bi. Then it is integrable on any subinterval
hc, di, where a 5 c < d 5 b.
Thus the restriction of the domain of integration retains integrability.
Theorem 2.9 Let function f be defined on the interval ha, bi and a < c < b. Then function f is
integrable on the interval ha, bi if and only if it is integrable on both the interval ha, ci and the interval
hc, bi. Moreover, in case of integrability we have:
Z
b
Z
c
f (x) d x =
a
Z
f (x) d x +
a
b
f (x) d x.
(2.7)
c
The previous property is called additivity with respect to the domain of integration.
The previous theorem is useful in cases where the integrand is not given by one formula on the whole
interval ha, bi. Moreover, its statement can easily be generalized by induction. If a < c1 < c2 < · · · <
< cn < b, n ∈ N, then
Z
b
Z
f (x) d x =
a
c1
Z
c2
f (x) d x +
a
Z
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f (x) d x + · · · +
c1
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Integrability on the whole interval ha, bi is equivalent to integrability on all intervals occurring in the
integrals on the right-hand side of the previous equality.
R4
Example 2.10 Evaluate −2 f (x) d x , where


 2 for x ∈ h−2, 1i,
f (x) = −1 for x ∈ (1, 3),


1 for x ∈ h3, 4i.
Solution. Due to Theorem 2.9 and its generalization we have
Z 4
Z 1
Z 3
Z
f (x) d x =
f (x) d x +
f (x) d x +
−2
1
−2
Z 1
Z
2 dx +
=
−2
3
f (x) d x =
3
Z
4
1 d x.
(−1) d x +
1
4
3
Integrals on the right-hand side of the previous equality exist, as we proved in Example 2.3, so we can
apply the mentioned theorem. (Notice that in the second integral we silently changed the values of f at
end points to −1; given Theorem 2.6, and the explanations that follow, this has no impact on the existence
of the integral and its value.)
Another way to verify integrability is based on Theorem 2.5. The function involved is bounded and
continuous up to two points x = 1 and x = 3.
Altogether, using the result of Example 2.3, we have
Z 4
f (x) d x = 2 · (1 − (−2)) + (−1) · (3 − 1) + 1 · (4 − 3) = 5.
−2
The graph of the function f is displayed in Fig. 2.6. The result is the sum of the areas of three parallelograms (two rectangles and one square), where the area of the middle one is taken negative.
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Definite Integral
y
2
y = f (x)
1
x
−2
2
1
3
4
−1
Fig. 2.6: Graph of a piecewise constant function
Further we will discuss inequalities that are valid for the definite integral.
Theorem 2.11 Let f and g be integrable functions on the interval ha, bi and for each x ∈ ha, bi, let
the inequality f (x) 5 g(x) hold. Then
Z
b
Z
f (x) d x 5
a
b
g(x) d x.
Rb
0 d x = 0, the previous theorem implies that for any nonnegRa b
ative integrable function g the inequality a g(x) d x = 0 holds. This fact can be used to roughly check
the correctness of the result. If an integrand is evidently nonnegative, the result cannot be negative.
As by virtue of Example 2.3 we have
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a
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Definite Integral
Theorem 2.12 Let f be a function integrable on the interval ha, bi. Then function |f | is also integrable
on this interval. Moreover,
Z b
Z b
f (x) d x 5
|f (x)| d x.
a
a
In summary: the absolute value of the definite integral is less than or equal to the definite integral of the
absolute value.
Theorem 2.13 (Mean Value Theorem of Integral Calculus) Let f be a function integrable on the
interval ha, bi and let the inequality m 5 f (x) 5 M , where m and M are constants, hold for all
x ∈ ha, bi.
Then there exists a number c, where m 5 c 5 M , such that
Z
b
f (x) d x = c(b − a).
a
Moreover, if function f is continuous, the constant c can be set to a function value of f , that is, there
exists x0 ∈ ha, bi such that
Z b
f (x) d x = f (x0 )(b − a).
a
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The number c is called the mean value of function f on interval ha, bi.
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Rb
Rb
Rb
Proof. By Theorem 2.11, we have a m d x 5 a f (x) d x 5 a M d x , therefore the inequality m(b −
Rb
Rb
1
− a) 5 a f (x) d x 5 M(b − a) holds. Hence, m 5 b−a
a f (x) d x 5 M , and it is sufficient to set
Rb
1
c = b−a a f (x) d x .
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Definite Integral
y
y
M
y = f (x)
y=
f (x0 )
f (x0 )
1
b−a
Rb
a
f (x) d x
m
x
a
x0
b
x
a
b
Fig. 2.7: Geometrical meaning of the Mean Value Theorem of integral calculus
It the function f is continuous, then due to the Weierstrass Theorem f attains its greatest and smallest
values on the interval ha, bi, which means that there exist x1 , x2 ∈ ha, bi such that f (x1 ) 5 f (x) 5
5 f (x2 ) for any x ∈ ha, bi. Thus we can set m = f (x1 ) and M = f (x2 ), therefore f (x1 ) 5 c 5
5 f (x2 ). The Cauchy-Bolzano Theorem states that there exists x0 lying between x1 and x2 , for which
f (x0 ) = c.
The previous theorem has clear geometrical meaning. Assume for simplicity that the function f
is continuous and nonnegative. From the previous
exposition (geometrical motivation of the definition
Rb
of definite integral) we already know that a f (x) d x expresses the area of the figure bounded by the
graph of the function f , the x -axis, and straight lines parallel to the y -axis passing through the points
a and b. The Mean Value Theorem states that we can construct a rectangle over the interval ha, bi having
the same area (which is trivial) and height equal to the function value at an appropriate point x0 (see
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Definite Integral
Rb
1
Fig. 2.7). In fact, it is the graph of the constant function y = f (x0 ), where f (x0 ) = b−a
a f (x) d x . It
can be seen in the graph that
R bpoint x0 is not generally determined uniquely. In our case, the straight line
1
with the equation y = b−a
a f (x) d x intersect the graph of function f twice.
2.4 Evaluation of the Definite Integral
In the previous sections we established several properties of the definite integral, but with the exception
of a constant function (which is in fact the area of a rectangle) we were not able to evaluate any definite
integral. We will repair that now. The key tool is a formula named after Newton1 and Leibniz2 , two mathematicians who contributed substantially to the development of foundations of differential and integral
calculus of functions of a single variable. This Newton-Leibniz formula is the promised relation between
indefinite and definite integrals.
Theorem 2.14 (The Fundamental Theorem of Calculus) Let a function f be integrable on the interval ha, bi and let F be its antiderivative. Then
Z
b
f (x) d x = F (b) − F (a).
(2.8)
a
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Proof. We will show that for any partition D : a = x0 < x1 < · · · < xn = b of the interval ha, bi the
difference F (b) − F (a) equals the integral sum S (f, D, Ξ ), with an appropriate choice of representatives.
1 Isaac Newton (1643–1727) (read nju:t@n)—English mathematician, physicist and astronomer. He laid the foundations of
differential and integral calculus, which he needed for building up classical mechanics.
2 Gottfried Wilhelm Leibniz (1646–1716) (read laIbnIts)—German mathematician, philosopher, inventor, lawyer, and
linguist. He laid the foundations of differential and integral calculus.
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Definite Integral
Function F satisfies the assumptions of the Lagrange Mean Value Theorem on each interval hxi−1 , xi i,
i = 1, . . . , n. So we can find numbers ξi , ξi ∈ hxi−1 , xi i, such that F (xi ) − F (xi−1 ) = F 0 (ξi )(xi −
− xi−1 ). Because F 0 (x) = f (x) for any x ∈ ha, bi, we have F (xi ) − F (xi−1 ) = f (ξi )(xi − xi−1 ).
Adding up these equalities we get
F (b) − F (a) =
= [F (x1 ) − F (x0 )] + [F (x2 ) − F (x1 )] + · · · + [F (xn ) − F (xn−1 )] =
= f (ξ1 )(x1 − x0 ) + f (ξ2 )(x2 − x1 ) + · · · + f (ξn )(xn − xn−1 ) =
= S (f, D, Ξ ),
where Ξ = {ξ1 , . . . , ξn }.
As function f is integrable, after we refine the partitions,
R b integral sums with any representatives
approach some constant I (value of the definite integral a f (x) d x ). Partitions in the previous construction can be arbitrarily fine while the value of the corresponding integral sum is F (b) − F (a) each
time. This is only possible if F (b) − F (a) = I .
Comment 2.15
1. The difference F (b) − F (a) is usually denoted as [F (x)]ba , so that Formula (2.8) can be written as
Z b
f (x) d x = [F (x)]ba .
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a
2. From Section 1.1 we know that if F is an antiderivative of f , it is not unique. Thus at first sight it
seems that Formula (2.8) could give different results for different antiderivatives. But Theorem 1.2
implies that such a thing cannot happen and the result is always the same, and independent of the
choice of antiderivative. Indeed, if G is another antiderivative of f , there exists a constant c such that
G(x) = F (x)+c for any x ∈ ha, bi. Hence G(b)−G(a) = [F (b)+c]−[F (a)+c] = F (b)−F (a).
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Definite Integral
y
y
F (b)
y = F (x)
y = f (x)
F (b) − F (a)
F (a)
x
a
b
x
a
b
Fig. 2.8: Geometrical meaning of the Fundamental Theorem of Calculus
For this reason, in the following examples, where the definite integral is to be evaluated, the integration
constant will not be added to the indefinite integrals.
Rb
3. Fig. 2.8 shows the geometrical meaning of the Fundamental Theorem of Calculus. a f (x) d x equals
the increment of the antiderivative F on the interval ha, bi (both functions can be defined on a larger
interval than ha, bi as in this case). Notice that the fact that f is positive implies that F is increasing.
Indeed, because F 0 (x) = f (x) > 0 on the interval ha, bi, F must be increasing by the well-known
statement of differential calculus. This agrees with the expectation: for the fixed lower limit a and
increasing upper limit b the area below the graph increases, so the increment F (b) − F (a) must be
increasing as a function of b, which means that F must be increasing.
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Definite Integral
Example 2.16 Using the Fundamental Theorem of Calculus evaluate the definite integrals:
Z 2
Z 4
Z π
Z
√
2
a)
x d x,
b)
x d x,
c)
sin u d u,
d)
1
Z 1
e)
0
0
0
1
2
4
x
√
−
+ 2
+ 2
x2 + 3 x + 1 x + 2 x + 2
1
−2
t2
dt
,
+1
d x.
Solution. The above definite integrals exist, because their integrands are continuous. Evaluating indefinite integrals we use Formulae from Table 1.1 on page 19.
a) Using Formula 3 we get:
Z
1
2
1 3
x
x dx =
3
2
2
=
1
8 1
7
− = .
3 3
3
b) Using Formula 3 we get:
4
Z
√
x dx =
0
Z
0
4
x 1/2 d x =
x 3/2
3/2
4
=
0
√ 4
2
16
2√ 3
x3 =
4 −0=
.
3
3
3
0
c) Using Formula 7 we get:
Z π
sin u d u = [− cos u]π0 = − cos π − (− cos 0) = −(−1) + 1 = 2.
0
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d) Using Formula 9 we get (remember that inverse tangent is an odd function):
Z
Contents
1
−2
dt
π
= [arctan t]1−2 = arctan 1 − arctan(−2) = + arctan 2.
2
t +1
4
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199
e) First we transform the given integral to several definite integrals, using Formulae (2.5) and (2.6). Then
we evaluate these using the Fundamental Theorem of Calculus and Formulae 11, 4, 9, and 14.
Z 1
0
4
x
1
2
√
−
+
+
dx =
x2 + 3 x + 1 x2 + 2 x2 + 2
Z 1
Z 1
Z 1
Z 1
dx
dx
dx
x
√
=
−2
+4
+
dx =
2
2
2
x +3
0
0 x +1
0 x +2
0 x +2
1
p
1
1
1
1
x
1
2
= lnx + x + 3 0 − 2 ln |x + 1| 0 + 4 √ arctan √
+ ln(x 2 + 2) 0 =
2
2 0 2
√
1
1
1
1
= ln 3 − ln 3 − 2(ln 2 − ln 1) + 4 √ arctan √ − √ arctan 0 + (ln 3 − ln 2) =
2
2
2
2
√
5
1
= ln 3 − ln 2 + 2 2 arctan √ .
2
2
Instead of separating the definite integral into the sum of four integrals, it is possible to directly find
the antiderivative. Then the evaluation looks as follows:
Z 1
1
2
4
x
√
−
+
+
dx =
x2 + 3 x + 1 x2 + 2 x2 + 2
0
1
p
x
1
4
2
2
= ln x + x + 3 − 2 ln |x + 1| + √ arctan √ + ln(x + 2) =
2
2 2
0
√
5
1
= · · · = ln 3 − ln 2 + 2 2 arctan √ .
2
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2
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Definite Integral
Example 2.17 Using the Fundamental Theorem of Calculus evaluate the definite integrals:
Z π
Z 1 p
a)
(x − 1) sin x d x,
b)
x 1 − x 2 d x.
0
−π/2
Solution. The given definite integrals exist, because their integrands are continuous. This time we cannot
find the antiderivatives as easily as in the previous examples. Thus we first evaluate the indefinite integrals
separately.
a) Integrating by parts we get:
Z
u = x − 1 u0 = 1
=
(x − 1) sin x d x = 0
v = sin x v = − cos x Z
= −(x − 1) cos x + cos x d x = (1 − x) cos x + sin x.
So
Z
π
−π/2
π
(x − 1) sin x d x = (1 − x) cos x + sin x −π/2 =
h
π π π i
1− −
cos −
+ sin −
=
2
2
2
π
· 0 − (−1) = π.
= (1 − π) · (−1) + 0 − 1 +
2
b) Using the substitution method we get:
1 − x 2 = u2
Z p
Z
1
1p
2
x 1 − x d x = −2x d x = 2u d u = − u · u d u = − u3 = −
(1 − x 2 )3 .
3
3
x d x = −u d u = (1 − π) cos π + sin π −
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Definite Integral
So
Z
1
x
0
p
1−
x2
1p
dx = −
(1 − x 2 )3
3
1
0
1
1
=0− −
= .
3
3
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With the procedure in the previous example it was necessary to first use integration by parts or the
substitution method for the indefinite integral. This is not overly efficient. A better approach is to modify
the above mentioned methods for the definite integral. Then the evaluation is usually much faster. These
modifications will be explained in the following sections. But before we start with these we present an
example illustrating the situation where the use of the Fundamental Theorem of Calculus is often wrong.
Z 2π
dx
Example 2.18 Evaluate
.
2 − cos x
0
1
Solution. The above integral exists, because its integrand 2−cos
is continuous. The antiderivative on
x
the interval (−π, π) was found in Example 1.65. There, we found that
√
2
x
F (x) = √ arctan 3 tan ,
2
3
x ∈ (−π, π).
(2.9)
But if we want to use the Fundamental Theorem of Calculus, we need the antiderivative, which is defined at least on the closed interval h0, 2πi. Therefore, we have to apply the construction described in
Section 1.6.2.
Z 2π
2π
dx
= F (x) 0 =
2 − cos x
0
√
√
2
2π
2
= √ arctan 3 tan π + √ − √ arctan 3 tan 0 =
3
3
3
2
2π
2
2π
2π
= √ arctan 0 + √ − √ arctan 0 = 0 + √ − 0 = √ .
3
3
3
3
3
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Definite Integral
√
Compare the result with Fig. 1.4. We see at once that F (0) = 0 and F (2π) = 2π/ 3.
If we were not careful enough and also “blindly” used Formula (2.9) for the value 2π, we would
obtain
√
√
2
2
√ arctan 3 tan π − √ arctan 3 tan 0 = 0 − 0 = 0.
3
3
Thus we would use the function G from Fig. 1.4, coinciding with F only on (−π, π), and would evaluate
1
G(2π) − G(0) instead of F (2π) − F (0). The result is evidently wrong, because the integrand 2−cos
x
is positive (for any x ), hence the integral must assume a positive value.
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2.4.1 Integration by Parts for the Definite Integral
Theorem 2.19 Let functions u and v have integrable derivatives on the interval ha, bi, a < b. Then
Z
a
b
b
u(x)v 0 (x) d x = u(x)v(x) a −
Z
b
u0 (x)v(x) d x.
(2.10)
a
Proof. The existence of derivatives implies that u and v are continuous. By means of the text on page 189,
the functions uv 0 and u0 v are integrable, so, due to Theorem 2.7, the function uv 0 + u0 v is integrable. Its
antiderivative is uv . Using the Fundamental Theorem of Calculus we have
Z
a
b
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b
[u(x)v (x) + u (x)v(x)] d x = u(x)v(x) a .
0
0
After a short manipulation using Theorem 2.7 we get this statement.
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Definite Integral
Later in Theorem 2.30 we will formulate the integration by parts method under more general assumptions.
The application is analogous to that for the indefinite integral. Namely, the hints are valid, for which
functions this method is appropriate (see page 41). As opposed to the procedure described in Example 2.17 a, the advantage consists in running substitution of limits into the partially determined antiderivative, which needn’t be always copied till the end of the evaluation. The evaluation is shorter and more
readable, as the following examples show.
Z 2
Example 2.20 Evaluate
(x 2 + 1) ln x d x .
1
Solution. Integrating the polynomial x 2 + 1 we get
Z 2
u = ln x
u0 = x1
2
(x + 1) ln x d x = =
1 3
0
2
v =x +1 v = 3x +x 1
2 Z 2 1 3
1 1 3
=
x + x ln x −
x + x dx =
3
3
1 x
1
Z 2
8
1
1 2
=
+ 2 ln 2 −
+1 ·0−
x + 1 dx =
3
3
3
1
2
14
1 3
=
ln 2 −
x +x =
3
9
1 8
1
14
16
14
=
ln 2 −
+2 −
+1
=
ln 2 −
.
3
9
9
3
9
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The function (x + 1) ln x is positive on the interval h1, 2i (except the point x = 1), thus the result must
be positive. Using a calculator you can verify that the value is approximately 1.46.
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Definite Integral
Z
Example 2.21 Evaluate
π
x 2 cos x d x .
0
Solution. Differentiating the polynomial x 2 , and using the method twice, we get (beware of changes of
signs)
Z π
Z π
0
2
u = x2
u
=
2x
2
= x sin x π −
2x sin x d x =
x cos x d x = 0
0
v = cos x v = sin x 0
0
u = 2x
u0 = 2
=
= 0
v = sin x v = − cos x Z π
π
2
= π · 0 − 0 − −2x cos x 0 +
2 cos x d x =
0
π
= − −2π · (−1) − 0 − 2 sin x 0 = −2π − (0 − 0) = −2π.
First try to evaluate the complete antiderivative to x 2 cos x and then use the Fundamental Theorem
of Calculus. Compare how much longer such an evaluation takes.
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2.4.2 Substitution Method for the Definite Integral
Before formulating the corresponding statement we have to extend the definition of the definite integral.
Until now we have assumed that the integration domain is an interval ha, bi, that is a, b ∈ R and a < b.
Now we remove this assumption and admit that a = b is allowed. We set
Z a
f (x) d x = 0,
(2.11)
b
Z
f (x) d x = −
a
a
f (x) d x
b
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Z
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for a > b.
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Definite Integral
If a > b, the function must be integrable on the interval hb, ai. Recall that the interchange of limits
changes the sign of the definite integral.
Notice that Equality (2.8) from the Fundamental Theorem of Calculus is still valid for this extended
definition of the definite integral. Theorem 2.7 and the method of integration by parts remain valid as
well.
Theorem 2.22 Let function f (t) be continuous on the interval ha, bi, a < b. Let function ϕ(x) have
the derivative ϕ 0 (x) on the interval hα, βi, α < β , which is integrable on this interval. Further, let
a 5 ϕ(x) 5 b for x ∈ hα, βi. (So ϕ maps interval hα, βi into interval ha, bi). Then
Z
α
β
f [ϕ(x)] ϕ 0 (x) d x =
Z
ϕ(β)
f (t) d t.
(2.13)
ϕ(α)
Formula (2.13) resembles the first substitution method for an indefinite integral (compare to (1.8)). Of
course the assumptions are different. Both integrals are definite, but in general they have different limits.
Along with choosing the right substitution ϕ(x) = t and evaluating the differentials ϕ 0 (x) d x = d t , this
time, we must find new limits. “Old” limits α and β are for the original variable x , “new” limits ϕ(α)
and ϕ(β) are for the new variable t . In concrete cases it can happen that ϕ(α) = ϕ(β), which is solved
by extensions (2.11) and (2.12).
The use and the notation are analogous to those for the indefinite integral, we only add finding new
limits. We denote it in the auxiliary table as α ; ϕ(α) (old lower limit α is replaced by new lower
limit ϕ(α)) and β ; ϕ(β) (old upper limit β is replaced by new upper limit ϕ(β)).
Compared to Example 2.17 b the advantage is that after the substitution we do not have to revert to
the old variable, which is sometimes rather unpleasant as we saw earlier (see for example Example 1.30).
Formula (2.13) can be used in both directions. In the next example we will use it from left to right.
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Definite Integral
Example 2.23 Evaluate the following definite integrals:
Z 1
Z 2π
3
2
a)
x(x − 1) d x,
b)
esin x cos x d x,
Z
c)
π
0
0
π/2
sin x cos2 x
√
d x.
4
1 + cos3 x
Solution.
a) We will use the substitution x 2 − 1 = t . Old limits are for the variable x , thus we put them into this
equation in succession, for x , and obtain the values of new limits for variable t . We get 02 − 1 = −1
for the new lower limit and 12 − 1 = 0 for the new upper limit. Then we apply the Fundamental
Theorem of Calculus to the transformed integral. The whole evaluation looks like this:
2
x −1=t
Z 0
Z 1
2x d x = d t
1 3
1 t4 0
1
3
2
=
t dt =
x(x − 1) d x = =− .
1
2 4 −1
8
−1 2
0
x dx = 2 dt 0 ; −1, 1 ; 0 b) This time we use the substitution sin x = t . We get sin π = 0 for the new lower limit and sin 2π = 0
for the new upper limit. Hence the evaluation will be very short:
Z 0
Z 2π
sin x = t
sin x
e
cos x d x = cos x d x = d t =
et d t = 0.
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Each time this happens (that is when the new limits coincide), it is not necessary to arrange the new
integral since the result equals zero—which keeps us happy. In this case it is evident that the evaluation
is much shorter, compared to Example 2.17 b.
3
3
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3
c) We choose the substitution 1 + cos x . We get 1 + cos 0 = 1 + 1 = 2 for the new lower limit and
1 + cos3 π2 = 1 + 03 = 1 for the new upper limit, thus the new lower limit is greater then the new
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Definite Integral
upper limit. So we use the extension from (2.12) and interchange the limits. Therefore,
1 + cos3 x = u
Z 1
Z π/2
−3 cos2 x sin x d x = d u
sin x cos2 x
−1/3
=
√
√
d x = du =
1
2
4
4
cos x sin x d x = − 3 d u u
1 + cos3 x
0
2
0 ; 2, π/2 ; 1
Z 2
2 4 √
1 u3/4 2 1 4 √
1
4
4
−1/4
u
du =
= ·
u3 1 =
8−1 .
=− −
3
3 3/4 1 3 3
9
1
b
b
We used the evident fact that for any constant c there is cF (x) a = c F (x) a . Later, we will silently
use this, because the evaluation is usually much simpler.
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In the next example we demonstrate the use of Formula (2.13) from right to left. This corresponds to
the second substitution method for the indefinite integral (compare (1.11)). This time we know the values
ϕ(α) and ϕ(β) (these are “old limits” now) and must correctly choose α and β so that the assumptions
of Theorem 2.22 are satisfied.
When Formula (2.13) is used this way, typically the function ϕ(x) is strictly monotonic on the interval
hα, βi, α < β , and maps it onto the interval ha, bi, a < b. Thus ϕ(α) = a , ϕ(β) = b for an increasing
function ϕ and ϕ(α) = b, ϕ(β) = a for a decreasing function ϕ . To avoid problems with the notation
of the lower and upper limits a, b depending on whether ϕ is increasing or decreasing, it is reasonable
to admit that α > β (thus the function ϕ is defined on hα, βi if it is increasing, and on hβ, αi if it is
decreasing).
Then, under the above additional assumptions Formula (2.13) can be simplified and written as follows:
Z
Z
b
β
f [ϕ(t)] ϕ 0 (t) d t,
f (x) d x =
a
where ϕ(α) = a , ϕ(β) = b.
α
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(2.14)
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Definite Integral
Notice that we interchanged the notation of independent variables in integrands. The variable in the
integral on the left (the one to be evaluated) is usually denoted by the letter x .
Example 2.24 Evaluate the following definite integrals:
Z 4√
Z 1
x−1
dx
√
√
,
a)
d x,
b)
x+1
x2 + 1 − x
1
0
Z
c)
1
p
x 2 + 1 d x.
0
Solution.
a) The integrand is continuous on the interval h1, 4i, so the integral exists. The indefinite integral is of
Type (1.26), thus we choose the substitution x = t 2 . The function ϕ(u) = t 2 , having a parabola as the
graph, is not monotonic on R. We must find α and β such that ϕ(α) = α 2 = 1 and ϕ(β) = β 2 = 4.
If we only consider the interval h0, +∞)√on which the function ϕ(t) is increasing we get α = 1,
β = 2. Moreover, for t ∈ h1, 2i we have t 2 = |t| = t . Thus
Z 2
x = t2
Z 2 2
Z 4√
t −1
x−1
2t − 2t
√
=
d x = d x = 2t d t
2t d t =
dt =
x+1
t +1
1 t +1
1
1
1 ; 1, 4 ; 2 Z 2
2
4
d t = t 2 − 4t + 4 ln |t + 1| 1 =
=
2t − 4 +
t +1
1
= (4 − 8 + 4 ln 3) − (1 − 4 + 4 ln 2) = 4 ln 3 − 4 ln 2 − 1.
The improper rational function
2t 2 −2t
t+1
was reduced to the proper one:
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2
2t − 2t
4
= 2t − 4 +
.
t +1
t +1
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Definite Integral
Another possibility was to consider the interval (−∞, 0i where the
√ function ϕ(t) is decreasing,
and set α = −1, β = −2. Of course, then for t ∈ h−2, −1i we have t 2 = |t| = −t .
b) The integrand is continuous on the interval h0, 1i, so the integral exists. The indefinite integral is
of Type (1.32). This can be quickly evaluated using a substitution, which we find from the equation
√
x 2 + 1 − x = t (Euler’s second substitution). Rearranging and raising to two this relation we get
p
x2 + 1 = x + t
⇒
x 2 + 1 = x 2 + 2tx + t 2
⇒
x=
1 − t2
.
2t
Continuing, we prepare the derivative:
ϕ(t) =
1 − t2
2t
⇒
−2t · 2t − (1 − t 2 ) · 2
t2 + 1
=
−
.
4t 2
2t 2
ϕ 0 (t) =
At
√ which hold for the variable x , into the
√ relation
√ last we find new limits. Substituting the old limits,
x 2 + 1 − x = t we get t = 1 for x = 0 and t = 2 − 1 for x = 1. Thus α = 1, β = 2 − 1 in
Formula (2.14) and α > β .
The function ϕ(t) is not defined at t = 0. The formula for ϕ 0 (t) implies that evidently ϕ 0 (t) < 0
for
√ t 6 = 0. Hence the function ϕ(t) is decreasing on the interval (0, +∞) and maps the interval
h 2 − 1, 1i onto the interval h0, 1i. We have (notice the interchange of limits, which changes the
sign):
x = 1−t 2
Z √
2
Z 1
2t
2−1
dx
1
t
+
1
2
√
= d x = − t +1
· −
dt =
=
dt
2t 2 √
t
2t 2
x2 + 1 − x
0
1
0 ; 1, 1 ; 2 − 1 Z
=
1
√
2−1
2
t +1
1
dt =
3
2t
2
Z
1
√
2−1
1
1
+ 3
t
t
1
1 1
dt =
ln |t| − 2 √
=
2
2t
2−1
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Definite Integral
√
1
1
1
1
ln 1 −
−
ln 2 − 1 − √
2 =
2
2
2
2 2−1
√
√
√
1 1
1
1+ 2 1
√ =
− ln 2 − 1 .
= − − ln 2 − 1 +
4 2
2
2
4 3−2 2
=
c) The integrand is continuous on the interval h0, 1i, so the integral exists. The indefinite integral is of
Type (1.32). The recommended substitution was x = tan v . Likewise we can use the substitution
x = cot v , which will be more suitable in our case.
We find new limits. The function ϕ(v) = cot v is decreasing on the interval (0, π). As we need
cot α = 0, cot β = 1 we can set α = π/2, β = π/4. Then the interval hπ/4, π/2i is mapped by
the function ϕ(v) onto the interval h0, 1i. Because sin v > 0 on the interval hπ/4, π/2i, we have
| sin v| = sin v .
After the substitution we get (again, the order of limits is changed, so the sign will be changed):
s
x = cot v
Z
Z 1p
π/4
−1
cos2 v
x 2 + 1 d x = d x = − sin12 v d v =
+ 1 · 2 dv =
2
sin v
sin v
π/2
0
0; π, 1; π 2
Z
π/2
=
π/4
4
1
1
· 2 dv =
| sin v| sin v
Z
π/2
π/4
1
d v.
sin3 v
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The indefinite integral is of Type (1.18), where m = 0 and n = −3. We use the substitution
method. Although the recommended substitution is t = cos v , we prefer the universal substitution
t = tan v2 (see (1.20)), which will be faster this time, because it gives a simpler rational function. This
is the reason why we chose the substitution x = cot v at the beginning; verify that the substitution
x = tan v gives the integral of 1/ cos3 v , the evaluation of which is a bit more laborious.
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Definite Integral
When using Formula (2.13), do not forget that the differential must be evaluated from the equality
v = 2 arctan t . The function tan v2 is increasing on the interval (−π, π), thus new limits are easy to
evaluate. Substituting for v we get tan π8 for the lower limit and tan π4 = 1 for the upper limit. Due to
(1.20) we get:
tan v = t
2
Z π/2
v = 2 arctan t Z 1
1
2
1
·
dv = dt =
=
2
3
3
2t
d v = 1+t 2 d t
1 + t2
π/4 sin v
tan π8
π
t 2 +1
; tan π , π ; 1 4
8
2
Z
Z 1 4
2
t + 2t + 1
1
1 1
1 1
2
1 t2
+ 2 ln |t| − 2
=
dt =
t + + 3 dt =
=
4t 3
4 tan π8
t
t
4 2
2t tan π
tan π8
8
1 1
1
1 1
π 1
2 π
2 π
=
+ 2 ln 1 −
−
tan
+ 2 ln tan − cot
=
4 2
2
4 2
8
8
2
8
=
π 1
π 1
π
1
cot2 − tan2 − ln tan .
8
8
8
8
2
8
The result looks quite “wild”, so we rearrange it. Using the trigonometric identity
r
x
1 − cos x
tan =
,
2
1 + cos x
valid for x ∈ h0, π), we get
π
tan =
8
s
1 − cos π4
1 + cos π4
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v
u
u1 −
=t
1+
√1
2
√1
2
s √
s√
2
√
2−1
2−1
= √
=
= 2 − 1.
2−1
2+1
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Definite Integral
Using this result we have:
Z
0
1
√
2
√
2−1
1
− ln 2 − 1 =
√
2 −
8
2
8 2−1
√
√
2 1
=
− ln 2 − 1 .
2
2
p
x2 + 1 dx =
1
If we chose substitution t = cos v to evaluate the integral of 1/ sin3 x , we would obtain a more
complicated rational function, but would avoid the evaluation of tan π8 . It is possible to try Euler’s
substitution as well, like in part b) of this example. Moreover, notice that
√
p
1
x2 + 1 + x
√
= √
√
= x 2 + 1 + x.
x2 + 1 − x
x2 + 1 + x
x2 + 1 − x
As both integrals b) and c) of this example have the same integration domain h0, 1i, their results must
1
R1
differ by 0 x d x = 21 x 2 0 = 12 , which can be easily verified.
This example clearly shows that the evaluation of the integral of a seemingly simple function can
be rather technically complicated and time consuming. Choosing the easiest possible method is a
matter of experience. In cases like this, suitable computer programs can help us a lot.
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At the end of this section we mention some generalization of the Fundamental Theorem of Calculus. In
Theorem 2.14 the existence of an antiderivative of the integrand was assumed on the whole integration domain
ha, bi. Sometimes, this assumption is too strong. When, for example, integrand f is discontinuous at some
interior point and has different one-sided limits, the antiderivative cannot exist (see for example the function sign
in Fig. 1.2 on page 16). If this is the only point of discontinuity (or f has finitely many points of discontinuity), the
definite Riemann integral exists by Theorem 2.5. This situation can be solved by dividing the integration domain
into several parts (see Theorem 2.9). Sometimes it is more useful to apply the generalization of the Fundamental
Theorem of Calculus.
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Definite Integral
Definition 2.25 Function F is called the generalized antiderivative of function f on the interval I if
• F is continuous on the interval I ,
• F 0 = f holds on the interval I with the possible exception of at most finitely many points.
Thus the generalized antiderivative needn’t have the derivative at some points or its derivative can be different from f at some points. But the substantial property is its continuity. This guarantees that the generalized
antiderivative is determined uniquely up to an additive constant (thus has the same property like the “ordinary”
antiderivative).
Theorem 2.26 (The Generalized Fundamental Theorem of Calculus) Let function f be integrable on the
interval ha, bi and let F be its generalized antiderivative. Then
Z
b
f (x) d x = F (b) − F (a).
(2.15)
a
The proof is analogous to that of Theorem 2.14. Only the partitions containing all points x , at which F 0 (x) =
= f (x) is not valid are considered. This is possible since we assume that there are only a finite number of these
points.
Example 2.27 Evaluate the integral from Example 2.10 using the Generalized Fundamental Theorem of Calculus.
Solution. The graph of the integrand f is displayed in Fig. 2.6. It is a function that is constant on any of the
three contiguous intervals, but the constants are different. Finding the generalized antiderivative F is quite easy.
It is sufficient to find an “ordinary” antiderivative on each interval, then fix one of them and shift the others in the
direction of the y -axis (thus to set properly integration constants) to obtain a continuous function. In our case we
have:




 2x + c1 for x ∈ h−2, 1i,
 2 for x ∈ h−2, 1i,
f (x) = −1 for x ∈ (1, 3),
⇒
F (x) = −x + c2 for x ∈ (1, 3),




1 for x ∈ h3, 4i,
x + c3 for x ∈ h3, 4i.
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Definite Integral
It only remains to determine integration constants c1 , c2 , c3 to get
a continuous function F . For example, let us set c1 = 0. The
formulae must give the same values at end points of the contiguous
intervals. Substituting x = 1 we get 2 · 1 = −1 + c2 , thus c2 = 3,
and then substituting x = 3 we get −3+3 = 3+c3 , thus c3 = −3.
Hence,


2x for x ∈ h−2, 1i,

F (x) = −x + 3 for x ∈ (1, 3),


x − 3 for x ∈ h3, 4i.
Any other generalized antiderivative differs from this by a constant.
The function F is graphed in Fig. 2.9. Therefore, Theorem 2.26
gives
Z
y = F (x)
y
2
F (4) = 1
x
−22
1
3
4
−4 = F (−2)
4
f (x) d x = F (4) − F (−2) = 1 − (−4) = 5,
Fig. 2.9
−2
which is the same result like in Example 2.10.
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2.4.3 The Definite Integral as the Function of Limits
Let us assume that a function f is Riemann integrable on an interval ha, bi and choose an arbitrary
R xc ∈ ha, bi,
which will be fixed in the following considerations. Then, for any x ∈ ha, bi the definite integral c f (t) d t is
correctly defined. For c < x it is the consequence of Theorem 2.8, for c = x it is necessary to take into account
Relations (2.11) and (2.12). Notice also that because we used the letter x for the upper limit, we had to use a
different letter for the independent variable of integrand t . As we know it has no impact on the value of the definite
integral.
The value obtained this way depends on the choice of x . Thus we get a new function, let us denote it F , which
assigns to the number x from the interval ha, bi the value of the definite integral of the function f over the interval
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Definite Integral
with the end points c and x ; c is considered to be the lower limit and x the upper limit. So the function F is defined
by the equality
Z
x
f (t) d t,
F (x) =
x ∈ ha, bi.
(2.16)
c
This integral is called function of the upper limit.
Function F can be defined by Equality (2.16) not only for the bounded closed interval but for any interval I
(open, closed, half open, bounded, unbounded) if we assume that the function f is Riemann integrable on any
bounded closed subinterval hd, ei ⊂ I .
If f (x) = cos x , we can take I = (−∞, ∞), because the function cosine is continuous and thus integrable
on any bounded closed interval. From Equality (2.16) we obtain that in this case
Z x
x
F (x) =
cos t d t = sin t c = sin x − sin c,
x ∈ (−∞, ∞).
c
Notice that the resulting function F is one of the antiderivatives of cosine, that is of integrand. Just the one for
which F (c) = 0.
The following important result can be proved:
Theorem 2.28 Let function f be defined on the interval I and be Riemann integrable on any bounded closed
subinterval of I . Let c ∈ I . Then:
1. Function F defined by Equality (2.16) is continuous on the interval I .
2. Moreover, if f is continuous at some point x0 ∈ I , function F has the derivative at this point and the equality
F 0 (x0 ) = f (x0 ) holds.
If x0 is the end point of I , one-sided derivative is meant in the previous theorem.
The second statement of Theorem 2.28 immediately implies the next result.
Corollary 2.29 If function f is continuous on the interval I , then function F defined by Equality (2.16) is its
antiderivative.
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Definite Integral
In fact we proved Theorem 1.3 on the existence of an antiderivative to a continuous function. Notice that the
definite integral was used to construct an antiderivative, that is the indefinite integral. This result is existential, as
we mentioned after Theorem 1.3, and does not generally enable us to find an antiderivative. To evaluate the definite
integral we have only the Fundamental Theorem of Calculus. But its use assumes knowledge of an antiderivative
to the integrand, so we get into a vicious circle.
By analogy it is possible to define the definite integral as the function of the lower limit.:
Z c
G(x) =
f (t) d t,
x ∈ I.
x
Function G has properties similar to function F . Only at points of continuity of the integrand f do we have
G0 (x) = −f (x). Thus if the integrand f is continuous on I , the function −G is its antiderivative.
The definite integral, which depends on its upper or lower limits, will be necessary in the following Chapter 3,
which concerns the improper integral. But it has important applications in many parts of mathematics, such as
the theory of multidimensional integrals, which are the generalization of our definite integral for functions of
several variables. Some of you encounter it also in the study of the Laplace transform, where convolution is
introduced(see [11]). As well, it is used in the following generalization of the integration by parts method, which
is often needed in proofs from integral transforms theory.
Theorem 2.30 Let functions f and g be integrable on the interval ha, bi and A, B ∈ R be constants. Let us
denote
Z x
Z x
F (x) =
f (t) d t + A,
G(x) =
g(t) d t + B.
a
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Then
Z
a
b
b
F (x)g(x) d x = F (x)G(x) a −
Z
b
f (x)G(x) d x.
(2.17)
a
For the proof see [9, p. 195]. Formula (2.17) is really the generalization of the integration by parts method for the
definite integral, from Theorem 2.19. Keeping in mind the assumptions of this theorem and using its notation we
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Definite Integral
can set f (x) = u0 (x), g(x) = v 0 (x). If we choose A = u(a), B = v(a), using the Fundamental Theorem of
Calculus we have
Z x
x
F (x) =
u0 (t) d t + u(a) = u(t) a + u(a) = u(x) − u(a) + u(a) = u(x)
a
and analogously G(x) = v(x). Substituting into Equality (2.17) we have at once the well-known integration by
parts formula for the definite integral (2.10).
Comment 2.31 Besides the Riemann integral the Newton integral is introduced (see [21]). Its definition assumes
among other things the existence of an antiderivative. The Fundamental Theorem of Calculus then states that if a
function is both Riemann and Newton integrable, then the values of these two integrals are the same.
The integration by parts method from Theorem 2.30 concerns solely the Riemann integral while in the assumptions of Theorem 2.19 the existence of the Newton integral of some function is indirectly needed.
Exercises
1. Evaluate:
Z π/2
a)
sin x d x,
Z
0
Z
π
5 sin 4φ dφ,
d)
e)
0
Z
2 sin2 2x d x,
−π/2
Z 2
1
0.5
3
Z
m)
2
Z
cos x d x,
−π/2
Z 3
u
e 3 d u,
x2
Z
h)
Z
1
dV ,
V2 +V
(x 3 − 3x 2 + 1) d x,
−1
Z 3
f)
0
4
0
d x,
3
c)
0
π/2
g)
j)
π/2
b)
1
k)
−1
2
Z
n)
−2
Z
x−1
d x,
x+1
i)
2x 3 d x,
l)
6
d t,
8 + 3t 2
5
2
Z
1
−1
π
12
d z,
z2 + 9
4
d x,
x
1
d x,
x−4
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Z
2 cos y sin y dy.
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2. Evaluate:
Z π/4
a)
4 sin2 x d x,
−π/4
Z 4
√
3
d)
−1
Z 10
x d x,
Z
6
dp,
9p
f)
2 − V2
V e
4x 2 e−2x d x,
Z
tan β dβ,
−2
dV ,
c)
4. Evaluate:
Z 1 √
2 x
a)
d x,
1
+x
0
Z 2
2(1 + ln Q)
d)
dQ,
Q
1
Z π
p
g)
sin t 1 + cos2 t d t,
x
xe− 2 d x,
x 2 sin x d x,
0
1
1
Z
4y arctan 2y dy,
h)
0
π
f)
0
arctan w dw,
n2
dn.
n2 + 1
0
Z
x cos x d x,
Z
3
Z
π
e)
−1
Z 2
0.5
−0.5
Z 2
0
1
d)
0
1
b)
0
g)
c)
1
3. Evaluate:
Z π
a)
2ω sin2 ω dω,
Z
1
dα,
cos2 α
e)
1
Z
1
Z
b)
6 arcsin
i)
0
−1
Z
3
b)
2
Z
4
e)
2
Z
1
2
dx
h)
1
Z
eu
d u,
u2
√
S
√
dS,
S−1
x
p
1 − ln2 x
c)
f)
,
i)
t
d t.
2
3
3r
d r,
4r
+4
0
Z 4 r
1
12 x + d x,
4
0
Z 4
1
√ d x.
0 1+ x
√
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Definite Integral
5. Evaluate:
Z π
a)
8 cos2 φ sin2 φ dφ,
0
Z
√
12 s
0
π/2
g)
r
4s + 1
ds,
4s
e)
cos2 α sin 2α dα,
h)
0
π/2
Z
4 sin φ cos3 φ dφ,
i)
k)
3 sin3 x d x,
m)
0
1
p)
−1
2(1 + r 2 )
d r,
1 − r2
Z
2
d x,
x2 − 4
6. Evaluate:
Z 1
m−1
a)
dm,
m
+1
0
Z 9
3(z − 1)
d z,
d)
√
z+1
4
Z 2
2x
g)
d x,
4
0 1+x
Z π/2
10 dδ
j)
,
2
cos
δ+3
0
q)
Z
−1
Z 3
2
o)
1
2
T /2
e)
0
Z
h)
1
Z
k)
0
3
c)
0
Z
2
2πt
d t,
π sin
T
2
dK,
K + K3
2π
2 dw
,
5 + 3 cos w
5
r)
Z
(1 + tan 2β) dβ,
d x,
16
d x,
8 − 4x 2
2 dR
,
16 − 4R 2
√
3 x + 1 d x,
0
Z
π/8
b)
4 − x2
√
l)
1
dk,
3
0 (5 + 4k)
Z 2 e−t
et 1 +
d t,
t
1
n)
x
1
Z
0
1/2
1
Z
16 sin4 x d x,
0
π
Z
0
Z
√
f)
0
π/2
j)
Z
1 − cos β dβ,
6
d x,
6x − 1
1
Z
0
0
Z
c)
1
2π p
Z
2
Z
2(1 − cos α)3 dα,
0
1
d)
Z
π
Z
b)
Z
f)
i)
2 ln y
dy.
y
6p
dp,
p2 − 1
π/2
dε
,
1 + cos ε
0
Z 1 q
2 2p + p2 dp,
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Z
l)
0
π
2 sin ω
dω,
5 + 4 cos ω
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Definite Integral
Z
π/2
m)
0
π
Z
cos α
dα,
5 + sin α
n)
0
7. Find the generalized antiderivative and using it evaluate:
Z 2
Z 4
a)
sgn x d x,
b)
f (x) d x,
−1
where
2
(4 − y 2 ) dy.
−2
3
Z
2
o)
c)
−1


x
f (x) = x 2


1
Z
10
d t,
4 + cos2 t
Z
e−|x| d x,
5
g(x) d x,
d)
−2
−3


2 for x ∈ h−3, −1i,
g(x) = 4 for x ∈ (−1, 2),


1 for x ∈ h2, 5i.
for x ∈ h−1, 0i,
for x ∈ (0, 1),
for x ∈ h1, 4i,
Answers to Exercises
1. a)
1,
b)
2,
c)
−4,
d)
0,
f)
π,
g)
π,
h)
4 − 2 ln 5,
i)
4 ln
k)
2. a)
3. a)
d)
g)
0,
l)
−2 + π,
ln
3
,
5
b)
1 2
π ,
2
−5 e−2 + e2 ,
1
2 arctan 2 − ln 5,
2
2 tan 1,
m)
−3 ln 2 + 2 ln 3,
c)
0,
d)
√
6 arctan
n)
14,
5
,
2
e)
2
ln 10,
3
e)
3e − 3,
j)
1, 5,
o)
0.
√
6
,
2
f)
Contents
4 − 2 arctan 2.
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b)
−26 e−1/2 + 16,
c)
−10 e−3/2 + 4,
e)
−2,
f)
π2 − 4,
i)
√
π + 6 3 − 12.
h)
5 arctan 2 − 2,
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Definite Integral
4 − π,
b)
ln2 2 + 2 ln 2,
√
√ 2 + ln 1 + 2 ,
c)
4,
e)
−e1/3 + e1/2 ,
√
√
6 − 2 ln 2 − 1 − 2 2,
f)
√
17 17 − 1,
h)
arcsin ln 2,
i)
4 − 2 ln 3.
π,
√
5 5 − 1,
1
,
2
b)
c)
e)
5π,
√
4 2,
h)
3π,
i)
1,
k)
4,
l)
π
,
3
m)
−1 + 2 ln 3,
n)
18
,
4225
o)
14,
p)
− ln 3,
q)
e2 + ln 2 − e,
r)
ln2 5.
4. a)
d)
g)
5. a)
d)
g)
j)
c)
9 ln 2 − 3 ln 3,
e)
π 1
+ ln 2,
8
4
T,
f)
arctan 4,
√
√
5
4 5 arctan
,
5
h)
3 ln 2 − ln 5,
i)
1,
√
√ 2 3 − ln 2 + 3 ,
k)
π,
l)
ln 3,
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ln 6 − ln 5,
n)
√
5 π,
o)
512
.
15
J
1 − 2 ln 2,
b)
d)
23,
g)
6. a)
j)
m)
f)
11
,
5√
2 − 3,
√
√ 2 ln 3 + 2 2 ,
ln
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Definite Integral
7. a)
c)
a)
|x|,
(
ex
H (x) =
2 − e−x
1,
b)
for x 5 0,
d)
for x > 0,
b)
17
,
6
c)
2−e
F (x) =
 2
x


2
x3
3


x − 2
3


2x

G(x) = 4x + 2


x+8
−2
−3
−e
,
for x ∈ h−1, 0i,
for x ∈ (0, 1),
for x ∈ h1, 4i,
for x ∈ h−3, −1i,
for x ∈ (−1, 2),
for x ∈ h2, 5i,
d)
19.
2.5 Applications of the Definite Integral
In the final section of this chapter we present a few applications of the definite integral from geometry
and physics.
2.5.1 Geometric Applications
Consider the evaluation of lengths, areas and volumes. Each of you has some idea of what these concepts
mean for some simple shapes. For example, the length of a line segment or circle, the area of a square,
rectangle, trapezium or spherical surface, or the volume of a cuboid, cone or sphere, etc. Likewise you
have an intuitive idea of what is the length, for example, of a space spiral and are able to imagine how
it could be measured using a flexible measuring tape. Analogously you have an idea that an ellipse has
some area, although you probably do not know how to evaluate it. But if you were asked for example,
what is the length of the set of all rational numbers lying between zero and one, you might hesitate a lot
with the answer. The problem is that the concepts length, area and volume were not defined exactly.
With respect to the extent and determination of this text, it is not possible to define these concepts
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Definite Integral
exactly. This would represent a quite complicated and long exposition of measure theory and further
advanced mathematical topics. For our needs we will do without these exact mathematical definitions,
because we will restrict ourselves to simple objects, where it will be intuitively clear that they have a
length, area or volume. The formulae below will tell us how the respective values can be evaluated.
If A is a set, we denote the corresponding value m(A), where the letter m is used in place of the word
measure. But we must distinguish whether we have in mind length, area or volume. For this purpose we
use the subscript corresponding to the units in which the quantity is measured. Thus m1 (A) means the
length (measured in length units), m2 (A) means the area (measured in area units) and m3 (A) means the
volume (measured in volume units) of the set A. It is taken as given that the corresponding quantity has
a reasonable meaning for the given set A. Therefore, we will evaluate length for curves, area for surfaces
and volume for solids. But what a curve, surface and solid is will be understood only intuitively, we do
not have exact definitions.
Area of a Plane Figure
The evaluation of the area of a plane figure belongs to the most important applications of the definite
integral. By the way, it was used as the main motivation (see page 175).
Let f be a nonnegative function defined on a bounded closed interval ha, bi. The plane set (figure)
described by the relation
Contents
A = {(x, y) ∈ R2 | a 5 x 5 b, 0 5 y 5 f (x)}
is usually called subgraph of function f on the interval ha, bi. Thus the figure is bounded by the x -axis,
parallels with the y -axis having the equations x = a and x = b, and the graph of function f . The
function need not be continuous (see, Fig. 2.10 a), for example).
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Definite Integral
y
y = f (x)
y
x=a
x=b
x
y = f (x)
a
b
B
y = g(x)
A
−c
x
a
b
x=a
a)
y = −c
x=b
b)
Fig. 2.10: Evaluation of the area of a set
Theorem 2.32 Let f be a nonnegative function integrable on the interval ha, bi. Then the area of set A
is given by the formula
Z b
m2 (A) =
f (x) d x.
(2.18)
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a
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Let us emphasize that the function must be nonnegative
on the interval ha, bi. On the other hand,
Rb
it is evident that for function f to be nonpositive a f (x) d x equals the area of the set bounded by the
graph of function f , the x -axis and straight lines x = a and x = b (lying this time below the x -axis),
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Definite Integral
but multiplied by minus one. To prove this it is sufficient to replace function f by −f , which will be
nonnegative (set A is reflected about the x -axis) and take out the number −1.
The previous consideration and the additivity of
the definite integral with respect to the integration domain imply that in the general case,
R b when function f
y
can change the sign arbitrarily, a f (x) d x equals the
y = f (x)
area of the set bounded by the graph of f , the x -axis,
and the straight lines x = a and x = b, where the parts
+
+
+
being above the x -axis are taken with the positive sign
x
while the parts being below the x -axis are taken with
a
b
−
−
the negative sign (see Fig. 2.11).
Thus, for example, from theRshape of the graph of
2π
sine or cosine it is evident that 0 sin x d x = 0 and
R 2π
Fig. 2.11
cos x d x = 0 hold. Sketch the corresponding pic0
tures.
Theorem 2.32 can be easily generalized to the case of the set displayed in Fig. 2.10 b). Let us assume
that the graph of function f is above the graph of function g on the interval ha, bi (equality is allowed,
so g(x) 5 f (x) must hold for x ∈ ha, bi). We denote
Contents
B = {(x, y) ∈ R2 | a 5 x 5 b, g(x) 5 y 5 f (x)}.
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Hence B is the set bounded by the straight lines x = a and x = b and the pair of graphs of functions
f and g . Sometimes the name curvilinear rectangle or curvilinear trapezoid is used for such figures in
mathematical jargon.
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Definite Integral
Theorem 2.33 Let functions f and g be integrable on the interval ha, bi and g(x) 5 f (x) for any
x ∈ ha, bi. Then the area of set B is given by the formula
Z
m2 (B) =
b
[f (x) − g(x)] d x.
(2.19)
a
The correctness of the formula is almost clear. It is sufficient to move the set B up by a suitable
constant so that the function g + c (and, therefore, also the function f + c) is nonnegative on the interval
ha, bi. It can be done, because the function g is integrable, and thus lower bounded. The area does not
change. The straight line y = −c in Fig.. 2.10 b) plays the role of the new x -axis.
Now it is evident, that the moved set B is the set difference of the subgraph of f and the subgraph
of g
Rb
(hashed part). Thus its area is the difference of the areas of these subgraphs. So m2 (B) = a [f (x) +
Rb
Rb
+ c] d x − a [g(x) + c] d x = a [f (x) − g(x)] d x . Theorem 2.32 is the special case for g(x) = 0 on
ha, bi.
Example 2.34 Evaluate the area of set K bounded by the graphs of the functions g : y = x 2 + x − 3
and f : y = −x 2 − 2x + 2.
Solution. In examples of this type we usually need to sketch a picture. First we determine the limits. To
do so we need to find the intersection points of the graphs of the functions involved, that is we solve the
equation f (x) = g(x). In our case, we get x 2 + x − 3 = −x 2 − 2x + 2, so 2x 2 + 3x − 5 = 0 and
(
−3 ± 7
− 52 ,
x1,2 =
=
4
1.
As both functions are quadratic polynomials, their graphs are parabolas. The signs at x 2 tell us which
parabola is turned up and which one down. The result is displayed in Fig. 2.12. Thus f (x) = g(x) on
the interval h−5/2, 1i. Otherwise, we would have to change the roles of f and g in Formula (2.19).
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Definite Integral
y
f : y = −x 2 − 2x + 2
K
1
x
− 52
g : y = x2 + x − 3
Fig. 2.12
Thus the area of the set K is:
Z 1
m2 (K) =
(−x 2 − 2x + 2) − (x 2 + x − 3) d x =
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−5/2
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Z
1
2
3
(−2x 2 − 3x + 5) d x = − x 3 − x 2 + 5x
=
3
2
−5/2
2 3
250 75 25
343
= − − +5 −
−
−
=
.
3 2
24
8
2
24
1
J
=
−5/2
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Definite Integral
In elementary and secondary school you learned the formulae for the length of a circle (circumference), the area of a disc and sphere, and the volume of a ball. But you did not have the tools necessary
to prove these. In this section we will prove all these formulae. The first is the area of the disc.
Example 2.35 Evaluate the area of disc K with radius r > 0.
Solution. The centre of the disc can be placed at the origin, because
it has no impact on the area. Then
√
2
2
2
2
the equation of the boundary circle is x + y = r , so y = ± r − x 2 . Let us denote
p
p
f (x) = r 2 − x 2
and
g(x) = − r 2 − x 2 ,
x ∈ h−r, ri,
see Fig. 2.13. Thus the area of disc is
Z
r
m2 (K) =
=
[f (x) − g(x)] d x =
−r
Z r p
y
f:y=
p
r 2 − x2 − − r 2 − x2 dx =
√
r 2 − x2
−r
Z
r
=2
K
p
r 2 − x 2 d x.
−r
x
r
−r
We have already evaluated a similar indefinite integral (compare to Example 1.30). It matches Type (1.31). Therefore we use the substitution
method (again applied from right to left, see Formula (2.14)). We set
ϕ(t) = r sin t , thus x = r sin t . The function sin t is one-to-one mapping of the interval h−π/2, π/2i onto the interval h−1, 1i, so that
Contents
√
g : y = − r 2 − x2
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Fig. 2.13
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Definite Integral
function ϕ(t) maps the interval h−π/2, π/2i onto the interval h−r, ri. We obtain:
x = r sin t
Z rp
2
2
=
m2 (K) = 2
r − x d x = d x = r cos t d t
−r
−r ; − π , r ; π 2
2
Z π/2 p
Z π/2
=2
r 2 − r 2 sin2 t · r cos t d t = 2
r| cos t| · r cos t d t =
= 2r
−π/2
Z π/2
2
−π/2
−π/2
cos2 t d t = 2r 2
Z
π/2
−π/2
1
(1 + cos 2t) d t =
2
sin 2t π/2
2
=r t+
=
2 −π/2
π sin π
π sin(−π)
2
=r
+
− − +
= πr 2 .
2
2
2
2
We used the fact that cos t = 0 on the interval h−π/2, π/2i and applied Formula (1.19).
With respect to the symmetry it was possible to evaluate only the area of one quarter of the disc lying
in the first quadrant and multiply the result by four. Then the lower
R r √ function would be g(x) = 0 and the
N
domain of integration would be h0, ri, therefore m2 (K) = 4 0 r 2 − x 2 d x .
Comment 2.36 Sometimes the set C , the area of which is to be evaluated, is not bounded by two appropriate graphs of functions of the independent variable x and we cannot apply Formula (2.19). But it may
have the appropriate shape if we rotate the figure by 90◦ , that is interchange x and y . In other words,
there exist functions x = h(y), x = k(y), h(y) 5 k(y) for y ∈ hc, di, such that
C = {(x, y) ∈ R2 | c 5 y 5 d, h(y) 5 x 5 k(y)}.
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Definite Integral
Thus the set C is bounded by the graphs of the functions h(y) and k(y) and parallels with the x -axis
having the equations y = c and y = d (see Fig. 2.14). Then its area is:
Z
m2 (C) =
d
[k(y) − h(y)] dy.
(2.20)
c
In the case of even more complicated sets it is usually possible to break them into simpler pairwise
disjoint parts (more exactly, they overlap by boundary curves), the area of which can be evaluated using
Formula (2.19) or (2.20).
y
y=d
d
x = h(y)
C
x = k(y)
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y=c
c
x
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Fig. 2.14
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Definite Integral
Length of a Curve
Another important application of the definite integral is the evaluation of the length of the plane curve
(for the definition of the length see, for example [12, p. 22]). Initially we will restrict ourselves to the case
where the curve is represented by the graph of the function y = f (x).
Theorem 2.37 Let function f be defined on the interval ha, bi and have the continuous derivative there.
Then the length of its graph G is given by
b
Z
m1 (G) =
p
1 + [f 0 (x)]2 d x.
(2.21)
a
Example 2.38 Evaluate the length of graph G of the function f : y = ln x , where x ∈
√ √ 3, 15 .
Solution. Function f is differentiable and f 0 (x) = 1/x . Due to Formula (2.21) we have:
√
Z
m1 (G) =
√
3
15
r
1
1 + 2 dx =
x
√
15
Z
√
√
3
2
Z 4
x + 1 = t2
=
x
d
x
=
t
d
t
=√
√
2
3 ; 2, 15 ; 4 Z √15 √ 2
x2 + 1
x +1
dx = √
· x dx =
x
x2
3
√
Z 4
t2
t2
·
t
d
t
=
d t.
2
t2 − 1
2 t −1
√
2
2
x 2 + 1. This function is
We used the substitution
given
by
equality
x
+
1
=
t
,
that
is
t
=
√ √ increasing on the interval
3, 15 and maps it onto the interval h2, 4i. We obtained the definite
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Definite Integral
integral of the improper rational function. Dividing it we have
t2
t2 − 1 + 1
1
=
=1+ 2
.
t2 − 1
t2 − 1
t −1
We need to decompose the proper rational function that we obtained into partial fractions. The denominator has simple real roots ±1, so
t2
1
A
B
=
+
−1
t −1 t +1
⇒
1 = A(t + 1) + B(t − 1).
Substituting the roots we find the constants A and B :
t =1:
1 = 2A
⇒
t = −1 :
1 = −2B
⇒
1
,
2
1
B=− .
2
A=
Hence
Z 4
1/2
1/2
−
m1 (G) =
1+
dt =
t −1 t +1
2
4
1
1
= t + ln |t − 1| − ln |t + 1| =
2
2
2
1
1
1
1
= 4 + ln 3 − ln 5 − 2 + ln 1 − ln 3 =
2
2
2
2
= 2 + ln 3 −
1
ln 5.
2
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Definite Integral
Now we tackle a more general case, when the curve need not be the graph of a function. With respect
to the extent of this text we cannot introduce the concept of the curve. We recommend that readers
interested in this topic review, for example [12, p. 15]. For our purposes it will be sufficient to understand
that the curve is in fact the trajectory drawn by a point moving continuously on the plane through the time.
Therefore we must describe the position of the point on the plane at a given moment. This can be done
with the help of two continuous functions ϕ(t) and ψ(t), determining the x -coordinate and y -coordinate
of the moving point. This way we obtain the parametric equations of the curve having the form
x = ϕ(t),
y = ψ(t),
t ∈ hα, βi.
(2.22)
Variable t is called the parameter (it is not necessarily time it may, for example, represent length). The
special case (parametric equations of the segment) are well-known from analytic geometry. From a
physical point of view the length of the curve is the distance, which the point passes from moment α to
moment β . The following statement can be proved:
Theorem 2.39 Let curve C be described by the parametric equations (2.22) and functions ϕ and ψ
are continuously differentiable on the interval hα, βi. Then
Z
m1 (C) =
β
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 d t.
Contents
(2.23)
α
The graph of any function f (x), x ∈ ha, bi can be parameterized by the equations x = t , y = f (t),
t ∈ ha, bi, thus ϕ 0 (t) = 1, ψ 0 (t) = f 0 (t). Substituting into (2.23), we can see at once that it is really
the generalization of Formula (2.21).
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Definite Integral
Example 2.40 Evaluate the length of circle C with radius r > 0.
Solution. Without the loss of generality the circle can be
centred at the origin, this has no impact on the length. Its
y
equation is x 2 + y 2 = r 2 . Now we could find the function
r
corresponding
to the upper semicircle, which is f (x) =
√
C
A
= r 2 − x 2 , where x ∈ h−r, ri, then using Formula (2.21)
evaluate its length and multiply the result by two. The probr
r sin t
lem is that the derivative of this function is f 0 (x)0 = √ −x
,
2
2
t
r −x
hence it is not defined for x = −r and x = r (only one-sided
r
x
−r
O r cos t
improper derivatives exist at these points). Thus the assumptionsp
of Theorem 2.37 are not satisfied. Moreover, the function 1 + y 02 = √ 2r 2 is not bounded on the interval
r −x
(−r, r).
−r
Therefore, we find the parametric equations of circle C ,
which is easy. From the definition of functions sine and coFig. 2.15
sine (see Fig. 2.15) we know that the position of any point
A = (x, y) on circle C is A = (r cos t, r sin t), where t is
the angle that the radius vector of A makes with the positive part of the x -axis. If we change the angle t
from zero to 2π, then point A passes the whole circle C. Let us denote ϕ(t) = r cos t , ψ(t) = r sin t .
The parametric equations are
C:
x = r cos t,
y = r sin t,
t ∈ h0, 2πi.
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Definite Integral
As ϕ 0 (t) = (r cos t)0 = −r sin t and ψ 0 (t) = (r sin t)0 = r cos t , we have from Formula (2.23):
Z
2π
m1 (C) =
Z
p
2
2
(−r sin t) + (r cos t) d t =
0
Z
=
0
2π
p
r 2 sin2 t + r 2 cos2 t d t =
0
2π
2π
r d t = r t 0 = 2πr.
N
Comment 2.41
1. Analogously, it is possible to introduce the space curve, by simply appending a third coordinate. The
parametric equations of curve K are
x = ϕ(t),
y = ψ(t),
t ∈ hα, βi,
z = ω(t),
and its length (assuming the existence of continuous derivatives ϕ 0 , ψ 0 and ω0 ) is
Z
m1 (K) =
β
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 + [ω0 (t)]2 d t.
α
Contents
2. Considering the physical interpretation, where Equations (2.22) describe the position of the mass
point, the pair (ϕ 0 (t), ψ 0 (t)) is the vector of the instantaneous velocity at moment
p t . From analytic
geometry it is known that the size of this vector, which is the speed, equals [ϕ 0 (t)]2 + [ψ 0 (t)]2 .
Thus Formula (2.23) expresses the fact that the definite integral of speed over the interval hα, βi
equals the distance that this point passes from moment α to moment β . The same is true for space
curves.
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Definite Integral
3. The integral used to evaluate the length of the curve involves the square root. Therefore it is often that
case, that even with very simple functions the result cannot be expressed using elementary functions
and some numerical method must be used (see Chapter 4). This is the case with the ellipse, where it
is possible to prove that its length is expressed by the elliptic integral (see page 154).
Solid of Revolution: Volume and the Area of its Lateral Surface
Let us consider a continuous nonnegative function f defined on the interval ha, bi. This function determines the figure (subgraph of function f )
P = {(x, y) ∈ R2 | a 5 x 5 b, 0 5 y 5 f (x)}.
(2.24)
Revolving P about the x -axis we obtain the solid of revolution V (Fig. 2.16). The surface of V consists
of lateral surface Q and two side discs. Our goal is to evaluate the volume of V and the area of Q.
The first statement concerns the volume of the solid of revolution.
Theorem 2.42 Let function f be continuous and nonnegative on the interval ha, bi. The volume of the
solid of revolution V obtained by revolving figure P described in (2.24) is given by the formula
m3 (V ) = π
Z
b
f 2 (x) d x.
(2.25)
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a
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Regarding the area of the lateral surface, the continuity of function f is not sufficient, the assumptions
must be strengthened to guarantee the existence of the two-dimensional measure of the lateral surface.
The next statement holds.
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Definite Integral
y
y = f (x)
P
a
b
x
O
z
Fig. 2.16: Solid of revolution
Theorem 2.43 Let function f be nonnegative and have the continuous derivative on the interval ha, bi.
The area of the lateral surface Q of the solid of revolution V obtained by revolving figure P described
in (2.24) is given by the formula
Z
m2 (Q) = 2π
b
p
f (x) 1 + [f 0 (x)]2 d x.
(2.26)
a
Notice that the previous formula for the area of the lateral surface involves the square root of the same
expression as Formula (2.21) for the evaluation of the length of the curve.
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Definite Integral
If we want to evaluate the area of the surface, it is sufficient to add to the area of the lateral surface
the area of both side “lids”, which are the discs having the radii f (a) and f (b).
Now we demonstrate the use of both formulae. While the volume can be evaluated for quite complicated functions determining the figure P , in the case of the area of the lateral surface we can encounter
problems with the integration of the expression involving the root even for very simple functions (the
situation is similar to that of the length of the curve, compare with Comment 2.41). Then the result may
only be approximated (see Chapter 4).
Comment 2.44 In the following examples a simple but very useful property will come in handy, that
is valid for the definite
R rintegrals of even
R r functions. If the function f is even on the symmetric interval
h−r, ri, r > 0, then −r f (t) d t = 2 0 f (t) d t . The statement is evident—the right and left halves of
the graph are symmetric about the y -axis. The exact proof is performed by separating into two integrals
and the substitution in the first one:
t = −s
Z r
Z 0
Z r
=
f (t) d t =
f (t) d t +
f (t) d t = d t = −d s
−r
−r
0
−r ; r, 0 ; 0 Z 0
Z r
=−
f (−s) d s +
f (t) d t =
r
0
Z r
Z r
Z r
=
f (s) d s +
f (t) d t = 2
f (t) d t,
0
0
0
because f (−s) = f (s) and the notation of the
R r independent variable is not important.
Analogously, for odd functions we have −r f (t) d t = 0. Geometrically, the right half of the graph
is obtained from the left one by mirroring first about the x -axis and then about the y -axis.
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Definite Integral
y
y =1+
1
2
sin 3x
P
a
b
x
O
z
Fig. 2.17
Example 2.45 Evaluate the volume of the solid of
V which is obtained by revolving the
revolution
subgraph P of function f (x) = 1 + 12 sin 3x , x ∈ π3 , 13π
, about the x -axis.
6
. Due to Formula (2.25) we have
Solution. The solid V is shown in Fig. 2.17, where a = π3 and b = 13π
6
for the volume:
2
Z 13π/6 1
m3 (V ) = π
1 + sin 3x d x =
2
π/3
Z 13π/6 1 2
=π
1 + sin 3x + sin 3x d x =
4
π/3
Z 13π/6 1
=π
1 + sin 3x + (1 − cos 6x) d x =
8
π/3
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Definite Integral
13π/6
1
1
1
=
= π x − cos 3x + x −
sin 6x
3
8
48
π/3
13π 1
13π 13π
1
=π
− cos
+
−
sin 13π −
6
3
2
48
48
π 1
π
1
−π
− cos π +
−
sin 2π =
3
3
24 48
13π 13π π 1
π
33π2 π
=π
+
− − −
=
− .
6
48
3
3 24
16
3
In the evaluation, the formula sin2 3x = 21 (1 − cos 6x) was used.
N
Example 2.46 Evaluate the volume of the solid of revolution V and the area of its lateral surface Q.
The solid is obtained by revolving the subgraph P of function f (x) = 2| sin x|, x ∈ h0, 2πi, about the
x -axis.
Solution. The solid V is shown in Fig. 2.18. With respect to the shape of the function sine, it is evident
that it is sufficient to consider the interval h0, πi, where sin x = 0, that is | sin x| = sin x , and multiply
the result by two.
Given Formula (2.25), for the volume of the solid V , we have:
Z π
Z π
1
m3 (V ) = 2 · π
4 sin2 x d x = 8π
(1 − cos 2x) d x =
0
0 2
π
1
= 4π x − sin 2x = 4π(π − 0) − 4π(0 − 0) = 4π2 .
2
0
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Definite Integral
y
y = 2| sin x|
P
0
2π
x
z
Fig. 2.18
Further, using Formula (2.26), we have for the area of the lateral surface Q:
2 cos x = t
Z π
p
−2 sin x d x = d t
m2 (Q) = 2 · 2π
2 sin x 1 + 4 cos2 x d x = 0
2 sin x d x = −d t
0 ; 2, π ; −2
Z
= −4π
2
−2
Z
p
2
1 + t d t = 8π
2
=
p
1 + t 2 d t.
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0
In the previous evaluation, the integral obtained after the substitution was rearranged. The integration
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Definite Integral
√
limits were interchanged, which caused the change of the sign. As function 1 + t 2 is even on the interval
h−2, 2i, given Comment 2.44 it is possible to integrate over the interval h0, 2i and multiply the result by
two.
Evaluating the resulting integral we can use the same substitution as in Example 2.24 c. To present
a different approach, we separate it into two integrals and use the integration by parts to the second one
(compare with the evaluation of the indefinite integral of a similar integrand in Example 1.15). Using
Formula 11 from Table 1.1 we have
Z 2p
Z 2
Z 2
Z 2
1
t
1 + t2
2
√
√
dt =
dt +
t·√
dt =
1 + t dt =
2
2
1+t
1+t
1 + t2
0
0
0
0
u=t
u0 = 1
√
= 0 √t
2 =
v
=
v
=
1
+
t
2
1+t
h i2 Z
p
i2 h p
= ln t + 1 + t 2 + t 1 + t 2 −
0
= ln 2 +
0
√ √
5 − ln 1 + 2 5 − 0 −
Z
2
p
1 + t2 dt =
0
2
p
1 + t 2 d t.
0
Contents
From this equation we find:
Z
2
0
2
Z
p
√ √
2
1 + t d t = ln 2 + 5 + 2 5 ⇒
0
2
p
√ √
1
1 + t 2 d t = ln 2 + 5 + 5.
2
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Definite Integral
Thus the complete result is
Z
m2 (Q) = 8π
2
p
√ √
1 + t 2 d t = 4π ln 2 + 5 + 8π 5.
0
N
At the end we prove the last two promised formulae—the volume of the ball and the area of the
sphere.
Example 2.47 Evaluate the volume of the ball and the area of the sphere having radius r > 0.
Solution. The equation of the circle centred at the origin and having the radius r is x 2 + y 2 = r 2 .
Revolving the upper
√ semicircle P about the x -axis we obtain a ball (see Fig. 2.19). The equation of the
upper arc is y = r 2 − x 2 , x ∈ h−r, ri.
y
y=
√
r 2 − x2
P
−r
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r
x
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Fig. 2.19
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244
In virtue of Formula (2.25) the volume of ball V is:
Z r
Z r p
2
1 3 r
2
2
2
2
2
=
(r − x ) d x = π r x − x
m3 (V ) = π
r − x dx = π
3
−r
−r
−r
1 3
1 3
4
3
3
= π r − r − π −r + r = πr 3 .
3
3
3
Before the next evaluation we prepare the expression 1 + y 02 :
y0 =
1 2
−x
x2
r2
(r − x 2 )−1/2 (−2x) = √
⇒ 1 + y 02 = 1 + 2
=
.
2
r − x2
r 2 − x2
r 2 − x2
Now, in virtue of Formula (2.26) the area of the lateral surface Q, thus the area of the sphere is
Z rp
Z r
r
r
2
2
m2 (Q) = 2π
r −x · √
d x = 2πr
d x = 2πr x −r = 4πr 2 .
r 2 − x2
−r
−r
The previous evaluation was not quite correct. The derivative y 0 is not defined for ±r (one-sided improper
derivatives exist at these points). Cancelling the integrand we obtained the constant 1, defined also at
these points, nevertheless the assumptions of Theorem 2.43 were not satisfied. But we can evaluate the
integral on the interval h−r + δ, r − δi, where δ > 0 is a small number (in fact we cut off two small
spherical caps). Its value is 4πr(r − δ). Passing to the limit for δ → 0+ we come to the same result.
But the above evaluation is sufficient for our purposes.
N
Here are some possible generalizations based on the previous results.
1. Let 0 5 g(x) 5 f (x), x ∈ ha, bi. Consider the plane figure determined by the two functions f and g , lying
above the x -axis (compare Fig. 2.10 a)). Revolving it about the x -axis we obtain the ring solid V , having the
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Definite Integral
lateral surface Q. Let us denote Vf , Vg the solid of revolution determined by the subgraph of the functions
f , g and Qf , Qg its lateral surface, respectively. Imagine, for example, the inner tube of a tyre, which can be
obtained by revolving the circle; f corresponds to the upper semicircle and g to the lower semicircle. We have
m3 (V ) = m3 (Vf ) − m3 (Vg ),
m2 (Q) = m2 (Qf ) + m2 (Qg ).
If we want to find the area of the whole surface of solid V, we must add the area of the two side circular rings to
the area of the lateral surface.
2. It is quite common to have a situation where the graph of a nonnegative function f (x), x ∈ ha, bi, is given by
the parametric equations x = ϕ(t), y = ψ(t), t ∈ hα, βi. We will explain how Formulae (2.18), (2.25) and
(2.26) can be modified for this situation.
We will assume that the function ψ is continuous and nonnegative and the function ϕ is strictly monotonic
and continuously differentiable on hα, βi. Then the function ϕ has the inverse t = ϕ −1 (x), x ∈ ha, bi.
Excluding parameter t we obtain the explicit formula of the function f (x) = ψ[ϕ −1 (x)]. Then we use the
substitution x = ϕ(t), d x = ϕ 0 (t) d t in the mentioned formulae (in the case of the last formula it is necessary
to assume that, moreover, ϕ 0 (t) 6 = 0 and ψ is continuously differentiable, because in the evaluation of f 0 the
formula for the derivative of the inverse function must be used).
After some manipulations we gain the following generalizations of the formulae for the evaluation of the area
of the subgraph P of function f , the volume of the solid of revolution V and the area of its lateral surface Q,
where the solid V is obtained by revolving subgraph P about the x -axis:
Z β
m2 (P ) =
ψ(t) · |ϕ 0 (t)| d t,
Contents
α
m3 (V ) = π
Z
β
ψ 2 (t) · |ϕ 0 (t)| d t,
α
Z
m2 (Q) = 2π
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β
ψ(t)
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 d t.
α
Using these formulae we evaluate once more the area of the disc, the volume of the ball and the area of the
sphere. We will see that the evaluation is faster. The upper semicircle of the disc K with the centre at the origin
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Definite Integral
and the radius r > 0 (see Example 2.40) has the parametric equations x = r cos t , y = r sin t , t ∈ h0, πi.
Subgraph P is the upper semidisc. As sin t = 0 on the interval h0, πi we have |ϕ 0 (t)| = |−r sin t| = r sin t .
Therefore,
Z π
Z π
2
m2 (K) = 2 m2 (P ) = 2
r sin t · r sin t d t = 2r
sin2 t d t =
0
0
π
1
2
2
(1 − cos 2t) d t = r t − sin 2t = πr 2 ,
=r
2
0
0
Z π
Z π
m3 (V ) = π
r 2 sin2 t · r sin t d t = πr 3
sin3 t d t =
π
Z
0
0
cos t = u
Z π
− sin t d t = d u = πr 3
(1 − cos2 t) sin t d t = =
0
sin t d t = −d u 0 ; 1, π ; −1 3
= −πr
Z
1
Z
m2 (Q) = 2π
−1
4
1 3 1
= πr 3 ,
(1 − u ) d u = πr u − u
3
3
−1
π
r sin t
2
p
3
(−r sin t)2 + (r cos t)2 d t = 2πr 2
0
π
= 2πr 2 − cos t 0 = 4πr 2 .
Z
π
sin t d t =
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Definite Integral
Exercises
1. Find the area of a plane figure bounded by given curves:
a)
y = 0, x = −1, y = x 2 ,
b)
y = ex , y = e−x , x = 1,
c)
y = 4 − x 2 , y = 0,
d)
yx = 1, x = 1, x = 3, y = 0
e)
y 2 = 2x + 1, x − y − 1 = 0,
f)
y(1 + x 2 ) = 1, y =
g)
y = ln x, x = 5, x = 7, y = 0,
h)
y = | log x|, x =
i)
y = −x 2 + 4x − 2, x + y = 2,
j)
k)
y = x 3 − 4x 2 − x + 4, x = −1, x = 2, y = 0,
l)
m)
y = ln x, y = ln 9, y = ln 3, x = 0,
n)
x2
,
2
1
, x = 10, y = 0,
10
y = arcsin x, x = 0, x = 1,
4
x = , y = 1, y = 4, x = 0,
y
y = x sin x, x ∈ hkπ, (k + 1)πi, y = 0.
2. Find the area of a plane figure bounded by given curves:
a)
y = 1 − x, y 2 + x 2 = 1, 0 5 x, y > 0,
b)
x 2 = y, y 2 = x,
c)
y = x 2 − x − 6, y = −x 2 + 5x + 14,
d)
yx = 4, x + y = 5,
e)
sin x, x ∈ h0, πi,
1
y = | ln x|, x = , x = e2 , y = 0,
e
3
2
y = x + x − 6x, y = 0, x ∈ h−3, 3i,
f)
y = ln2 x, y = ln x,
2
, y = x2,
y=
1 + x2
4x 2 + 9y 2 = 36,
g)
i)
y = 0, y = e
−x
h)
j)
l)
y = 6x − x , y = 0,
m)
x 2 − 10x + 34
10 − 3x 2 + 18x
y=
, y=
,
5
5
x 2 + y 2 = 16, y 2 = 6x, x = 0,
n)
y = x 2 + 4x, y = x + 4,
o)
y 2 = 2x + 1, x − y − 1 = 0,
p)
y 2 = x 3 , y = 8.
k)
2
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Definite Integral
3. Find the length of the arc of a plane curve:
c)
5(ex/5 + e−x/5 )
, x ∈ h0, 10i,
2
p
√
y = x − x 2 − arcsin x, x ∈ h0, 1i,
e)
x = a(t − sin t), y = a(1 − cos t), a > 0, (cycloid),
f)
x = r(cos t + t sin t), y = r(sin t − t cos t), r > 0, t ∈ h0, πi,
g)
k)
x = a cos3 t, y = a sin3 t, a > 0, (astroid),
h) y 2 = (x + 1)3 , x 5 4, (semicubical parabola),
x
π π
e +1
, x ∈ h1, 2i,
j) y = ln sin x, x ∈
,
,
y = ln x
e −1
3 2
x = 2a(1 + cos t) cos t, y = 2a(1 + cos t) sin t, t ∈ h0, 2πi, a > 0, (cardioid),
l)
x=
a)
i)
y=
b)
y 3 = x 2 , x ∈ h0, 1i,
d)
y = arcsin e−x , x ∈ h0, 1i,
t4
t6
, y = 2 − , between intersection points with the coordinate axes.
6
4
4. Find the length of the arc of a space curve:
a)
b)
d)
x = a cos t, y = a sin t, z = bt, t ∈ h0, 2πi, a, b > 0, (one turn of the helix),
1p 3
1
t
x = t, y =
8t , z = t 2 , t ∈ h0, 1i, c) x = t − sin t, y = 1 − cos t, z = 4 sin , t ∈ h0, πi,
3
2
2
√
x = et , y = e−t , z = t 2, t ∈ h0, 1i.
5. Find the volume of a solid obtained by revolving the subgraph of the function k or the figure P about the x -axis:
a x/a
a) k : y =
e + e−x/a , a > 0, y = 0, x ∈ h−4, 4i, (revolution of the catenary),
2
b) P : xy = 4, x = 1, x = 4, y = 0,
c) P : y = −x 2 + 1, y = −2x 2 + 2,
d)
P : b2 x 2 + a 2 y 2 = a 2 b2 , a, b > 0, y = 0,
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e)
k : x = a(t − sin t), y = a(1 − cos t), a > 0, t ∈ h0, 2πi, (cycloid),
f)
P : x 2/3 + y 2/3 = a 2/3 , y = 0, (astroid),
i)
1
, x = −1, x = 1,
1 + x2
k : x = t 2 − 1, y = t − t 3 , t ∈ h0, 1i,
k)
k : x 2 + y 2 = 25, y = 0,
g)
k: y =
h)
P : y 2 = 5x, x = 8,
j)
k : y = sin x, x ∈ h0, πi,
l)
P : y 2 = x, y = x 2 , y = 0.
6. Find the area of the lateral surface of a solid obtained by revolving the subgraph of function k or figure P about
the x -axis:
a)
P : y 2 = 4ax, y = 0, x = 3a, a > 0,
b)
P : y 2 = x, y = x 3 ,
c)
k : y = 4 + x, x ∈ h−4, 2i,
d)
k: y =
e)
P : (y − 1)2 + x 2 = 1, (torus),
f)
P : 9ay 2 = x(3a − x)2 , a > 0, y = 0, between intersection points with the x -axis.
1 x
(e + e−x ), x ∈ h0, 1i,
2
7. Find the area of the lateral surface of the following solids of revolution:
a)
a right-circular cylinder with base radius r > 0 and height v > 0,
b)
a right-circular cone with base radius r > 0 and height v > 0,
c)
a frustum of a right-circular cone with radii of the bases r1 > r2 > 0 and height h > 0,
d)
a segment of a sphere with radius r > 0 having height v > 0, 0 < v < 2r ,
e)
a hollow right-circular cylinder with outer and inner radii r1 , r2 , r1 > r2 > 0, and height v > 0,
f)
a torus (obtained by revolving the disc with the radius r and the centre [0, R], where R > r > 0, about
the x -axis).
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Definite Integral
Answers to Exercises
1. a)
e)
i)
m)
1
,
3
16
,
3
9
,
2
6,
b)
f)
j)
n)
1
− 2,
e
π 1
− ,
2
3
π
− 1,
2
e+
c)
32
,
3
d)
ln 3,
g)
7 ln 7 − 5 ln 5 − 2,
h)
9.9 ln 10 − 8.1
,
ln 10
k)
9
,
4
l)
8 ln 2,
(2k + 1)π.
π−2
,
4
b)
1
,
3
c)
343
,
3
e)
1 + e−π
,
2
f)
3 − e,
g)
2−
i)
18,
j)
6π,
k)
4 √
3 + 4π ,
3
n)
2. a)
m)
5 2
(e − e−2 ) ,
2
b)
e)
8a,
f)
i)
ln
3. a)
4. a)
e2 + 1
,
e
p
2π a 2 + b2 ,
j)
125
,
6
√
8 13 13
−1 ,
27
8
2
π r
,
2
1
ln 3,
2
b)
3
,
2
o)
c)
2
+ e2 ,
e
50
,
3
16
,
3
c)
2,
d)
g)
6a,
h)
k)
16a,
l)
2π,
d)
d)
15
− 8 ln 2,
2
h)
π−
l)
36,
p)
19.2.
ln e +
670
,
27
13
.
3
e − e−1 .
2
,
3
p
e2 − 1 ,
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Definite Integral
πa 3 8/a
e − e−8/a + 4πa 2 ,
4
b)
12π,
c)
e)
5π2 a 3 ,
f)
32a 3
π,
105
g)
i)
π
,
12
j)
π2
,
2
k)
5. a)
6. a)
d)
7. a)
b)
c)
d)
56πa 2
,
3
π 2
(e − e−2 + 4),
4
16
π,
15
π
(π + 2),
4
500
π,
3
d)
4
πab2 ,
3
h)
160π,
l)
3
π.
10
b)
√
√
π
20 10 + 45 5 − 11 ,
54
c)
√
36 2π ,
e)
4π2 ,
f)
3πa 2 .
f (x) = r , x ∈ h0, vi, m2 (Q) = 2πrv , m3 (V ) = πr 2 v ,
p
p
r
1
f (x) = x , x ∈ h0, vi, m2 (Q) = πr r 2 + v 2 = πrs , where s = r 2 + v 2 , m3 (V ) = πr 2 v ,
v
3
p
r1 − r2
f (x) =
x + r2 , x ∈ h0, vi, m2 (Q) = π(r1 + r2 )s , where s = v 2 − (r1 − r2 )2 ,
v
πv 2
m3 (V ) =
(r + r1 r2 + r22 ),
3 1
p
1
f (x) = r 2 − v 2 , x ∈ hr − v, ri, m2 (Q) = 2πrv , m3 (V ) = πv 2 (3r − v),
3
e)
f (x) = r1 , g(x) = r2 , x ∈ h0, vi, m2 (Q) = 2πv(r1 + r2 ), m3 (V ) = πv(r12 − r22 ),
f)
f (x) = R +
p
r 2 − x 2 , g(x) = R −
p
r 2 − x 2 , m2 (Q) = 4π2 rR , m3 (V ) = 2π2 Rr 2 .
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For a better sense of the shapes we are discussing pictures of some curves and of one surface, which appear in
the previous exercises but are not known from secondary school, are shown on the following pages (Fig. 2.20 and
2.21). The notation in the figures corresponds to the equations in exercises.
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Definite Integral
y
y
a
2a
x
0
x
−aa
2πa
O
a
−a
a) cycloid
b) astroid
y
y
2a
x
O
x
4a
Contents
−11
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−2a
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c) cardioid
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253
Definite Integral
R
z
z 0
2πb
x
a
−R
−R
y 0
y
R
a) helix
r
−r 0 x
b) torus
Fig. 2.21
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254
Quiz—Length of a Curve
Choose the correct answer (only one is correct).
1. (1 pt.) Find the length of the arc of the semicubical parabola y 2 = x 3 between the points [1, 1] and
[4, 8].
34
10
√ √
√ 1
1 √
9 73 80 10 − 13 13
2 73 + ln1 +
27
2
4
2
1
2. (1 pt.) Find the length of graph of the function f : y = x 2 − ln x between the points [1, 1] and
8
[3, f (3)].
1
2
ln 3
√ + 8 + ln 81
8−
8
8
2
ln 3
8
8
3. (1 pt.) Find the length of graph of the function f : y = 1 + 6x 3/2 , x ∈ h0, 1i.
√
2
7
82 82 − 1
243
8+
41
2
81
243
4. (1 pt.) Find the length of graph of the function f : y = 2 − 3x , x ∈ h−2, 1i.
√
√
√
9
30
10 3
3 10
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Definite Integral
5. (1 pt.) Find the length of graph of the function f : y =
π
π
2
6
6. (1 pt.) Find the length of graph of the function f : y =
8+
ln 4
4
p
4 − x 2 , x ∈ h0, 2i.
π
3
x 2 ln x
−
, x ∈ h2, 4i.
2
4
12
ln 2
31
4
16
7. (1 pt.) Find the length of graph of the function f : y = x 2 , x ∈ h0, 3i.
√
√ 3 37 1
16
+ ln 6 + 37
2
4
√
√ 1
6
3 37 + ln 6 + 37
4
8. (1 pt.) Find the length of graph of the function f : y = ln | sin x|, x ∈ hπ/3, 2π/3i.
1
ln 3
ln 3 −
2
√
2 3
π
6+
Contents
Correct Answers:
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256
Quiz—Area of a Plane Figure
Choose the correct answer (only one is correct).
1. (1 pt.) Find the area of a plane figure bounded by the curves y = x 2 and y = 4x , for x between 0
and 1.
3
5
5
1
−
5
3
3
3
3
2. (1 pt.) Find
√ the area of a plane figure lying in the first quadrant and bounded by the curves y = x
and y = x .
5
5
1
1
+ C, C ∈ R
4
12
12
12
3. (1 pt.) Find the area of a plane figure lying in the first quadrant and bounded by the curves y = sin x ,
y = cos x , and the y -axis.
√
√
√
2−1
2− 2
2−2
0
4. (1 pt.) Find the area of a plane figure bounded by the curves y = x 2 − 2, and y = 2.
4
8
16
32
−
3
3
3
3
5. (1 pt.) Find the area of a plane figure bounded by the curves y = x , y = 1/x 2 , and x = 2.
1
1
0
2
2
6. (1 pt.) Find the area of a plane figure bounded by the curves x = 4 − y 2 , and x = 0.
64
16
32
256
3
3
3
3
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Definite Integral
7. (1 pt.) Find the area of a plane figure bounded by the curves y = sin x and y = x , for x between 0
and π/4.
√
1 π2
π2
2
+
+
−1
2 32
32
2
√
√
π2
2
2
1−
−
+1
32
2
2
8. (1 pt.) Find the area of a plane figure bounded by the curves y = x 2 , y = x 3 , and x = 3.
135
40
12
136
12
137
12
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Definite Integral
Quiz—Volume of a Solid of Revolution
Find the volume of solid obtained by revolving the figure bounded by given curves and lines about the
x -axis. Choose the correct answer (only one is correct).
1. (1 pt.) y = 3x − 1 and 5 5 x 5 8
111 2
111
π
π
2
2
2. (1 pt.) y = x , x = 2, and y = 0
4π2
2π
3. (1 pt.) y = 2x 2 , x = 1, and y = 0
4π
2π
5
3
4. (1 pt.) y = x , y = 2x , x = 2, and x = 5
21π
123π2
2
5. (1 pt.) xy = 2, x = 1, x = 2, and y = 0
14π
7π
3
√
6. (1 pt.) y = x , y = 2 − x , and y = 0
5π
5
6
6
1 047π
1 047
8π
3
π
2
2π
5
π
√
3 2
125π
3
117π
2π
8π2
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5π2
6
16π
+4
3
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Definite Integral
7. (1 pt.) y = cos x , x = 0, x =
π
, and x -axis
2
π2
4
π
4
π
1
π
1
π2
4
π arctan
π
1
, x = 0, x = , and y = 0
8. (1 pt.) y = √
2
4
x +1
π
4
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Definite Integral
2.5.2 Physical Applications
As mentioned at the beginning of this chapter it is possible to give hundreds of examples, when using
the definite integral some extensive quantities are evaluated of intensive quantities (for example, mass of
density). In the general case, when intensive quantities depend on two or three coordinates, a double or
triple integral is necessary, which you will study later. Therefore, we only restrict to simple examples
from mechanics and explain how the mass and coordinates of the centre of mass can be evaluated in
special cases.
Likewise we cannot define the needed physical quantities, it is the task of other subjects. We will work
with the quantities and concepts which are well-known to students from the secondary school. Further
examples of the use of the definite integral will be presented in other subjects.
Mass and Coordinates of the Center of Mass of a Plane Curve
Imagine that we have a piece of generally nonhomogeneous wire. Our goal is to evaluate its mass and
the coordinates of the centre of mass. The mathematical model of the wire is a curve. We assume that
a nonnegative function ρ is defined on the curve assigning to each point the linear density at this point.
Let us denote T = [ξ, η] the centre of mass of this curve.
First we tackle the case, where the curve is described by parametric equations:
C:
x = ϕ(t),
y = ψ(t),
Contents
t ∈ hα, βi.
The function ρ(t) is the linear density at point [ϕ(t), ψ(t)] of curve C .
(2.27)
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261
Definite Integral
Theorem 2.48 Suppose that the functions ϕ and ψ have continuous derivatives on the interval hα, βi
and the function ρ is continuous and nonnegative on this interval.
Then the curve C having the parametric equations (2.27) and linear density ρ(t) has the mass
β
Z
M(C) =
ρ(t)
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 d t.
(2.28)
α
The coordinates of its centre of mass are
Sy (C) Sx (C)
T =
,
,
M(C) M(C)
(2.29)
where
Z
β
Sx (C) =
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 d t,
(2.30)
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 d t.
(2.31)
ψ(t)ρ(t)
α
Z
Sy (C) =
β
ϕ(t)ρ(t)
α
In technical mechanics and statics the quantities Sx (C) and Sy (C) are sometimes called the system
moment of curve C with respect to the x and y -axis, respectively.
If we denote 1t = ti −
pti−1 the interval used in the construction of the definite integral, this construction
where ti−1 5 t 5 ti , expresses approximately the length
implies that the expression [ϕ 0 (t)]2 + [ψ 0 (t)]2 1t , p
of a small piece of the curve. Thus the expression ρ(t) [ϕ 0 (t)]2 + [ψ 0 (t)]2 1t involved in the integral for mass
is the product
p of density and length, and therefore approximates the mass of this piece. Therefore, the expression
ψ(t)ρ(t) [ϕ 0 (t)]2 + [ψ 0 (t)]2 1t involved in the formula for Sx (C) is the product of the mass of this piece and
its distance from the x -axis (ψ(t) is the y -coordinate of a point of the curve, that is the directed distance of this
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262
Definite Integral
point from the x -axis). This explains why the quantity Sx (C) is called the system moment with respect to the
x -axis. The analogous conclusion holds for the formula for Sy (C) (ϕ(t) is the x -coordinate of a point of the curve,
that is the directed distance of this point from the y -axis). See page 266 for similar considerations and further
details.
In the special case where curve C is the graph of a function f (x) and ρ(x) is its linear density at
point [x, f (x)], the previous result can be simplified as follows:
Corollary 2.49 Suppose that the function f has the continuous derivative on the interval ha, bi and the
function ρ is continuous and nonnegative on this interval.
Then the coordinates of the centre of mass of graph G of function f having the linear density ρ are
given by Formula (2.29), where
Z
b
M(G) =
ρ(x)
p
1 + [f 0 (x)]2 d x,
(2.32)
a
Z
b
Sx (G) =
f (x)ρ(x)
p
1 + [f 0 (x)]2 d x,
(2.33)
a
Z
Sy (G) =
b
xρ(x)
p
1 + [f 0 (x)]2 d x.
(2.34)
Contents
a
Example 2.50 Evaluate the mass and coordinates of the centre of mass of the homogeneous upper semicircle K : x 2 + y 2 = r 2 , y = 0, r > 0.
Solution. Parametric equations of the semicircle K are x = r cos t , y = r sin t , t ∈ h0, πi (see Example 2.40). Since the curve is homogeneous the density is constant, that is ρ(t) = c, p
c > 0. We use the formulae from Theorem 2.48. To simplify the evaluation we prepare the expression [ϕ 0 (t)]2 + [ψ 0 (t)]2
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263
Definite Integral
beforehand. We have
p
√
[−r sin t]2 + [r cos t]2 = r 2 = r . Thus
Z π
π
M(K) =
cr d t = rc t 0 = πrc,
Z0 π
π
Sx (K) =
cr sin t · r d t = cr 2 − cos t 0 = 2r 2 c,
Z0 π
π
Sy (K) =
cr cos t · r d t = cr 2 sin t 0 = 0,
0
so the coordinates of the centre of mass are
0 2r 2 c
2r
,
= 0,
.
T =
πrc πrc
π
N
Example 2.51 Evaluate the mass and coordinates of the centre of mass of curve G, which is the graph
of function y = 21 x 2 , x ∈ h0, 1i, if its linear density is ρ(x) = x .
Solution. This time the curve involved is thep
graph of a function
(arc of parabola), so we use the formulae
√
0
02
2.49. We have y = x , thus 1 + y = 1 + x 2 . Using the substitution x 2 + 1 = t 2 ,
from Corollary
√
2
that is x + 1 = t , we obtain for the mass and the first system moment
1 + x2 = t 2
Z √2
Z 1 p
M(G) =
x 1 + x 2 d x = x d x = t d t√ =
t · t dt =
0
1
0 ; 1, 1 ; 2 √
1 3 √2 2 2 − 1
=
t 1 =
,
3
3
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Definite Integral
1 + x2 = t 2
Z √2
1
x2 p
Sx (G) =
x·
1 + x 2 d x = x d x = t d t√ =
(t 2 − 1)t · t d t =
2
2
0
1
0 ; 1, 1 ; 2 √
√ √
√
1 t5 t3 2 1 4 2 2 2
2+1
1 1 1
=
=
−
−
−
−
=
.
2 5
3 1
2
5
3
2 5 3
15
Z
1
Finding the second system moment will be more difficult. The evaluation will be almost identical to
that of Example 2.24 c on page 210, and we will therefore
not comment on the particular steps in detail.
√
We use the equality derived there, namely tan π8 = 2 − 1. We have
x = cot v
Z 1
Z 1 p
p
Sy (G) =
x · x 1 + x2 dx =
x 2 1 + x 2 d x = d x = − sin12 v d v =
0
0
0; π, 1; π 2
4
s
Z π/4
Z π/2
2
cos v
cos2 v
−1
cos2 v
1
=
·
+
1
·
d
v
=
·
dv =
2
4
sin2 v
sin2 v
π/2 sin v
π/4 | sin v| sin v
tan v = t
2
Z
Z π/2
1−t 2 2
1
v = 2 arctan t
cos2 v
2
t 2 +1
=
·
dv = dt =
= √
2
5
5
2t
d v = 1+t 2 d t
1 + t2
π/4 sin v
2−1
2
√
t +1
π ; 2 − 1, π ; 1 4
2
Z 1
Z 1 1 − 2t 4 + t 8
1
1
2
3
= √
dt =
− + t dt =
16t 5
16 √2−1 t 5
t
2−1
√
√
1
1
t4 1
3 2 1
=
− 4 − 2 ln t +
=
+ ln 2 − 1 .
√
16
4t
4 2−1
8
8
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265
Definite Integral
The coordinates of the centre of mass are:
√
√
√
Sy (C) Sx (C)
2−1
3 3 2 + ln 2 − 1 1
√
T =
, · √
.
,
=
·
M(C) M(C)
8
5 2 2−1
2 2−1
N
Comment 2.52 We can proceed analogously with space curves. Using the notation from Comment 2.41
we have
Syz (K) Sxz (K) Sxy (K)
T =
,
,
,
M(K) M(K) M(K)
where
Z
β
M(K) =
ρ(t)
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 + [ω0 (t)]2 d t,
α
Z
β
Syz (K) =
ϕ(t)ρ(t)
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 + [ω0 (t)]2 d t,
α
Z
β
Sxz (K) =
ψ(t)ρ(t)
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 + [ω0 (t)]2 d t,
α
Z
Sxy (K) =
β
ω(t)ρ(t)
p
[ϕ 0 (t)]2 + [ψ 0 (t)]2 + [ω0 (t)]2 d t.
α
Contents
The quantities Syz , Sxz , and Sxy are called system moments with respect to coordinate planes x = 0,
y = 0, and z = 0, respectively.
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Mass and Coordinates of the Center of Mass of a Plane Figure
Analogously we can now describe how to find the coordinates of the centre of mass T = [ξ, η] of a
plane figure (mathematically the model of a plate of negligible thickness). In general it cannot be done
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Definite Integral
with the one-dimensional integral we have at our disposal. We are restricted to the special case when the
surface density ρ at the point [x, y] depends only on the x -coordinate. Thus at points having the same
x -coordinate, that is lying on parallels with the y -axis, the density is constant.
Further, we will suppose that the figure has the form (see Fig. 2.10 b))
B = {(x, y) ∈ R2 | a 5 x 5 b, g(x) 5 y 5 f (x)}.
(2.35)
Theorem 2.53 Assume the functions f (x), g(x), and ρ(x) are continuous on the interval ha, bi
andf (x) = g(x) for x ∈ ha, bi.
Then the mass of figure B described in (2.35) with surface density ρ(x) is
b
Z
ρ(x) f (x) − g(x) d x.
M(B) =
(2.36)
a
Its centre of mass T has coordinates
Sy (B) Sx (B)
T =
,
,
M(B) M(B)
(2.37)
where
Contents
Z
1 b
Sx (B) =
ρ(x) f 2 (x) − g 2 (x) d x,
2 a
Z b
Sy (B) =
xρ(x) f (x) − g(x) d x.
(2.38)
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(2.39)
a
The quantities Sx (B) and Sy (B) are called system moments of figure B with respect to the x -axis
and y -axis, respectively.
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267
We will explain from a physical point of view how the previous formulae can be obtained. If we denote
1x = xi − xi−1 the subinterval of the partition used in the construction of the definite integral, the value f (x) −
− g(x) 1x , where xi−1 5 x 5 xi , expresses the approximate area of a narrow curvilinear rectangle, bounded
from above by the graph of function f , from below by the graph of function g , and having the base 1x . As this
rectangle is homogeneous vertically and
narrow horizontally the surface density is approximately constant on it.
Then the expression ρ(x) f (x)−g(x) 1x (that is, the product of density and area) is approximately mass. Using
a limit process (evaluating integral sums and refining partitions unlimitedly) we formally obtain the formula for the
mass of B .
We say that a straight line is directed if its normal vector is chosen, which then directs to the positive half-plane
determined by the line; the opposite half-plane is called negative. The directed distance of the point from the line
is the ordinary (nonnegative) distance, multiplied by +1 if the point is in the positive half-plane, and by −1 if the
point is in the negative half-plane.
Imagine that we replace this curvilinear rectangle by its centre of mass, to which we concentrate its total mass.
The directed distance from the y -axis of any point of this curvilinear rectangle,
thus alsoof its centre of mass,
is approximately x , because the rectangle is narrow. This means that xρ(x) f (x) − g(x) 1x is the product of
the mass and the directed distance of the centre of mass from the y -axis. The system moment of a point from the
directed straight line is defined as the product of the mass concentrated at this point and the directed distance from
the line. This consideration explains the formula for the system moment Sy (B).
In the case of the x -axis we must proceed in a different way. The curvilinear rectangle is homogeneous vertifrom
cally therefore the centre of mass is approximately in the middle, that is, in the directed distance f (x)+g(x)
2
f (x)+g(x)
1
2
the x -axis. The product of the mass and this distance is
· ρ(x) f (x) − g(x) 1x = 2 ρ(x) f (x) −
2
− g 2 (x) 1x , which explains the formula for the system moment Sy (B).
In physics and other disciplines we proceed in a similar way. From physical laws formulae are derived, which
are approximately valid for “small” dimensions. Then the result is extended globally by integration. From a
mathematical point of view, we use the limit process in the integral sum, which leads to the corresponding integral.
The symbol of differential d x takes the meaning of some “infinitely small” increment. This is the way in which
the creators of integral calculus Newton and Leibniz proceeded. Later on the whole construction was delivered
from all secret “infinitely small quantities” and built exactly using limits. Nevertheless, these considerations are
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Definite Integral
precious from a motivation point of view and we used them at the beginning of this chapter as the motivation for
the definition of definite integral.
The information about the birth and history of integral and other related concepts can be found in [28, 29, 30].
Example 2.54 Find the mass and coordinates of the centre of mass of the subgraph of function y =
= 4x(1 − x) if the surface density is ρ(x) = x 2 .
Solution. We denote B for the subgraph. It is the segment of a parabola (see
Fig. 2.22). Using the notation from Theorem 2.53 we have f (x) = 4x(1 − x),
g(x) = 0. Therefore,
Z
1
Z
2
x · 4x(1 − x) d x = 4
M(B) =
0
B
3
4
(x − x ) d x =
1
2
2
x
0
ξ
1
Fig. 2.22
Z
x · [4x(1 − x)] d x = 8
0
T
η
0
Further we evaluate the system moments:
Z
y = 4x(1 − x)
1
1
4
x5 1
1 1
1
x
−
−
=4
= .
=4
4
5 0
4 5
5
1
Sx (B) =
2
y
1
x 4 (1 − 2x + x 2 ) d x =
0
5
x
x6 x7 1
1 1 1
8
=8
(x 4 − 2x 5 + x 6 ) d x = 8
−
+
=8
− +
=
,
5
3
7
5
3
7
105
0
0
5
Z 1
Z 1
x
x6 1
1 1
2
Sy (B) =
x · x 2 · 4x(1 − x) d x = 4
(x 4 − x 5 ) d x = 4
−
=4
−
=
.
5
6 0
5 6
15
0
0
Z
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1
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269
Hence the coordinates of the centre of mass T = [ξ, η] are:
"
# 8
2
2 8
15
105
T = 1 , 1 =
,
.
3 21
5
5
Note that the centre of mass is moved to the right from the vertical axis of symmetry of B having the
equation x = 1/2. This is due to the fact that B is not homogeneous. Otherwise, the result would be
ξ = 1/2.
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Exercises
1. Find the mass and coordinates of the centre of mass of the given plane curves with linear density ρ :
a)
half of an astroid x = a cos3 t , y = a sin3 t , a > 0, lying above the x -axis, ρ(t) = 1,
b)
upper semicircle with radius r > 0 and centre at the origin, ρ(x) = 1,
c)
d)
e)
f)
1 x
e + e−x between points x = −1, x = 1, ρ(x) = 1,
2
arc of a cycloid x = a(t − sin t), y = a(1 − cos t), a > 0, t ∈ h0, 2πi, ρ(t) = 1,
arc of a catenary y =
1
x2
− ln x , 1 5 x 5 2, ρ(x) = 1,
4
2
y = x 2 , x ∈ h−4, 4i, ρ(x) = |x|,
y=
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g)
arc of the astroid x = a cos3 t , y = a sin3 t , a > 0, x, y = 0, with linear density ρ(t) at the point
[x(t), y(t)] directly proportional to the x -coordinate of this point,
h)
x 2 + y 2 = r 2 , r > 0, x = 0, y = 0, when the linear density at point [x, y] equals the product of the
coordinates.
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270
Definite Integral
2. Find the mass and coordinates of the centre of mass of the given space curves with linear density ρ :
a)
one turn of helix x = a cos t , y = a sin t , z = bt , t ∈ h0, 2πi, a, b > 0, ρ(t) = 1,
b)
one turn of helix x = a cos t , y = a sin t , z = bt , t ∈ h0, 2πi, a, b > 0, ρ(t) = 2π − t .
3. Find the mass and coordinates of the centre of mass of the given homogeneous plane figure enclosed by:
b)
x2
, the x -axis and the line x = 8,
8
the curves y 2 = 4x , x 2 = 4y ,
c)
the curve y = 4 − x 2 and the x -axis,
d)
the curves y 2 = x , y = x 3 .
a)
the curve y =
4. Find the mass and coordinates of the centre of mass of the given plane figure A having density ρ :
a)
A : 0 5 y 5 sin x , 0 5 x 5 π, ρ(x) = | cos x|,
b)
A : x 2 + y 2 5 4, 0 5 x 5 y , ρ(x) = x ,
c)
A : x 2 + y 2 = ay , a > 0, ρ(y) = y ,
d)
A : − 1 5 x 5 |y − 1|, 0 5 y 5 2, ρ(y) = y 2 .
Contents
Answers to Exercises
1. a)
c)
2a
M = 3a, T = 0,
,
5
1
e4 + 4e2 − 1
M = e − , T = 0,
,
e
4e(e2 − 1)
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b)
d)
2r
M = πr, T = 0,
,
π
4
M = 8a, T = πa, a ,
3
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e)
f)
g)
h)
2. a)
b)
3. a)
c)
4. a)
c)
271
3 1
20
27 − 4 ln2 2 − 16 ln 2 .
+ ln 2, T =
,
= [1.52; 0.40],
4 2
6 ln 2 + 9
16 ln 2 + 24
√
√
65 65 − 1
6 175 65 − 1 .
√
= [0; 9.52],
M=
, T = 0,
6
650 65 − 10
3ka 2
5a 15πa
M=
,T =
,
, ρ(t) = ka cos3 t, k > 0,
5
8 256
p
2r 2r
r3
,T =
,
M=
, ρ(x) = x r 2 − x 2 .
2
3 3
p
M = 2π a 2 + b2 , T = [0, 0, πb],
p
2
2
2
2
a + b , T = 0, 0, bπ .
M = 2π
3
64
12
16
9 9
M=
, T = 6,
,
b) M =
,T =
, ,
3
5
3
5 5
32
8
5
12 3
M=
, T = 0, ,
d) M =
,T =
, .
3
5
12
25 7
√
π 1
3(π − 2)
8−4 2
3
√ ,
√ ,
M = 1, T =
, ,
b) M =
,T =
2 3
3
8 2− 2 4 2− 2
πa 3
5a
25
24 39
M=
, T = 0,
,
d) M =
,T = −
,
.
8
8
6
125 25
M=
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272
Definite Integral
Quiz—Definite Integral
Answer Yes or No to the following questions.
Z
2π
cos x d x = 0
1. (2 pts.)
0
No.
Yes.
Z
47
Z
x 17 d x >
2. (2 pts.)
23
Yes.
Z
3. (2 pts.)
47
x 55/3 d x
23
No.
4
x2 dx +
1
Z
6
x2 dx =
4
Z
6
(x 2 + x 2 ) d x
1
Yes.
No.
4. (2 pts.) Suppose that functions f (x) and g(x) are continuous at any x ∈ R.
Z b
Z b
Z b
Then
f (x) d x +
g(x) d x =
f (x) + g(x) d x for any a, b ∈ R.
a
a
a
Yes.
No.
Contents
5. (2 pts.) Suppose that a function f (x) is continuous at any x ∈ R. If f (x) 5 M on the interval
Z b
ha, bi, where a < b and M is a constant, then
f (x) d x 5 M(b − a).
a
Yes.
Z
6. (2 pts.)
No.
π/2
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sin x d x = cos(π/2) − cos 0.
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0
Yes.
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No.
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273
Definite Integral
7. (2 pts.) The area of a figure bounded by the curves y = 5 and y = x − sin x on the interval
Z 7
x ∈ h−2, 7i can be evaluated as
|5 + x − sin x| d x .
−2
Yes.
No.
8. (2 pts.) Revolving a figure P , bounded by the graphs of functions y = 3x − 1 and y = 0, x ∈ h5, 8i,
and the linesZx = 5 and x = 8, about the x -axes we obtain a solid of revolution. Its volume can be
8
evaluated as
π(3x − 1) d x .
5
Yes.
No.
9. (2 pts.)
The length of arc of a plane curve x/3 + y/4 = 1, where x ∈ h0, 3i, can be evaluated as
Z 3r
4 2
dx.
1+ −
3
0
Yes.
No.
10. (2 pts.) The area of a figure A bounded by the graphs of functions y = f (x) and y = g(x), where
x ∈ ha, biZ, a < b, and the lines x = a and x = b is given by the formula
b
m2 (A) =
|f (x) − g(x)| d x .
a
Yes.
No.
Contents
Correct Answers:
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Points Gained:
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Success Rate:
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274
Definite Integral
Self-Test
1. Evaluate the following definite integrals:
Z 1
2
a)
xe2x d x,
π
4
0
Z
π
c)
0
sin t
√
d t,
1 + cos2 t
2. Evaluate the following definite integrals:
Z 1
a)
arcsin x d x,
Z
dx
.
x2 + x
5
x−1
√
d x,
4x − 2
1
Z
b)
2
1
Z
sin3 x cos x d x,
3
d)
0
Z
π
2
Z
b)
√
3
x
d x.
4 − x2
0
0
x2
x
3. Find the area of the plane figure enclosed by the curves y =
and y = + 2.
4
2
3
12
4. Find the length of the arc of a plane curve y = ln x on the interval 5 x 5
.
4
5
5. Find the volume of the solid obtained by revolving the subgraph of function y = x 3 /3, x ∈ h0, 1i,
about the x -axis.
c)
ln (x + 1) d x,
d)
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6. Find the area of the lateral surface of the solid obtained by revolving figure P enclosed by the curves
y 2 = 2x and 2x = 3, about the x -axis.
7. Find the mass and coordinates of the centre of mass of the homogeneous plane figure enclosed by
the parabola y = 2x − x 2 and the x -axis.
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275
Definite Integral
Answers to Self-Test
1. a)
1 2
(e − 1) ,
4
2. a)
π
− 1,
2
3.
9,
4.
3
b)
,
16
√
3 2
b)
,
2
27
+ ln 2 ,
20
5.
π
,
63
c)
√2 + 1 ln √
,
2−1
c)
6.
14π
,
3
ln
d)
4
,
e
ln
d)
7.
M=
3
.
2
1.
4
2
, T = 1, .
3
5
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276
Chapter 3
Improper Integral
In the definition of the Riemann definite integral the function to be integrated (integrand) has to satisfy
two substantial restrictions:
• the integration domain is a bounded closed interval,
• the integrand is a function bounded (both above and below) on this interval.
Our goal will be to somewhat loosen these restrictions and generalize the concept of the definite integral.
This will be accomplished in two directions. First we will admit that the integration domain is a closed
interval unbounded from one side, that is, (−∞, bi or ha, +∞). Then we will consider the case when
the interval is bounded and half open and the integrand is unbounded. Finally, we will describe a generalization which will be the combination of the previous two cases. These generalized definite integrals
are called improper. In the remaining sections of this chapter we will deal with tests of convergence and
the problem of absolute and conditional convergence.
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Improper Integral
3.1 Improper Integral on Unbounded Intervals
Consider
function f defined on the interval ha, +∞), a ∈ R, such that for any c > a the definite integral
Rc
a f (x) d x exists. Thus we can define function F as follows:
Z
F (c) =
c
f (x) d x,
c = a.
a
In virtue of Theorem 2.28 this function is continuous on the interval ha, +∞), but this fact is not important in the following definition.
Now assuming that upper limit c increases unlimitedly we will study the behaviour of the value F (c).
Definition 3.1 Under the above assumptions suppose that the limit lim F (c) = I , I ∈ R, exists.
c→+∞
R +∞
Then the improper integral a f (x) d x converges and assumes the value I . Hence
Z
+∞
c
Z
f (x) d x = lim F (c) = lim
a
c→+∞
c→+∞
If lim F (c) is infinite or does not exist, the improper integral
c→+∞
f (x) d x.
(3.1)
a
R +∞
a
f (x) d x diverges.
Contents
Rc
The situation is shown in Fig. 3.1. The gray area is the value of a f (x) d x . As c increases unlimitedly we are interested if the value of this integral, depending on c, approaches some finite number I (the
finite limit exists), increases or decreases unlimitedly (the limit is ±∞) or oscillates (the limit does not
exist). In the first case we say that the integral is convergent (has the finite value, namely the number I ),
in the remaining two cases we say that it is divergent (does not have a finite value).
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278
Improper Integral
y
y = f (x)
F (c)
x
a
c → +∞
Fig. 3.1: Definition of the improper integral on a unbounded interval
Example 3.2 Investigate the following improper integrals:
Z +∞
Z +∞
dx
dx
a)
,
b)
,
2
x +1
x
0
1
Z
+∞
sin x d x.
c)
0
Solution. Following Definition 3.1 first we will find the formula of the auxiliary function F (c), which
depends on the upper integration limit. Then we will evaluate its limit for c → +∞.
a) We have
Z
F (c) =
0
c
c
dx
= arctan x 0 = arctan c − arctan 0 = arctan c,
+1
Contents
x2
therefore
Page 278 from 358
π
lim F (c) = lim arctan c = .
c→+∞
c→+∞
2
The integral is convergent and
Z
0
+∞
dx
π
= .
+1
2
x2
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279
Improper Integral
y
y
arctan c
1
y=
0
ln c
1
y=
1
x 2 +1
1
x
x
x
1
c → +∞
y
c → +∞
y
π/2
y = ln x
y = arctan x
arctan c
arctan c → π/2
0
ln c
ln c → +∞
x
c → +∞
x
1
c → +∞
a)
b)
Fig. 3.2
b) This time we have
Contents
Z
F (c) =
1
c
c
dx
= ln x 1 = ln c − ln 1 = ln c,
x
J
thus
lim F (c) = lim ln c = +∞.
c→+∞
The integral is divergent.
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c→+∞
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280
Improper Integral
y
1 − cos c
y = sin x
1
+
c → +∞
π
0
x
2π
−
y
2
y = 1 − cos x
1 − cos c
1 − cos c
x
π
0
c → +∞ 2π
a)
Fig. 3.3
c) In the last case we have
Contents
Z
F (c) =
0
c
c
sin x d x = − cos x 0 = − cos c + cos 0 = 1 − cos c,
and
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lim F (c) = lim (1 − cos c)
c→+∞
The integral is also divergent.
c→+∞
I
I
does not exist.
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281
Improper Integral
The behaviour of both functions, f and F , is shown in Figs. 3.2 and 3.3. In each pair the upper
picture displays integrand f and the lower picture displays function F , which assumes the value of the
definite integral of f depending on the upper integration
limit.
Rc
The gray area corresponds to the value of a f (x) d x . Note that in the first two cases, when integrand f is the positive function, the value F (c) increases as c increases while in the third case this value
oscillates.
In Fig. 3.2 a) we see that as c increases the value F (c) = arctan c increases and approaches the
number π/2, which is the value of this improper integral.
The situation in Fig. 3.2 b) is analogous, but this time the value F (c) = ln c increases unlimitedly
(although very slowly).
In Fig. 3.3 the integrand f (x) = sin x changes its sign. Therefore F (c) equals the difference between
the gray area lying
For c = 0 we have
R 0 above the x -axis and the gray area lying below the xR-axis.
π
F (0) = 0 = 0 sin x d x . Then this value increases up to F (π) = 2 = 0 sin x d x . Next it starts to
diminish because
the area of the region lying below the x -axis is subtracted. It decreases up to the value
R 2π
F (2π) = 0 = 0 sin x d x . Then the whole process repeats. Thus the value 1 − cos c oscillates for c
tending to +∞.
N
Similarly the improper integral on the interval (−∞,
R b bi, where b ∈ R, is introduced. Function f
must be such that for any c < b the definite integral c f (x) d x exists. We denote
Z b
G(c) =
f (x) d x,
c 5 b,
c
and consider the limit lim G(c). The terminology is the same as in Definition 3.1. The notation is
c→−∞
Z
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I
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f (x) d x.
−∞
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282
Improper Integral
Z
0
Example 3.3 Evaluate the improper integral
ex d x .
−∞
Solution: We will find function G(c), which depends on the lower integration limit, and then evaluate its limit for c → −∞. We have in succession:
Z 0
0
G(c) =
ex d x = ex c =
c
0
= e − ec = 1 − ec ,
y
1 − ec
1
y = ex
−∞ ← c
x
0
therefore
y
lim G(c) = lim (1 − ec ) =
c→−∞
c→−∞
= 1 − 0 = 1.
The integral is convergent and
Z 0
ex d x = 1.
y = 1 − ex
1
1 − ec
1 ← 1 − ec
−∞ ← c
x
0
−∞
The situation is shown in Fig. 3.4. As the inteFig. 3.4
grand f (x) = ex is positive the value G(c) = 1−ec
increases as c decreases, approaching the number 1, which is the value of the improper integral.
Contents
N
It is clear from the previous examples of improper integrals, that in addition to knowledge of the
definite integral, it is necessary to master the evaluation of limits. In particular, having reliable knowledge
of the graphs of common elementary functions is important because from these the necessary limits can
be usually found. The use of l’Hospital’s rule may be necessary in more complex cases.
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Improper Integral
R +∞
Comment 3.4 Consider the improper integral a f (x) d x and suppose d > a . As for c > d we
Rc
Rd
Rc
Rc
have a f (x) d x = a f (x) d x + d f (x) d x , lim a f (x) d x exists, and is finite, if and only
c→+∞
Rc
R +∞
if lim d f (x) d x exists and is finite. This implies that a f (x) d x is convergent if and only if
c→+∞
R +∞
Rd
f (x) d x is convergent. Of course, their values differ by a f (x) d x .
d
R +∞
This consideration implies that for the convergence or divergence of a f (x) d x the behaviour of
function f on any finite initial interval ha, di is not important, while the behaviour as x → +∞ is.
(Naturally function f is supposed to satisfy all the assumptions mentioned before Definition 3.1.)
For example, if we change function
R +∞ f on the interval ha, di (keeping its integrability on this interval),
the convergence or divergence of a f (x) d x is not changed.
Rb
A similar statement holds for −∞ f (x) d x . The behaviour of function f for x → −∞ decides its
convergence or divergence.
R0
R2
R −3
In Example 3.3 we learned that −∞ ex d x converges. Thus integrals −∞ ex d x and −∞ ex d x are
also convergent. But their values are different.
The following example will be important in connection with tests of convergence (see section 3.4.1).
Z +∞
dx
Example 3.5 Decide for which k ∈ R is
convergent.
xk
1
Contents
Solution. In Example 3.2 b) we learned that the integral diverges for k = 1. Thus suppose k 6 = 1. Then
−k+1 c
Z c
Z c
dx
x
1
1 − c−k+1
−k
−k+1
F (c) =
=
x
d
x
=
=
(c
−
1)
=
.
k
−k + 1 1 −k + 1
k−1
1 x
1
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We need to evaluate lim c−k+1 . The function involved is a power function with the exponent s =
c→+∞
= −k + 1. Let us remember the graphs of power function y = x s depending on the exponent s .
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284
Improper Integral
y
s>1
s=1
0<s<1
s=0
1
s<0
O
1
x
Fig. 3.5: Graphs of functions y = x s , s ∈ R, x > 0
Therefore, the limit equals zero for any negative exponent s , that is −k + 1 < 0, and equals +∞
otherwise, that is for −k + 1 > 0. All together we have
1 − c−k+1
1−0
1
=
=
c→+∞
c→+∞
k−1
k−1
k−1
−k+1
1−c
1−∞
lim F (c) = lim
=
= +∞
c→+∞
c→+∞
k−1
k−1
lim F (c) = lim
Taking into account the case where k = 1 we obtain
(
Z +∞
dx
converges
k
x
diverges
1
for k > 1,
for k < 1.
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for k > 1,
for k 5 1.
(3.2)
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In the convergent case k > 1 we have
Z
1
+∞
dx
1
=
.
k
x
k−1
But the last result is not so important as the fact that the value k = 1 is the boundary between the
convergence and divergence of this integral.
Moreover, Comment 3.4 implies that the answer is the same if we replace the lower limit 1 by any
positive number d in (3.2).
Later on we will see that the comparison test (see Corollary 3.19) implies that the statement concerning the divergence is trivial in the case of k 5 0, that is when the integrand f (x) = 1/x k is positive and
nondecreasing (for k < 0 even increasing). Thus only case k > 0 is interesting; then the integrand is
positive and decreasing.
N
3.2 Improper Integral of Unbounded Functions
Consider function
R cf defined on the interval ha, b), a, b ∈ R, a < b, such that for any c ∈ (a, b) the
definite integral a f (x) d x exists. Moreover, we suppose that function f is not bounded on the interval
ha, b). Then the point b is the singular point of function f . Thus function f is not bounded in any left
δ -neighbourhood (b − δ, b) of the point b, 0 < δ < b − a .
Thus we can define function F by
Z c
F (c) =
f (x) d x,
a 5 c < b,
a
and study what happens with the value F (c) as c approaches point b unlimitedly from the left. (see
Fig. 3.6).
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Definition 3.6 Under the above assumptions suppose that the limit lim F (c) = I , I ∈ R exists. Then
c→b−
Rb
the improper integral a f (x) d x converges and assumes the value I . Hence
Z
a
b
c
Z
f (x) d x = lim− F (c) = lim−
c→b
f (x) d x.
c→b
If lim F (c) is infinite or does not exist, the improper integral
(3.3)
a
Rb
c→b−
a
f (x) d x diverges.
As the situation is analogous to that of the improper integral on an unbounded interval, as introduced
in Definition 3.1, we will continue more quickly in the following explanation.
y
y = f (x)
Contents
F (c)
x
a
c → b−
b
Fig. 3.6: Definition of the improper integral of an unbounded function
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Improper Integral
Z
Example 3.7 Evaluate the improper integral
0
1
x dx
√
.
1 − x2
Solution. The integrand is a function continuous on the interval h0, 1). It is not defined at point x = 1.
Moreover,
x
1
lim− √
=
= +∞.
2
+
x→1
0
1−x
Therefore, it is the improper integral of an unbounded function. (The function has the vertical asymptote
x = 1.) First we evaluate the definite integral on the interval h0, ci, where 0 5 c < 1:
1 − x2 = t
Z c
Z
2
−2x d x = d t
1 1−c d t
x dx
√
√ =
=−
F (c) =
=
x d x = − 12 d t 2 1
t
1 − x 2 0
0 ; 1, c ; 1 − c2 2
Z
2
p
√ 1
1 t 1/2 1−c
1 1−c −1/2
=−
t
dt = −
=
t 1−c2 = 1 − 1 − c2 .
2 1
2 1/2 1
Then we evaluate the limit for c → 1− :
lim− F (c) = lim− 1 −
c→1
p
1 − c2 = 1 − 0 = 1.
Contents
c→1
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The integral is convergent and its value is
Z
0
The situation is shown in Fig. 3.7 a).
1
J
x dx
√
= 1.
1 − x2
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y
y
y=
√
1−
√x
1−x 2
y=
1
x
1 − c2
ln 2 − ln c
x
0
c→
1−
x
1
0+ ←
0
c
2
y
y
1
√
y = 1 − 1 − x2
y = ln 2 − ln x
G(c)
F (c)
√
1−
0
c→
1−
1 − c2
x
Contents
ln 2 − ln c
x
1
0
a)
0+ ←
c
b)
2
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Fig. 3.7
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Improper Integral
We proceed analogously if function f is defined on the interval (a, bi, a, b ∈ R, a < b, and is
integrable on any interval hc, bi, where c ∈ (a, b). Again we will suppose that function f is unbounded
on the interval (a, bi. Thus a is its singular point. We define function
Z b
G(c) =
f (x) d x,
a < c 5 b,
c
+
and investigate the limit for c → a . The terminology and notation are the same as in Definition 3.6.
Z 2
dx
Example 3.8 Evaluate the improper integral
.
x
0
Solution. The integrand is the function continuous on the interval (0, 2i. It is not defined at point x = 0.
As
1
1
lim+ =
= +∞,
+0
x→0 x
the integral is improper. (The graph of the function is a rectangular hyperbola, which has the vertical
asymptote x = 0.)
First we evaluate the definite integral on the interval hc, 2i, 0 < c 5 2:
Z 2
2
dx
= ln x c = ln 2 − ln c.
x
c
Then we evaluate the limit for c → 0+ :
lim+ (ln 2 − ln c) = ln 2 − (−∞) = +∞.
c→0
The integral is divergent. The situation is displayed in Fig. 3.7 b). The value G(c) = ln 2−ln c increases
unlimitedly for c → 0+ .
N
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290
Comment 3.9
1. The improper integral of an unbounded function has properties similar to the improper integral on
an unbounded interval. Namely the behaviour of the integrand near the singular point decides the
convergence or divergence.
2. The improper integral of an unbounded function has one feature which is very unpleasant for students.
While the improper integral on unbounded interval is recognized at first sight, because its notation
involves the symbol +∞ or −∞, the notation of the improper integral of an unbounded function looks
the same as that of the ordinary definite integral. Students often overlook that the integral involved is
improper and proceed as if it were the ordinary definite integral. Such an evaluation can lead to a fatal
error. In the following section (see Example 3.13) we will demonstrate what can be caused by such a
lapse.
Rb
Therefore, on seeing the symbol a f (x) d x we must determine if function f is bounded on the
interval ha, bi and so the integral is the ordinary definite integral, or if it is unbounded, has a singular
point, and therefore the integral is improper. A typical sign is that the function is not defined at some
point. For example, f is a fraction with the denominator assuming zero value at some
But
R π sinpoint.
x
this does not automatically imply that the integral is improper. Compare the integral 0 x d x (see
Fig. 2.5 a) on page 186). Function sinx x is not defined for x = 0, but it is bounded as was proved on
page 186. Hence this integral is an ordinary definite integral. The same is true with the integral of the
function in Fig. 2.5 b).
3. What happens, if evaluating the ordinary definite integral we proceed by mistake as if it were an
improper integral. Fortunately, nothing happens due to the properties of the definite integral being the
function of the upper limit (see Section 2.4.3).
Rc
For example, suppose we mistake point b to be a singular point. The function F (c) = a f (x) d x
is continuous on the whole interval ha, bi by Theorem 2.28. As the limit of a continuous function
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Improper Integral
equals the function value we have
Z
lim− F (c) = F (b) =
c→b
b
f (x) d x,
a
which is the correct result.
R1
Sometimes it is even advantageous to proceed this way. A typical example is 0 x ln x d x . Function x ln x is continuous on the interval (0, 1i. Using l’Hospital’s rule we evaluate the right-hand limit
at the point x = 0. We get
lim+ x ln x = lim+
x→0
x→0
ln x
1
x
=
−∞ +∞
LH
= lim+
x→0
1
x
− x12
= − lim+ x = 0.
x→0
The function is bounded on the interval (0, 1i so it is Riemann integrable on the interval h0, 1i. The
value at point x = 0 can be chosen arbitrarily, it has no impact on the result (see examples in Fig. 2.5 on
page 186). If we want to use the Fundamental Theorem of Calculus on the whole integration domain
h0, 1i, we have trouble finding the antiderivative at point x = 0. It is possible to apply Theorem 1.29.
Another possibility is to use the approach described above. This way we get:
Z 1
u = ln x u0 = 1 x
=
G(c) =
x ln x d x = 0
1 2 v
=
x
v
=
x
c
2
Z
1 1 1
1 2
=
x ln x c −
x dx =
2
2 c
1 2 1
1
1 c2
1
= − c2 ln c −
x c = − c2 ln c − + .
2
4
2
4
4
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Improper Integral
Rearranging the expression and using l’Hospital’s rule we get:
1 2
1 c2
lim G(c) = lim+ − c ln c − +
=
c→0+
c→0
2
4
4
=−
1
1
lim+ c2 ln c − + 0 =
2 c→0
4
1
1 1
1 1
ln c
=− −
lim+ 1 = − −
lim+ c 2 =
4 2 c→0 c2
4 2 c→0 − c3
1 1
1
1
=− +
lim+ c2 = − + 0 = − ,
4 4 c→0
4
4
thus
Z
1
1
x ln x d x = − .
4
0
Let us emphasize once more that this integral is not improper.
The result of the following example will be important later in the section on tests of convergence.
Z 1
dx
Example 3.10 Decide for which k ∈ R, k > 0, is the integral
convergent.
k
0 x
Solution. The function 1/x k = x −k is continuous on the interval (0, 1i. From the graphs of power
functions in Fig. 3.5 it is clear that for k > 0 there is lim = x −k = +∞ and the integral is improper.
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x→0+
(On the other hand, for k 5 0 we get the ordinary definite integral.)
We know from Example 3.8 that the integral is divergent for k = 1. Suppose that k 6 = 1. We have:
−k+1 1
Z 1
Z 1
dx
x
1
1 − c−k+1
−k
−k+1
=
(1
−
c
)
=
.
G(c) =
=
x
d
x
=
k
−k + 1 c
−k + 1
1−k
c x
c
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Improper Integral
Now we evaluate the limit. With the help of Fig. 3.5 we easily obtain that
1−0
1
1 − c−k+1
=
=
c→0
c→0
1−k
1−k
1−k
−k+1
1−c
1−∞
lim+ G(c) = lim+
=
= +∞
c→0
c→0
1−k
1−k
lim+ G(c) = lim+
for 0 < k < 1,
for k > 1.
Including the case k = 1 we get
Z
0
1
dx
xk
(
converges for 0 < k < 1,
diverges
for k = 1.
(3.4)
In the convergent case we have
Z
1
1
dx
.
=
k
1−k
0 x
This result is not so important as the fact that the boundary between the convergence and divergence of
this integral is the value k = 1.
N
Moving the graph of function 1/x k to the right or to the left by a number α or eventually revolving
it about the line x = α we easily find that the integrals
Z α
Z d
dx
dx
,
d
<
α,
or
, d > α,
(3.5)
k
k
d (α − x)
α (x − α)
are also convergent for 0 < k < 1 and divergent for k = 1 (see Fig. 3.8 a) and 3.8 b)).
Summing up the results of Examples 3.5 and 3.10 we conclude that the improper integrals
Z d
Z +∞
dx
dx
and
,
k
x
xk
0
d
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Improper Integral
y
y=
1
(α−x)k
y=
1
(x−α)k
x
d
α
y=
1
xk
x
x
α
a)
0
d
b)
d
c)
Fig. 3.8
where d > 0, are both divergent for k = 1 and just one is convergent and one is divergent for k > 0,
k 6 = 1 (see Fig. 3.8 c)). In particular, we have:
Z
0
d
dx
xk
Z
d
+∞
dx
xk
0 < k < 1 convergent divergent
k=1
divergent
divergent
k>1
divergent
convergent
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295
3.3 Generalization of Improper Integral
When introducing the indefinite integral we assumed that the interval on which we integrated contained
just one “bad” point, which was in each instance an end point of the domain of integration. It was
represented as +∞ or −∞ or it was a singular point, near to which the integrand f was unbounded.
We cut off this “bad” point from the integration domain by the line x = c and integrated the function f
over the remaining bounded closed interval. Then using a limit process we made the cut part increasingly
small.
Now we allow the integration domain J (an interval in each instance) to contain more “bad” points,
but of finite number. Thus, for example, the interval can be unbounded on both sides, that is being
(−∞, +∞). Both end points can also be singular, or a singular point can be in the interior of the
integration domain. In this case by singular point we mean a point where the integrand f is not bounded
on any bilateral neighbourhood of it. At singular points the integrand is not usually defined, but this fact
has no impact. So J will be an interval with end points α and β , where −∞ 5 α < β 5 +∞.
We will proceed in such a way that we will insert an auxiliary point between any pair of consecutive “bad” points. Then these points and all finite “bad” points separate the integration domain J into
subintervals not containing any interior singular point. Thus all singular points will be end points of the
obtained subintervals. Moreover, each subinterval will have just one “bad” point.
Then we will investigate the integrals on particular intervals. We will suppose that for any bounded
closed interval ha, bi which is the subset
R b of the integration domain J and does not contain any singular
point, the ordinary definite integral a f (x) d x exist. Therefore, we will obtain improper integrals of
the two basic types introduced above.
We will explain the whole idea on a function f having the integration domain (−∞, +∞) shown
in Fig. 3.9. Evidently, “bad” points are ±∞ and the points a and b, which are singular (the lines x = a
and x = b are vertical asymptotes to the graph of f ). We insert an auxiliary point d1 between −∞ and
a and an auxiliary point d2 between a and b. It is not necessary to insert an auxiliary point between b
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Improper Integral
y
y = f (x)
x
d1
a
d2
b
Fig. 3.9: Generalization of Improper Integral
and +∞ because the integrand f is bounded on the right neighbourhood of b. We obtain five improper
integrals
Z d1
Z a
Z d2
Z b
Z +∞
f (x) d x,
f (x) d x,
f (x) d x,
f (x) d x,
f (x) d x.
−∞
d1
a
d2
Contents
b
Rβ
Definition 3.11 Under the above assumptions we say that the improper integral α f (x) d x converges
if and only if all particular improper integrals are convergent. Then its value is the sum of the values of
all the particular integrals.
Otherwise,
if at least one of the integrals involved is divergent, we say that the improper integral
Rβ
f
(x)
d
x
diverges.
α
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Improper Integral
R +∞Thus if all five particular integrals in our illustrative example were convergent, the improper integral
−∞ f (x) d x would be also convergent and
Z
+∞
Z
d1
Z
f (x) d x =
−∞
a
f (x) d x +
f (x) d x +
d1
−∞
Z
Z
b
d2
f (x) d x +
a
Z
f (x) d x +
+
d2
+∞
f (x) d x.
b
If we encounter a divergent integral when investigating particular improper integrals, the evaluation
is over and the original integral is also divergent. That is why, it is reasonable in a concrete case to
start with those particular integrals about which we think are divergent. If our estimate is correct the
calculation will be shorter.
Finally it is necessary to mention the role of auxiliary points. We would be in trouble if a different
choice of an auxiliary point could give a different result. Thus if the answer to the question whether the
integral is convergent or divergent, and its value in the convergent case depended on this choice. Additivity of the definite integral with respect to the integration domain (Theorem 2.9) and Comment 3.4 imply
that nothing like this can happen. Therefore, the auxiliary points can be chosen arbitrarily. Moreover,
it does not matter if we add unnecessary points causing some particular integrals not to be improper. If
possible, it is useful to choose them so an eventual symmetry of the integrand (even and odd function),
symmetry about some parallel with the y -axis, or other symmetry can be used to simplify the calculation.
We will demonstrate the whole procedure on a few examples.
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Improper Integral
Example 3.12 Evaluate the following improper integrals:
Z +∞
Z 1
x2
dx
√
a)
,
d x,
b)
6
1 − x2
−∞ x + 1
−1
Z +∞
Z 2 2
x −x+1
dx
√
c)
,
d)
d x.
x (x + 1)
x−1
0
0
Solution.
2
a) The integrand x 6x+1 is continuous on the whole real line so it is sufficient to insert one auxiliary point
separating −∞ and +∞. With respect to symmetry (the function is even) it is reasonable to choose
zero (see Fig. 3.10 a)). We obtain two improper integrals on unbounded intervals
Z +∞
Z 0
x2
x2
d
x
and
d x.
(3.6)
6
x6 + 1
0
−∞ x + 1
We will start with the second one. We “cut off ” the right end, that is, we evaluate for c = 0 the definite
integral using the substitution method:
x3 = t
Z c
Z
2
2
3x d x = d t
1 c3 d t
x
F (c) =
dx = 2
=
1
=
6
2
0 x +1
x dx = 3 dt 3 0 t + 1
0 ; 0, c ; c3 c3
1
1
1
1
= arctan t 0 = arctan c3 − arctan 0 = arctan c3 .
2
3
3
3
Then we find the limit for c → +∞:
1
1 π
π
arctan c3 = · =
c→+∞ 3
3 2
6
lim
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Z
⇒
0
+∞
x2
π
dx = .
6
x +1
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Improper Integral
As the integrand is an even function, due to the symmetry, the first integral must also be convergent
and assume the same value. So it is not necessary to calculate it. Altogether, the given integral is
convergent and
Z
+∞
−∞
b) The integrand √ 1
x2
dx =
x6 + 1
1−x 2
0
Z
−∞
x2
dx +
x6 + 1
x2
π π
π
dx = + = .
6
x +1
6
6
3
+∞
Z
0
is continuous on the interval (−1, 1). As
1
=
lim + √
x→−1
1 − x2
1
+0
= +∞,
1
=
lim− √
x→1
1 − x2
1
+0
= +∞
both ends are singular points (at which the vertical asymptotes occur). We insert one separating point
between them. The best choice is again zero because the integrand is an even function. We obtain two
improper integrals of unbounded functions (see Fig. 3.10 b)):
Z
0
−1
1
Z
dx
√
1 − x2
and
0
dx
√
.
1 − x2
(3.7)
Again we will start with the second one. “Cutting off ” the right end we evaluate for 0 5 c < 1 that
Z c
c
dx
√
F (c) =
= arcsin x 0 = arcsin c − arcsin 0 = arcsin c.
2
1−x
0
Then we find the limit for c → 1− :
lim arcsin c =
c→1−
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π
2
Z
⇒
0
1
dx
π
√
= .
2
2
1−x
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Improper Integral
y
y= √1
1−x 2
y
y=
x2
x 6 +1
1
x
x
−1
1
0
0
a)
b)
y
3
y=
1
y
y=
x 2 −x+1
x−1
√ 1
x (x+1)
x
x
0
0
−1
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1
c)
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d)
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Improper Integral
Due to the symmetry the first integral in (3.7) must also be convergent and have the same value.
Altogether, the given integral is convergent and
Z 1
Z 0
Z 1
dx
dx
dx
π π
√
√
√
=
+
= + = π.
2
2
1 − x2
1 − x2
1 − x2
−1
−1
0
c) The integrand
√ 1
x (x+1)
is continuous on the interval (0, +∞). As
1
1
lim+ √
=
= +∞,
+0
x→0
x (x + 1)
a singular point is to the left (at which the vertical asymptote occurs). So we insert a separating point
between zero and +∞, for example one. We obtain two improper integrals, the first of an unbounded
function, and the second on an unbounded interval (see Fig. 3.10 c)):
Z +∞
Z 1
dx
dx
√
√
and
.
(3.8)
x (x + 1)
x (x + 1)
1
0
We evaluate the first. “Cutting off ” the left end we calculate for 0 < c 5 1 the definite integral using
the substitution method:
x = t2
Z 1
dx
2t d t
=
√
G(c) =
= d x = 2t d t
=
√
2
√
x (x + 1) c
c t (t + 1)
c ; c, 1 ; 1 Z 1
1
√
dt
=2 √ 2
= 2 arctan t √c = 2 arctan 1 − 2 arctan c =
c t +1
Z
1
=2·
√
√
π
π
− 2 arctan c = − 2 arctan c.
4
2
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Then we find the limit for c → 0+ :
√
π
π
π
lim+
− 2 arctan c = − 2 arctan 0 =
c→0
2
2
2
Z
⇒
0
1
dx
π
√
= .
x (x + 1)
2
Now we evaluate the second integral in (3.8). “Cutting off ” the right end we calculate for c = 1 that
x = t2
Z √c
Z c
2t d t
dx
=
√
= d x = 2t d t
=
F (c) =
√ x (x + 1) t (t 2 + 1)
1
1
1 ; 1, c ; c
Z √c
√c
√
dt
arctan
t
=2
=
2
= 2 arctan c − 2 arctan 1 =
1
2
t +1
1
√
√
π
π
c − 2 · = 2 arctan c − .
4
2
Then we find the limit for c → +∞:
Z +∞
√
π
π π
π
dx
π
√
=2· − =
⇒
= .
lim 2 arctan c −
c→+∞
2
2
2
2
x
(x
+
1)
2
1
= 2 arctan
As the two particular integrals are convergent the given integral is also convergent and
Z +∞
Z 1
Z +∞
dx
dx
dx
π π
√
√
√
=
+
= + = π.
x (x + 1)
x (x + 1)
x (x + 1)
2
2
0
0
1
2
−x+1
d) The integrand x x−1
is continuous on the interval h0, 2i with the exception of point x = 1, at which
it is not defined. As
x2 − x + 1
1
x2 − x + 1
1
lim−
=
= −∞,
lim+
=
= +∞,
−0
+0
x→1
x→1
x−1
x−1
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this point is singular (at which the vertical asymptote occurs). So we separate the integration domain
at this singular point. We obtain two improper integrals of unbounded functions (see Fig 3.10 d)):
Z
0
1
x2 − x + 1
dx
x−1
Z
and
1
2
x2 − x + 1
d x.
x−1
(3.9)
We evaluate for example the second. “Cutting off ” the left end we calculate for 1 < c 5 2 that
2
2
Z 2
x2 − x + 1
x
1
G(c) =
dx =
dx =
+ ln |x − 1| =
x+
x−1
x−1
2
c
c
c
c2
c2
− ln |c − 1| = 2 −
− ln |c − 1|.
= 2 + ln 1 −
2
2
Z
2
Then we find the limit for c → 1+ :
c2
1
lim+ 2 −
− ln |c − 1| = 2 − − (−∞) = +∞.
c→1
2
2
As this particular integral diverges, the given integral is also divergent. The behaviour of the first
integral in (3.9)s not important (it is easy to verify that it also diverges).
N
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Improper Integral
In section 2.5.1 we studied geometric applications of the definite integration. It can be proved that the
formulae mentioned there are also valid if they lead to the improper integrals. For example, the first three
integrands shown in Fig. 3.10 are nonnegative. As the corresponding improper integrals are convergent
they express the area of their subgraphs. Similarly, Fig. 3.8 c) and the table following it state that the area
of subgraph of the function 1/x k , where k > 0, is never finite on the interval (0, +∞).
In√Example 2.40 we rejected the evaluation of the length of the semicircle based on the function
y = r 2 − x 2 because the integrand obtained was an unbounded function leading to
Z r
r
√
d x,
2
r − x2
−r
and instead used the parametric description of the semicircle. Now we know that it is an improper integral
with singular points at both ends of the integration domain h−r, ri. It is easy to verify that this integral
is convergent and gives the correct result πr for the length of semicircle. The integrand is displayed in
Fig. 3.10 b) for r = 1, therefore the result was π.
The improper integral also has important physical applications.
Finally we calculate a simple example in which we demonstrate a serious error that students unfortunately sometimes make.
Z 1
dx
Example 3.13 Evaluate the integral
.
2
−1 x
2
Solution: The integrand 1/x is continuous on the interval h−1, 1i with the exception of point x = 0,
where it is not defined. As
1
1
lim 2 =
= +∞
+0
x→0 x
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Improper Integral
the point is singular (at which the vertical asymptote occurs). We separate the
integration domain at this singular point and obtain two improper integrals of unbounded functions (see Fig. 3.11):
Z 1
Z 0
dx
dx
and
.
2
2
0 x
−1 x
We investigate the first integral. “Cutting off ” the right end we evaluate for −1 5
5 c < 0 that
Z c
Z c
dx
1
1 c
−2
=
= − − 1.
x
d
x
=
−
2
x −1
c
−1 x
−1
y
y=
1
x2
x
−11
0
1
Fig. 3.11
We find the limit for c → 0− :
1
lim− − − 1 = −(−∞) − 1 = +∞.
c→0
c
Therefore, the integral is divergent. We could quickly also determine this from Formula (3.4). Due to the
symmetry it is evident that the second integral gives the same result, but it is not important. The given
integral definitely diverges on the interval h−1, 1i.
Students sometimes ignore the fact that the integral is improper and use formally the Fundamental
Theorem of Calculus as if the integral were a standard definite integral. Their calculation then looks like
this:
Z 1
Z 1
dx
1 1
−2
=
x dx = −
= −1 − 1 = −2.
2
x −1
−1
−1 x
!
Of course, the result is absolutely wrong! Instead of the correct answer, that the integral is divergent,
the authors of such an “evaluation” come to the conclusion that the integral is convergent (to be exact,
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Improper Integral
they do not notice that this integral is improper and consider it to be the ordinary definite integral). They
should at least be surprised that they obtained the negative result from the clearly positive function 1/x 2 ,
the subgraph of which must have the area equal to a positive number or +∞.
N
In Definitions 3.1 and 3.6 the concept of divergent integral includes two possibilities: either the limit of some
expression is improper or does not exist. Sometimes these two possibilities are distinguished. Then in the first case
we say that the improper integral has the value +∞ or −∞ while in the second case it has no value.
As to the generalization from Definition 3.11 such an improper integral has a value if all partial integrals that
are divergent have the infinite value of the same sign. Then the value of such an improper integral is infinity of the
same sign.
Comment 3.14 In many important applications, for example the evaluation of the inverse Laplace or Fourier
transforms (see [11]) another extension of the improper integral is important, and is different from that in DefiniRβ
tion 3.11. It is called principal value Its notation is v.p. α f (x) d x . The symbol v.p. being the abbreviation of
French words valeur principale (read valœr prensipal).
We will introduce two versions of this concept. First we will consider a function f defined on the interval
(−∞, +∞) at which there is no singular point and integrable on any closed bounded interval ha, bi. For an
arbitrary c = 0 we define function
Z
c
H (c) =
f (x) d x.
−c
So we symmetrically “cut off ” both ends from the integration domain (−∞, +∞) (see Fig. 3.12 a)). At the same
time we move both ends one from the other in the same velocity, that is we evaluate the limit
Z c
lim H (c) = lim
f (x) d x.
c→+∞
c→+∞ −c
If this limit is finite and equal to I , we say that the principal value of the improper integral of function f on the
interval (−∞, +∞) exists and we write
Z +∞
v . p.
f (x) d x = I.
−∞
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Improper Integral
y
y = f (x)
y = f (x)
H (c)
H (c)
x
−∞ ← −cc
0
x
c → +∞
a
d − c → d− d
a)
d+ ← d + c
b
b)
Fig. 3.12: Principal value of improper integrals
Second we will consider a function f defined on the interval ha, bi containing one interior singular point d .
Function f is supposed to be integrable on each interval hα, βi ⊂ ha, bi not containing d . For an arbitrary small
c > 0 we define function
Z
Z
d−c
H (c) =
b
f (x) d x +
a
f (x) d x.
d+c
So we “cut off ” a symmetric neighbourhood of point d having length 2c (see Fig. 3.12 b)). Now at the same time
we move both ends of this neighbourhood one to the other in the same velocity, that is we evaluate the limit
Z d−c
Z b
lim H (c) = lim
f (x) d x +
f (x) d x .
c→0+
c→0+
a
d+c
If this limit is finite and equal to I , we say that the principal value of the improper integral of function f on the
interval ha, bi exists and we write
Z b
v . p.
f (x) d x = I.
a
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Improper Integral
It is easy to verify that the convergence of the improper integral (in the sense of Definition 3.11) implies the
existence of its principal value and both results are the same. But the converse statement is not generally correct,
some additional assumptions are needed, for example that the integrand does not change sign. Therefore, principal
value is useful for integrands changing the sign (for the argument tending to ±∞ in the first version and near a
singular point in the second version).
For example, the principal value of the improper integral of any odd function f defined on (−∞, +∞) exists.
Of course, we assume that f is integrable on any interval ha, bi. We have (see Comment 2.44)
Z
c
Z
f (x) d x = 0
H (c) =
⇒
−c
lim H (c) = 0
c→+∞
⇒
+∞
f (x) d x = 0.
v . p.
−∞
R +∞
But if f (x) = x , then the integral −∞ x d x diverges because it is easy to show that both particular integrals
R0
R +∞
x d x = +∞ are divergent.
−∞ x d x = −∞ and 0
Finally, let us try to find the principal value of the divergent improper integral from Example 3.12 d, whose
2 −x+1
1
= x + x−1
has an interior singular point x = 1 on the
integrand is shown in Fig. 3.10 d). The function x x−1
interval h0, 2i. For small c > 0 we have:
1−c Z 2 1
1
H (c) =
x+
dx +
x+
dx =
x−1
x−1
0
1+c
1−c 2
2
2
x
x
+
=
=
+ ln |x − 1|
+ ln |x − 1|
2
2
0
1+c
Z
=
1
1
(1 − c)2 + ln | − c| − 0 − 0 + 2 + 0 − (1 + c)2 − ln |c| = 2 − 2c.
2
2
Hence
Z
lim (2 − 2c) = 2
c→0+
⇒
v . p.
0
1
x2 − x + 1
d x = 2.
x−1
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Improper Integral
Rβ
Rβ
Comment 3.15 Let us consider two improper integrals α f (x) d x and α g(x) d x of any types, but on the same
integration domain, −∞ 5 α < β 5 +∞. If they are both convergent (this assumption is substantial) it is easy
to derive that the integrals of the sum f + g and the constant multiple cf , where c ∈ R, are also convergent and
Z
β
f (x) + g(x) d x =
α
Z
β
Z
f (x) d x +
α
Z
β
g(x) d x,
α
Z
β
cf (x) d x = c
α
β
f (x) d x.
α
Thus convergent improper integrals are additive and homogeneous with respect to integrands (compare Theo
R β
rem 2.7 for the definite integral). Let us observe that α f (x) + g(x) d x need not be improper.
Exercises
1. Evaluate the following improper integrals:
Z +∞
2
a)
d x,
b)
3
x
1
Z
+∞
d)
1
Z
+∞
g)
1
Z
j)
1
+∞
e)
2
d x,
2
x + 2x + 2
h)
+∞
1
Z
1
√
dw,
w 1 + w2
3
d x,
x+1
Z
+∞
0
Z
3
p dy,
y5
Z
k)
0
2t
d t,
t2 + 1
+∞
8
dx,
8 + 2x 2
0
Z
4
d x,
1 + x4
f)
3 e−0.3φ dφ,
i)
0
+∞
Z
0
Z
+∞
c)
+∞
4 e−2t sin 2t d t,
+∞
r e−r
2 /2
d r,
+∞
x sin x d x,
l)
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J
0
Z
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Improper Integral
Z
−1
m)
−∞
Z
0
p)
−∞
Z
0
s)
−∞
dx
,
(x − 1)(x 2 + 1)
Z
2
Z
dx
,
2
(x + 1)(x 2 + 4)
q)
x dx
,
(x 2 + 1)(x 2 + 3)
t)
2. Evaluate the following improper integrals:
Z +∞
2m
a)
dm,
b)
(1
+
m)3
0
Z +∞
9
d x,
e)
d)
1
+
x3
0
Z π
2
1
g)
dα,
h)
2
0 cos α
Z 2
3t 3
√
j)
d t,
k)
4 − t2
0
Z 1
1
dp,
n)
m)
2
0 p − 4p + 3
Z +∞
2 arctan z
p)
d z,
q)
z3
1
+∞
n)
+∞
1
Z
+∞
1
Z
+∞
1
Z
0
Z
2
1
r)
ln2 u
d u,
u2
u)
dy,
Z
1
1
∞
Z
1
Z
1
d x,
2−x
1
1
√ d x,
x
l)
0
Z
3x
√
dx,
x−1
+∞
4
x 2 (1 + x 2 )
o)
1
Z
d x,
r)
1
cos ln x
d x,
x
2
i)
0
2
2
2ρ 3 e−ρ dρ,
0
0
Z
+∞
c)
1
(2r − 1) ln2 r d r,
du
.
u ln2 u
2
f)
ln x d x,
Z
+∞
Z
Z
4 − y2
2z
d z,
+3
2z2
0
0
Z
+∞
Z
ln u
d u,
u2
2
p
1
d x,
x 2 (x + 1)
o)
4 arctan x
d x,
x2
2
+∞
Z
ln u
d u,
u
2
1
ds,
s ln s
+∞
cos ln x
d x.
x2
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Improper Integral
3. Evaluate the following improper integrals:
Z +∞
dx
√
a)
,
b)
x
x−1
1
Z
d)
+∞
√
2
Z
1
g)
−1
Z
e1/x
d x,
x2
+∞
j)
2/π
Z
m)
0
dx
√
,
x x2 − 1
+∞
sin 1/x
d x,
x2
dx
,
x2 − 1
Z
+∞
0
Z
+∞
e)
−∞
Z
+∞
h)
−∞
Z
2
k)
−2
Z
1
dβ,
cos2 β
x2
dx
,
+ 2x + 2
dx
,
x(x + 1)
4x 3
d x,
x4 − 1
+∞
n)
−∞
a3
d x,
a2 + x 2
Z
+∞
c)
−∞
Z
+∞
f)
−∞
Z
2x
d x,
x2 + 1
arctan2 x
d x,
1 + x2
+∞
e−|x| d x,
i)
−∞
Z
+∞
dx
√
,
3
x(x + 1)
+∞
|x|
d x.
x4 + 1
l)
0
Z
o)
−∞
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Improper Integral
Answers to Exercises
Symbol D in the answers means that the integral is divergent, but does not have the value ±∞.
1. a)
1,
e)
π
i)
√ 2 ,
b)
2,
c)
+∞,
d)
ln 1 +
f)
π,
g)
π − 2 arctan 2,
h)
10,
1,
j)
+∞,
k)
1,
l)
D,
m)
2 ln 2 − π
,
8
n)
+∞,
o)
− ln 2 + 1,
p)
π
,
12
q)
1,
r)
D,
s)
−
1
ln 3,
4
t)
2,
u)
−
2. a)
1,
b)
π + 2 ln 2,
c)
1,
d)
√
2π 3,
e)
π,
f)
D,
g)
+∞,
h)
2 ln2 2 − 2,
i)
+∞,
j)
16,
k)
−1,
l)
2,
m)
+∞,
n)
8,
o)
+∞,
p)
1,
q)
4 − π,
r)
1
.
2
√
2,
1
.
ln 2
π,
b)
not defined,
c)
D,
d)
π
,
4
e)
π,
f)
g)
+∞,
h)
D,
i)
j)
1,
k)
D,
l)
m)
D,
n)
πa 2 ,
o)
2,
π
.
2
3. a)
π3
,
12
2π
√ ,
3
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Improper Integral
Quiz—Improper Integral
Choose the correct answer (only one is correct).
Z∞
1. (1 pt.)
1
dx
x2
1
∞
Z∞
2. (1 pt.)
2
1
0
1
2
∞
∞
−
1
dx
x
1
0
Z0
3. (1 pt.)
π
2
1
dx
1 + x2
−∞
Z5
4. (1 pt.)
π
2
π
4
1
√
dx
x−2
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2
√
2 3
ln
3
4
3
√
3
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Improper Integral
Z∞
5. (1 pt.)
3
dx
x−1
1
−∞
Z∞
ln x
6. (1 pt.)
dx
x
ln 2
∞
0
∞
0
1
4
∞
e2
0
−∞
1
ln | − 5|
2
1
1
2
Z−1
7. (1 pt.)
e−2x d x
−∞
2
e
2
Z0
8. (1 pt.)
−
1
dx
2x − 5
−∞
1
ln 5
0
2
Z1
dx
√
9. (1 pt.)
1 − x 2 arcsin x
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1
2
ln 3
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Improper Integral
Z1
10. (1 pt.)
−1
dx
√
3
x2
0
∞
6
2
Correct Answers:
Points Gained:
Success Rate:
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Improper Integral
3.4 Convergence Tests of Improper Integrals
Given our current knowledge, if we want to determine if a given improper integral is convergent or
divergent, we have to proceed as follows:
— evaluate an auxiliary definite integral depending on the upper or lower limit c,
— find the limit of the obtained expression as c approaches some value.
This process has a substantial drawback. The only way to evaluate the definite integral that we know
of is the Fundamental Theorem of Calculus. But this assumes we are able to find an antiderivative of
the integrand. As we know from the previous text this can be hard work even in the case of a simple
integrand from the set of elementary functions, and, what is worse, the antiderivative need not exist in
this set at all. In such a case we do not have an “analytic” formula of the expression, the limit of which
we are to calculate. So we are unable to say anything about such an improper integral.
In this section we will study how to investigate an improper integral without the antiderivative of
its integrand. While using the approach described above we obtain, in the case of a convergent integral,
not only the answer that it is convergent, but also its value. This time we will be interested only in the
information if the integral converges or diverges. As we will see this is much easier to obtain.
In the case of a convergent integral it is natural to ask if this weaker information is useful—given
that the goal is usually to find the value of the integral. As we will explain below this information is very
important if we want toRapproximate such an integral. We will demonstrate it on an improper integral on
+∞
an unbounded interval a f (x) d x . The situation is the same for other types of improper integrals.
R +∞
In Comment 3.4 we showed that for d > a the convergence of the improper integral a f (x) d x
R +∞
also implies the convergence of d f (x) d x and
Z
+∞
Z
f (x) d x =
a
d
Z
f (x) d x +
a
+∞
f (x) d x.
(3.10)
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Improper Integral
Therefore, we get from the definition of an improper integral and the previous equality that
Z
lim
d→+∞
d
+∞
Z
f (x) d x =
a
Z
f (x) d x
⇒
lim
d→+∞
a
+∞
f (x) d x = 0.
d
R +∞
So choosing d sufficiently big we can make the integral d f (x) d x very small. Therefore, neglecting
it in (3.10) we get
Z +∞
Z d
.
f (x) d x =
f (x) d x.
a
a
Hence, the definite integral can be used to find the value of an improper integral, though with some error,
because the approximate value of the definite integral (without knowing the antiderivative) can be found
quite easily and exactly (see Chapter 4).
R +∞
But it is necessary to emphasize that the assumption of the convergence of the integral a f (x) d x
R +∞
is substantial. It guarantees that d f (x) d x is small when d is sufficiently great.
In the case of a divergent integral having the value ±∞ the equality (3.10) has the sense but
bothRimproper integrals involved are represented by the symbol ∞ of the same sign. Thus neglect+∞
ing d f (x) d x we would neglect in (3.10) an “infinitely great number” towards the finite number
Rd
dx.
a f (x)
R +∞
If a f (x) d x diverges, but has no value (the auxiliary function oscillates and has no limit) then
R +∞
f (x) d x also diverges and has no value and their symbols cannot be used in an equality of a similar
d
type.
In the following section we present a few tests which enable us to decide under some conditions
whether the given improper integral converges or diverges. It is necessary to point out that there is no
universal test. We will restrict ourselves to improper integrals on an unbounded interval ha + ∞). It is
easy to carry over the results to further types of improper integrals. To simplify the formulation of the
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Improper Integral
following theorems in the rest of this chapter we will suppose that integrands are functions integrable on
each interval ha, bi, where b > a .
3.4.1 Convergence Tests of Nonnegative Functions
The situation is simpler for integrands that do not change the sign on the interval ha, +∞). It is sufficient
to confine oneself to nonnegative functions (a nonpositive function
f can be replaced by function −f ).
Rc
If f is nonnegative, then the auxiliary function F (c) = a f (x) d x is nondecreasing on the interval
R +∞
ha, +∞), Therefore, it has a proper or improper limit for c → +∞ and a f (x) d x either converges
or diverges and has the value +∞. The third case, where the limit of function F (c) does not exist, cannot
occur, so the integral cannot be divergent and not having the value.
Theorem 3.16 (Comparison Test) Suppose that inequalities 0 5 f (x) 5 g(x) are valid on the
interval ha, +∞). Then the following statements hold:
R +∞
R +∞
i) If a g(x) d x converges, then a f (x) d x also converges.
R +∞
R +∞
ii) If a f (x) d x diverges, then a g(x) d x also diverges.
The situation is shown in Fig. 3.13. The subgraph of the greater function is gray, the subgraph of the smaller function is hatched. Simply said,
if the area of the greater figure (subgraph of function g ) is finite, the area
of the smaller figure (subgraph of function f ) must be also finite, and vice
verse, if the area of the smaller figure is infinite, the area of the greater
figure must be also infinite.
We can summarize that if the area of the smaller figure is finite nothing can be said about the area of the greater figure and, if the area of the
y
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y = g(x)
y = f (x)
x
O
a
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Improper Integral
greater figure is infinite, nothing can be said about the area of the smaller figure (they can be both finite
and infinite).
To successfully make use of the Comparison Test it is important to have a large stock of functions
whose integrals we know to be convergent or divergent. Then we can compare with these functions.
Students usually have problems correctly deciding whether or not the given integral is convergent or
divergent. If their decision is wrong they cannot successfully find a greater function with the convergent
integral or a smaller function with the divergent integral. Moreover, the fact that functions are to be
nonnegative must not be forgotten.
Z +∞ sin x
e
Example 3.17 Decide if the integral
d x converges or diverges.
x2 + 1
0
Solution. Using the Comparison Test we prove that the integral converges. The exponential function
assumes only positive values so the integrand is positive. For any x ∈ R we have −1 5 sin x 5 1. As
the exponential function eu is increasing and the function x 21+1 is positive we get:
sin x 5 1,
esin x 5 e,
0<
esin x
e
5 2
.
2
x +1
x +1
R +∞
. In Example 3.2 a) we found that the integral 0 x 21+1 d x is convergent. By
R +∞
virtue of Comment 3.15 the integral 0 x 2e+1 d x is also convergent which implies that the given integral
converges.
The process based on finding the antiderivative and thus the exact value would fail because the ansin x
tiderivative to the integrand xe2 +1 is not elementary.
N
We set g(x) =
e
x 2 +1
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The following limit version of the preceding test is usually easier to apply for students.
Theorem 3.18 (Limit Comparison Test) Suppose that functions f and g are nonnegative on the
interval ha, +∞) and the limit
f (x)
= L,
x→+∞ g(x)
lim
0 5 L 5 +∞,
(3.11)
exists. Then:
R +∞
R +∞
i) If L < +∞ and a g(x) d x is convergent, a f (x) d x is also convergent.
R +∞
R +∞
ii) If L > 0 and a g(x) d x is divergent, a f (x) d x is also divergent.
Thus if limit L from the previous theorem is positive and finite, the integrals of f and g are either
both convergent or both divergent.
Functions 1/x k , about which we know for which k their integrals are convergent (see (3.2)), are
often used to compare. Similarly, for integrals with unbounded integrands functions 1/|x − α|k are used
(see (3.5)). In this case the limit in (3.11) is considered at the singular point α and is one-sided.
Before demonstrating this on examples we will mention a useful consequence of the Limit Comparison Test.
R +∞
Corollary 3.19 (Necessary Condition of Convergence) Suppose that the integral a f (x) d x converges and the limit lim f (x) = L exists. Then L = 0.
x→+∞
Proof. Let us admit that L > 0.R By the definition of limit for sufficiently great d > a we have
+∞
f (x) > 0, x = d . The integral d f (x) d x is also convergent (Comment 3.4). As the integrand
is positive on the interval hd, +∞) the Limit Comparison Test can be applied. We set g(x) = 1.
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R +∞
R +∞
Then lim f (x)/g(x) = lim f (x) = L. But a g(x) d x = a d x = +∞ so the integrand
x→+∞
R +∞ x→+∞
f
(x)
d
x
is
divergent,
which
is the contradiction.
a
If L < 0 we consider the function −f (x) and arrive at the same contradiction. This proves the
statement.
R +∞
The integrand of a convergent integral a f (x) d x need not have the limit for x → +∞ as the
following example shows.
Z +∞
Example 3.20 Evaluate
f (x) d x , where
0
(
0
f (x) =
x
for x = 0, x ∈
/ N,
for x ∈ N.
Solution. For any c > 0, the function f is bounded and continuous upto a finite number of nonnegative
integers on the interval h0, ci. Therefore, its definite integral exists and by virtue of Theorem 2.6 we
have:
Z c
Z c
F (c) =
0 dx = 0
f (x) d x =
0
so
⇒
0
lim F (c) = 0,
c→+∞
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Z
+∞
f (x) d x = 0.
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0
Hence the integral is convergent. The limit lim f (x) evidently does not exist, moreover, the function f
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Improper Integral
Example 3.21 Decide about the convergence or divergence of the following improper integrals:
Z +∞
Z +∞
Z π/2
x2 + 1
sin x
pπ
√
a)
d x,
b)
arccot x d x,
c)
d x.
3
2
x (x + x + 1)
−x
1
0
0
2
Solution. In all cases we use the Limit Comparison Test.
a) For x = 1 all terms of the numerator and denominator are positive so the integrand is positive. The
degree of polynomial in the numerator is 2. The degree of polynomial in the denominator is 3 and
√
moreover, this polynomial is multiplied by x = x 1/2 , therefore, the highest degree in the denominator is 3 + 12 = 72 . So the difference of the highest degrees in the denominator and numerator is
7
− 2 = 32 . Therefore, we set g(x) = x −3/2 and obtain:
2
L=
2
√ x3 +12
x (x +x +1)
lim
x→+∞
√1
x3
= lim
x→+∞
1+
1+
1
x
1
x2
+
1
x3
x3 + x
=
x→+∞ x 3 + x 2 + 1
= lim
=
1+0
= 1.
1+0+0
R +∞
x −3/2 d x is convergent hence the given integral is also convergent.
b) The function arccot x is positive for all x ∈ R. We will try to use the function g(x) =
l’Hospital’s rule we get:
− x 21+1
0 LH
arccot x
L = lim
=
= lim
=
1
x→+∞
x→+∞ − 12
0
x
x
By (3.2)
By (3.2)
1
R +∞
1
1
x
. Applying
x 2 LH
2x
= lim 2
= lim
= 1.
x→+∞ x + 1
x→+∞ 2x
R +∞
R +∞
1
d
x
is
divergent
hence
arccot
x
d
x
and
also
arccot x d x are divergent.
1
0
x
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c) The integrand is an unbounded function because
sin x
lim − p π
=
x→π/2
−x
2
Thus x =
π
2
is the singular point. Setting g(x) =
lim
x→π/2−
By (3.5)
R π/2
0
√ dx
π/2−x
√ sin x
π/2−x
√ 1
π/2−x
1
+0
√ 1
π/2−x
= +∞.
we get:
= lim − sin x = 1.
x→π/2
N
is convergent hence the given integral is also convergent.
3.4.2 Absolute and Conditional Convergence
TheR situation in the general case is much more complicated, where the limit of an auxiliary function F (c) =
c
= a f (x) d x need not exist for c → +∞ and sometimes it can be very difficult to decide about the convergence
or divergence of such an integral.
First we will prove an important result.
Theorem 3.22 Suppose that
R +∞
a
|f (x)| d x is convergent. Then
R +∞
a
f (x) d x is also convergent.
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Proof. For x = a we have 0 5 |f (x)| − f (x) 5 2|f (x)|. By virtue of the Comparison Test
− f (x) d x is convergent. Then
Z +∞
Z +∞ f (x) d x =
|f (x)| − |f (x)| − f (x) d x =
a
a
Z
+∞
Z
|f (x)| d x −
=
a
R +∞
a
|f (x)| −
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+∞
|f (x)| − f (x) d x
a
is the difference of two convergent integrals, and is therefore convergent due to Comment 3.15.
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Improper Integral
R +∞
a
f (x) d x
R +∞
a
|f (x)| d x
Name for
R +∞
a
f (x) d x
C
C
converges absolutely
C
D
converges conditionally
D
C
cannot happen due to Theorem 3.22
D
D
diverges
Tab. 3.1: Absolute and conditional convergence of improper integrals
By investigating the integral, the integrand of which changes sign, we can try to study the integral of |f (x)|.
As this function is nonnegative, tests from the Rprevious section canR be used.
+∞
+∞
The relation between the convergence of a f (x) d x and a |f (x)| d x plays an important role. In the
following table all logical combinations of convergence and divergence that can occur are listed (C means convergence, D divergence).
As we are not interested in the case where both integrals are divergent, only two combinations remain. This
leads to the following definition.
R +∞
R +∞
Definition 3.23 a f (x) d x is said to converge absolutely if a |f (x)| d x is convergent.
R +∞
R +∞
f (x) d x is said to converge conditionally or relatively if it is convergent but a |f (x)| d x is divergent.
a
The existence of absolutely convergent integrals is evident. If the integrand is nonnegative, it equals its absolute
value, so the integrals coincide and convergence, in this case, means absolute convergence. A similar conclusion
holds for nonpositive integrands. Therefore, both concepts introduced in the preceding definition are important for
functions changing sign. Later we show that conditionally convergent integrals exist.
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Improper Integral
Z
+∞
Example 3.24 Prove that the integral
1
cos ln x
d x converges absolutely.
x2
Solution. We must investigate the integral of nonnegative function | cosx 2ln x| . As for any u | cos u| 5 1 holds we
have
1
| cos ln x|
5 2
for x = 1.
0 5 | cos ln x| 5 1
⇒
05
2
x
x
The integral of g(x) = x12 is convergent by (3.2), therefore when we do the Comparison Test we determine that
R +∞ | cos ln x|
d x is convergent, which means that the given integral converges absolutely.
N
1
x2
Until now we have not had an appropriate tool to prove conditional convergence. We will formulate such a
test.
Theorem 3.25 (Dirichlet Test) Let functions f and g satisfy the following assumptions on the interval ha, +∞):
R b
1) A constant K > 0 exists such that f (x) d x 5 K for any b = a .
a
2) Function g is monotonic and lim g(x) = 0.
x→+∞
R +∞
Then a f (x)g(x) d x converges.
Z
+∞
Example 3.26 Prove that
0
sin x
d x converges conditionally.
x
Solution. First let us recall that x = 0 is not a singular point (see Fig 2.5 a)). Thus it is the basic type of improper
integral on an unbounded interval.
In the Dirichlet Test we set f (x) = sin x , g(x) = x1 and verify the assumptions. We have:
Z
0
b
b
sin x d x = − cos x 0 = − cos b + cos 0 = 1 − cos b.
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Improper Integral
Using this result we get
−1 5 cos b 5 1
⇒
1 = − cos b = −1
So |1 − cos b| 5 2 and we can set K = 2.
Function x1 is decreasing on the interval (0, +∞) and lim
1
x→+∞ x
2 = 1 − cos b = 0.
⇒
= 0. Therefore, all assumptions are satisfied
and the integral converges.
R +∞ | sin x|
To prove that the convergence is conditional we have to verify that 0
x d x is divergent. This is a bit
more difficult and will be proved by contradiction.
R +∞ | sin x|
Let us assume that this integral and hence also the integral 1
x d x is convergent. As for any x ∈ R
the inequalities 0 5 | sin x| 5 1 hold after multiplying by a nonnegative expression | sin x| we get 0 5 | sin x|2 =
R +∞ sin2 x
= sin2 x 5 | sin x|. Using the Comparison test we show that the integral 1
x d x must be convergent. We
will prove that this is impossible.
R +∞ cos 2x
As sin2 x = 21 (1 − cos 2x) using the Dirichlet Test once more we can prove, as above, that 1
x d x is
convergent. In virtue of Comment 3.15
Z
1
+∞ sin2 x
x
+
cos 2x
2x
Z
dx =
1
+∞ 1
cos 2x
cos 2x
−
+
2x
2x
2x
is also convergent. This contradicts (3.2). Therefore, the assumption that
R +∞
0
dx =
| sin x|
x
1
2
Z
1
+∞
1
dx
x
d x is convergent was wrong.
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R +∞ sin x
π
It is possible to prove that the value of this integral is 0
x d x = 2 , but this result cannot be found using
R sin x
elementary methods, because the indefinite integral
d x is a higher transcendental function (see Section 1.6).
R +∞ | sin x| x
The result is displayed in Fig. 3.14. 0
x d x diverges, which means that the area of the subgraph of
function | sinx x| in Fig. 3.14 a) is infinite on the interval h0, +∞).
R +∞ sin x
sin x
the sign on the
The situation is different in the case of the convergent 0
x d x . The function x changes
Rc
integration domain h0, +∞) (see Fig 3.14 b)). As a result the value of the function F (c) = 0 sinx x d x does not
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Improper Integral
y
1
y=
| sin x|
x
x
π
0
2π
3π
4π
a)
y
1
y=
sin x
x
x
0
π
2π
3π
4π
b)
Fig. 3.14: Relatively convergent integral
change monotonically as c increases. It increases on the interval h0, πi where the graph of sinx x is above the x -axis,
and therefore the area is added. It decreases on the interval hπ, 2πi where the graph of sinx x is below the x -axis,
and therefore the area is subtracted. It increases again on the interval h2π, 3πi, decreases again on the interval
h3π, 4πi etc. But its oscillation is smaller and smaller since, as we know from the Dirichlet Test, lim F (c)
exists. As we have already mentioned it can be proved that the value F (c) approaches the number
π
2
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c→+∞
.
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Improper Integral
Exercises
1. Determine the convergence or divergence of the following integrals:
Z +∞
Z +∞
arctan x
x+1
a)
d x,
b)
d x,
2
x
x +x+1
1
0
Z
0
d)
−∞
Z
+∞
g)
0
Z
j)
0
Z
x
d x,
3
x −1
1
2 + sin x
√
d x,
x3 + 1
sin x
√ d x,
x3
0
+∞
0
Z
+∞
h)
0
Z
k)
0
π/2
Z
tan x d x,
m)
Z
e)
1
x
√
d x,
1 + x4
√
4
x+1
√ d x,
(x + 1) x
ex
√
d x,
1 − x2
−1
Z
f)
−∞
cos x
d x,
x
1
ln x
d x,
x+1
0
−1
x 2 e−x
√
d x.
x+1
Z
+∞
0
Z
l)
0
o)
0
(x 2 + 1) d x
√
,
x2 x4 + 1
1
Z
i)
π/2 √
2. Determine the convergence or divergence of the following integrals:
Z 0
Z +∞ −x
x 2 e−x
e
√ d x,
a)
d x,
b)
2
x
−1 (x + 1)
0
Z π
Z π
1
1
√
d)
d x,
e)
d x,
sin
x
sin x
0
0
Z +∞
Z +∞
1
1
g)
d x,
h)
d x,
ln
x
2
+
sin x
1
−∞
x−2
d x,
x3 + 1
0
Z
tan x d x,
n)
+∞
Z
c)
c)
0
Z
f)
0
Z
π
e−x
d x,
x
1
d x,
sin2 x
+∞
i)
2
x 2 e−x d x.
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−∞
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329
Improper Integral
3. Prove that the following integrals are absolutely convergent:
Z +∞
Z +∞
sin x
e−|x| cos x d x,
a)
d
x,
b)
2
1
+
x
0
−∞
Z +∞
Z 1
sin x1
√
d)
sin x e−x d x,
e)
d x,
3
x
0
0
Z +∞
Z 0
cos 1 d x
sin ln x
√ x
√
,
h)
d x,
g)
2
1−x
x3 + 1
0
−1
4. Prove that the following integrals are convergent:
Z +∞
Z +∞
cos x
sin x
√ d x,
b)
d x,
a)
x
x
1
0
Z
+∞
sin x 2 d x,
d)
Z
0
Z
cos x 2 d x,
−∞
+∞
g)
1
(x − 1) sin x
d x,
x(x + 1)
+∞
f)
−∞
+∞
Z
i)
−∞
Z
sin3 x
d x,
x
+∞
cos x sin x
√
d x,
1 + x2
+∞
sin x ecos x
√
d x.
4
x2 + 1
0
Z
f)
0
+∞
h)
1
cos x d x
√
,
3
x3 + 1
Z
cos ex
d x,
1 + x4
√
cos |x| d x
.
x 4 + 2x 2 + 3
+∞
c)
0
Z
2
cos x 2 e−x d x,
Z
+∞
e)
+∞
Z
c)
i)
0
Hint: In d) and e) you must first apply the substitution x 2 = t .
5. Evaluate the principal value of the following integrals:
Z +∞
Z
a)
e−|x| + x 3 d x,
b)
−∞
Z
2
d)
−2
3
d x,
x(x + 3)
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1
−1
Z
Contents
dx
,
x3
Z
c)
0
∞
Z
sgn x d x,
e)
−∞
2
f)
0
x2
1
d x,
−1
x2
4
d x,
−4
3
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Improper Integral
Z
+∞
g)
Z
xe|x| + x 2 e−|x| d x,
+∞
Z
x cos x d x,
h)
−∞
−∞
1
−1
dx
.
x2
e)
C,
e)
0,
i)
Answers to Exercises
D indicates that the integral is divergent, C that it is convergent.
1. a)
D,
b)
D,
c)
C,
d)
C,
e)
D,
f)
C,
g)
C,
h)
C,
i)
D,
j)
C,
k)
C,
l)
C,
m)
D,
n)
C,
o)
C.
2. a)
D,
b)
C,
c)
D,
d)
D,
f)
D,
g)
D,
h)
D,
i)
C.
5. a)
2,
b)
0,
c)
−
− ln 5,
g)
4,
h)
0,
f)
1
ln 3,
2
d)
− ln 5,
i)
does not exist
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Improper Integral
Self-Test
1. Evaluate the following improper integrals:
Z 1
Z +∞
1
1
√ d x,
a)
b)
d x,
x
x
0
1
2. Evaluate the following improper integrals:
Z 6
dx
p
a)
d x,
b)
3
(4 − x)2
2
+∞
Z
c)
1
+∞
Z
xe
−x 2
Z
1
d x,
x2
1
d)
−1
Z
d x,
+∞
c)
0
dx
√
.
3
x2
e−x d x.
0
3. Evaluate the following improper integrals:
Z 1 2
Z +∞
x +3
dx
√ d x,
a)
b)
,
2
x
x +2
0
2
Z +∞
Z +∞
dx
dx
d)
,
,
e)
x
2
e
1
−∞ x + 2x + 2
Z
+∞
dx
,
x+2
+∞
a3d x
.
a2 + x 2
c)
0
Z
f)
−∞
Answers to Self-Test
1. a)
2,
2. a)
1
,
2
3. a)
d)
32
,
5
−1
,
e
b)
∞,
c)
b)
1,
√
3
6 2,
d)
6.
c)
1.
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b)
√
π arctan 2
√
−
,
2
2
c)
+∞,
e)
π,
f)
a 2 π sgn a.
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Chapter 4
Numerical Methods of Solving Definite
Integrals
The only method we have at our disposal to find the value of the definite integral is the Fundamental
Theorem of Calculus. As we already mentioned in previous chapters its usage has several drawbacks.
Let us recall the primary concerns.
Contents
• The antiderivative of the integrand does not exist in the set of elementary functions.
• The antiderivative can be theoretically found in the set of elementary functions but the calculation is
very complicated and time-consuming.
• The antiderivative of a very simple integrand need not exist (for example, this happens if at some point
different one-sided limits exist). This problem can sometimes be solved by separating the integration
domain, or using the generalized Fundamental Theorem of Calculus (Formula (2.15)).
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Numerical Methods of Solving Definite Integrals
Finally, it is necessary to take into account that in application the integrand is often obtained such
that we only have a finite table of values of the function to be integrated, as for example when we take
measurement at discrete values of the independent variable. Because of this we only know a finite set
of points of the graph of this function but have no analytic formula of it. The use of the Fundamental
Theorem of Calculus is thus not possible.
The goal of this chapter is to familiarize you with the methods of numerical integration. These will
enable us to find the value of the definite integral with some accuracy, when we do not have the formula
of the integrand or are not able to find the antiderivative. If the values of the integrand are obtained by
measurement, and therefore inexact, we do not mind that the numerical result is only an approximation.
Powerful computers and efficient programs, such as the above mentioned Maple, Mathematica, Matlab, Mathcad and others can now evaluate the approximate value of the definite integral with great accuracy. So it may seem an irrelevant topic. But we must remember that these programs involve some
algorithms, that help approximate the value of the definite integrals. Users should therefore have at least
a basic understanding of these methods so they understand what can be expected of the programs. Otherwise, serious problems can occur and erroneous results may be obtained. In general, this topic is rather
complicated and forms a large part of numerical mathematics. Three simple methods will be presented
in this text that can be used in manual calculations, or easily programmed. Users that need to create a
program which includes the integration of “wildly” behaving functions have to consult special textbooks.
Contents
Rb
The core of evaluating a f (x) d x is replacing function f by another function g having approximately the same function values as integrand f and being easier to use in integration. Geometrically
defiRb
nite integral expresses (for nonnegative functions) the area of subgraph so we may expect that a g(x) d x
gives approximately the same result, that is
Z b
Z b
f (x) d x =
g(x) d x + R,
a
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where the error R is a small number that can be neglected. In numerical mathematics, we must keep in
mind that the total error consists of several components:
• error of method (as with the above mentioned R )—we knowingly make it to simplify the problem,
• rounding error—calculator or computer rounding (unless they work symbolically),
• error of input data—if the input data were obtained by measurement.
Polynomials belong to functions that are easy to integrate. But if the integration domain ha, bi is
long, the request that the values of function g do not differ to much from those of f can be only satisfied in
general if the degree of polynomial g is very high. This has a negative impact namely on rounding errors.
To handle it we usually break up the interval ha, bi into a number
R b of small subintervals and replace f
separately on each subinterval by a different polynomial. Then a g(x) d x is broken up corresponding
to these subintervals. Formulae obtained this way are called composite.
Subsequently, we split the interval ha, bi into n subintervals of the same length, where n ∈ N. Such
a partition is called equidistant. We denote xi , i = 0, . . . , n the points of the partition and h > 0 their
common length. Further, we denote yi the value of the integrand at xi . Thus
h=
b−a
,
n
xi = a + ih,
yi = f (xi ),
i = 0, . . . , n.
We call xi tabular points or nodes, and yi nodal values. The formulae of numerical integration are called
quadrature formulae.
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4.1 Rectangle method
The simplest choice for function g is a constant. For example, we choose for this constant the function value of the integrand at the left end point of the interval hxi , xi+1 i, that is g(x) = f (xi ) = yi .
Integrating we get the rectangle formula
Z xi+1
Z xi+1
Z xi+1
.
f (x) d x =
g(x) d x =
yi d x =
xi
xi
x
i
xi+1
= yi x xi = yi (xi+1 − xi ) = yi h,
as xi+1 − xi = a + (i + 1)h − (a + ih) = h.
From this
Z b
Z x1
Z
f (x) d x =
f (x) d x +
a
x0
x2
Z
xn
f (x) d x + · · · +
x1
f (x) d x =
xn−1
= y0 h + y1 h + · · · + yn−1 h + R(h)
so rearranging the result and neglecting the error R(h) we obtain the composite rectangle formula
Z
b
.
f (x) d x = h(y0 + y1 + · · · + yn−1 ).
(4.1)
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The geometrical meaning for n = 6 is shown in Fig. 4.1. It is the sum of the areas of rectangles with
bases of the same length h and heights y0 , y1 , . . . , yn−1 (in general, some can be negative).
The rectangle method is convergent for any integrable function f . This means that for n → +∞,
that is h → 0 there is lim R(h) = 0. We obtain this from the fact that the right-hand side of (4.1) is an
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y
a = x0
y = f (x)
x1
x2
x3
x4
x5
x
x6 = b
Fig. 4.1: Rectangle method
Usually we want to know an error estimate. It can be proved that for a function f , with the continuous
first derivative, a number ξ ∈ ha, bi exists such that
R(h) =
h
(b − a)f 0 (ξ ).
2
(4.2)
We do not know the true value ξ . But the formula can be used to find an upper estimate of the error
replacing |f 0 (ξ )| by the maximal value of |f 0 (x)| on the interval ha, bi. In concrete examples finding
the maximum can be very difficult. Moreover, an estimate is often very pessimistic, and the true error
usually much smaller. Therefore, the value of such an estimate should not be too high.
Using Formula (4.2) for the error shows that the rectangle formula is exact for constant functions,
because for these f 0 (ξ ) = 0 for any ξ . This conclusion is evident even without this formula.
The use of rectangle formula is very easy but accuracy is naturally low and other more efficient
methods are used. These are presented below.
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4.2 Trapezoidal method
In this method we replace function f on the interval hxi , xi+1 i by linear function g the graph of which
is a line passing through points [xi , yi ] and [xi+1 , yi+1 ]. Using the slope form of the equation of the line
we obtain
yi+1 − yi
yi+1 − yi
(x − xi ),
that is
y = yi +
(x − xi ).
g : y = yi +
xi+1 − xi
h
Integrating it we get the trapezoidal formula
Z xi+1
Z xi+1
Z
.
f (x) d x =
g(x) d x =
xi
xi+1 yi+1 − yi
yi +
(x − xi ) d x =
h
xi
xi
x
yi+1 − yi 2 xi+1
+
= yi x xi+1
(x
−
x
)
= yi (xi+1 − xi ) +
i
xi
i
2h
yi+1 − yi 2 h
yi+1 − yi
(xi+1 − xi )2 = yi h +
h = (yi + yi+1 ).
+
2h
2h
2
For positive yi and yi+1 this result agrees with the secondary school formula for the area of a trapezium, which states that the area of a trapezium equals the sum of the lengths of bases multiplied by half
the height. We obtain this trapezium, by replacing the graph of f on the interval hxi , xi+1 i, by a line (see
Fig. 4.2).
From the previous formula we have
Z b
Z x1
Z x2
Z xn
f (x) d x =
f (x) d x +
f (x) d x + · · · +
f (x) d x =
a
x0
=
x1
xn−1
h
h
h
(y0 + y1 ) + (y1 + y2 ) + · · · + (yn−1 + yn ) + R(h)
2
2
2
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y
y = f (x)
a = x0
x1
x2
x3
x4
x5
x
x6 = b
Fig. 4.2: Trapezoidal method
so rearranging the result and neglecting the error R(h) we obtain the composite trapezoidal formula
Z
a
b
. h
f (x) d x = (y0 + 2y1 + · · · + 2yn−1 + yn ).
2
(4.3)
The geometrical meaning for n = 6 is shown in Fig. 4.2. If the values yi are positive, it is the sum of the
areas of those trapezia having the same height.
It is easy to prove again that the trapezoidal method is convergent for any integrable function f , that
is lim R(h) = 0, because the right-hand side of (4.3) is the arithmetic mean of two special integral sums
h→0
Contents
y0 h + y1 h + · · · + yn−1 h
and
y1 h + y2 h + · · · + yn h.
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As to the error estimate, it is possible to prove (see [3, p. 643]) that if the function f has a continuous
second derivative, there exists a number ξ ∈ ha, bi such that
R(h) = −
h2
(b − a)f 00 (ξ ).
12
(4.4)
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339
The same is true for the use of this formula as with the rectangle method. The previous formula shows
that the trapezoidal method is exact for linear functions f (x) = kx + q . For these functions f 00 (ξ ) = 0
for any ξ . With respect to the geometrical meaning of this method this conclusion is evident even without
this formula.
4.3 Simpson’s rule
Using this method, the number of subintervals must be even because a parabola replacing f is uniquely
determined by a triplet of points. Let us denote n = 2m, where m ∈ N. We will replace the function on
intervals hx0 , x2 i, hx2 , x4 i,. . . , hx2m−2 , x2m i. On each subinterval hxi−1 , xi+1 i with midpoint xi , where
i = 1, 3, . . . , 2m − 1, we will replace the function with a quadratic function g(x) = p + qx + rx 2
the graph of which is a parabola passing through points [xi−1 , yi−1 ], [xi , yi ], and [xi+1 , yi+1 ]. We will
prove that there exists a polynomial of degree at most two having this property, and is unique. In the
special case when the three points lie on a line we get r = 0, therefore it will not be a parabola but a line.
It is useful to express the polynomial we are looking for in powers x − xi (the evaluations that follow
are much shorter). This can be done using Taylor’s formula of a sufficiently high order so the remainder
in this formula is zero. In our case, this degree is two. We have
1 00
g (xi )(x − xi )2 =
2
= (p + qxi + rxi2 ) + (q + 2rxi )(x − xi ) + r(x − xi )2 .
g(x) = g(xi ) + g 0 (xi )(x − xi ) +
Denoting p̃ = p + qxi + rxi2 , q̃ = q + 2rxi , r̃ = r we will be looking for the polynomial in the form
g(x) = p̃ + q̃(x − xi ) + r̃(x − xi )2 . When its graph passes through points [xi−1 , yi−1 ], [xi , yi ], and
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[xi+1 , yi+1 ] we obtain the system of three equations for unknown coefficients p̃, q̃ , and r̃ :
p̃ − q̃h + r̃h2 = yi−1 ,
p̃
= yi ,
2
p̃ + q̃h + r̃h = yi+1 .
From these we easily find that there exists the unique solution
p̃ = yi ,
q̃ =
yi+1 − yi−1
,
2h
r̃ =
yi−1 − 2yi + yi+1
2h2
so
yi+1 − yi−1
yi−1 − 2yi + yi+1
(x − xi ) +
(x − xi )2 .
2
2h
2h
1
Integrating on the interval hxi−1 , xi+1 i we get Simpson’s rule
Z xi+1
Z xi+1
.
f (x) d x =
g(x) d x =
g : y = yi +
xi−1
xi−1
xi+1 yi+1 − yi−1
yi−1 − 2yi + yi+1
2
(x − xi ) +
(x
−
x
)
dx =
i
2h
2h2
xi−1
x
x
x
yi+1 − yi−1 yi−1 − 2yi + yi+1 = yi x xi+1 +
(x − xi )2 xi+1 +
(x − xi )3 xi+1 =
2
i−1
i−1
i−1
4h
6h
yi+1 − yi−1 2
yi+1 − 2yi + yi−1 3
= 2yi h +
(h − h2 ) +
(h + h3 ) =
4h
6h2
yi+1 − 2yi + yi−1
h
= 2yi h +
h = (yi−1 + 4yi + yi+1 ).
3
3
Z
=
yi +
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y
y = f (x)
a = x0
x1
x2
x3
x4
x
x6 = b
x5
Fig. 4.3: Simpson’s rule
Using the previous formula we have
Z b
Z x1
Z
f (x) d x +
f (x) d x =
x0
a
=
x2
Z
xn
f (x) d x + · · · +
x1
f (x) d x =
xn−1
h
h
(y0 + 4y1 + y2 ) + (y2 + 4y3 + y4 ) + · · · +
3
3
h
+ (y2m−2 + 4y2m−1 + y2m ) + R(h)
3
so rearranging the result and leaving aside the error R(h) we obtain the composite Simpson’s rule
Z b
. h
f (x) d x = (y0 + 4y1 + 2y2 + 4y3 + 2y4 · · · + 2y2m−2 + 4y2m−1 + y2m ) =
3
a
h
=
y0 + y2m + 4(y1 + · · · + y2m−1 ) + 2(y2 + · · · + y2m−2 ) .
(4.5)
3
The geometrical meaning for n = 6 is shown in Fig. 4.3.
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Again, it is easy to prove that Simpson’s rule is convergent for any integrable function f , that is
lim R(h) = 0, because the right-hand side of (4.5) is the arithmetic mean of three special integral sums
h→0
y0 h + y1 h + · · · + y2m−1 h,
y1 h + y2 h + · · · + y2m h,
y1 2h + y3 2h + · · · + y2m−1 2h.
The norm of partition in the first two integral sums is h, in the third is 2h.
As to the error estimate it is possible to prove (see [3, p. 645]) that if the function f has a continuous
fourth derivative, there exists a number ξ ∈ ha, bi such that
h4
(b − a)f (4) (ξ ).
(4.6)
180
The usage of Formula 4.6 entails difficulties because the forth derivative can be very complicated and
it would be even harder to find the maximum of its absolute value. But Formula (4.6) is worth noting
for another reason. It is not surprising that Simpson’s rule is exact for quadratic functions f (x) = p +
+ qx + rx 2 because on each subinterval the integrand f coincides with the replacement g , which is also
quadratic. But the error formula contains the fourth derivative. Therefore, Simpson’s rule is exact also
for cubic polynomials f (x) = p + qx + rx 2 + sx 3 because for these f (4) (ξ ) = 0 for any ξ .
This property of Simpson’s rule is displayed in Fig. 4.4. Function f represents an arbitrary cubic
polynomial and g is the corresponding quadratic function, which replaces f on one subinterval hx0 , x2 i.
Therefore, their graphs do not coincide. Using (4.6) we have
Z x2
Z x2
Z x2
f (x) d x =
g(x) d x,
⇒
[f (x) − g(x)] d x = 0,
R(h) = −
x0
therefore
x0
Z
x1
x0
Z
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[f (x) − g(x)] d x = −
x0
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[f (x) − g(x)] d x.
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y = g(x)
y = f (x)
x
x0
x1
x2
Fig. 4.4
This means that the regions between the graphs of f and g on the intervals hx0 , x1 i and hx1 , x2 i have the
same area.
Simpson’s rule is simple and relatively exact even if the number of subintervals is small. For these
reasons it is often used.
Analogously it would be possible to continue replacing the integrand by a cubic polynomial passing
through four points of its graph, etc. The quadrature formulae of this type, the three simplest of which
we already discussed, are called Newton-Cotes1 formulae. The derivation of these formulae is more and
more time-consuming. Usually Newton-Cotes higher order formulae are not used and Simpson’s rule
with a sufficient number of nodes is preferred.
There exist many other methods for numerically evaluating the definite integral. Among the most
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gral calculus and interpolation.
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important are the Gaussian quadrature formulae and Romberg’s method. One can learn about these and
other methods in [3, 23].
These methods for evaluating definite integrals enable us to also determine indefinite integrals. We know
that it is sufficient to find one antiderivative, others differ by a constant.
R x If the integrand f is continuous, due
to Corollary 2.29 the antiderivative F is given by formula F (x) = c f (t) d t , where c is a fixed point in the
integration domain. Hence values of F at particular points can be calculated using the definite integral. If we find
enough functional values, we can approximate F on the entire integration domain using one of the methods offered
by numerical mathematics.
Finally, let us remember that values of convergent improper integrals can also be approximated with the help
of the definite integral (see the beginning of Section 3.4).
Z
Example 4.1 Find an approximate value of
3
p
x 3 − 1 d x using the rectangle, trapezoidal, and Simp-
1
son’s methods. Break out the integration domain into eight subintervals.
Solution. Setting n = 8 we have nine nodes x0 , . . . , x8 . We denote a = 1, b = 3. The size of the step is
h = b−a
= 3−1
= 0.25. In the following table the necessary function values yi = f (xi ), i = 0, . . . , 8,
n
8 √
where f (x) = x 3 − 1, are given with accuracy to four decimal places.
i
0
1
2
3
4
5
6
7
8
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xi
1.0000 1.2500 1.5000 1.7500 2.0000 2.2500 2.5000 2.7500 3.0000
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yi
0.0000 0.9763 1.5411 2.0879 2.6458 3.2235 3.8242 4.4494 5.0990
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We denote InR , InT , and InS the approximate value of the given definite integral evaluated with rectangle, trapezoidal, and Simpson’s methods, respectively, using n + 1 nodes. After rounding to four decimal
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345
places we get from Formula (4.1):
.
I8R = h(y0 + y1 + y2 + y3 + y4 + y5 + y6 + y7 ) = 4.6870.
Analogously, from Formula (4.3) we get:
I8T =
h
.
(y0 + 2y1 + 2y2 + 2y3 + 2y4 + 2y5 + 2y6 + 2y7 + y8 ) = 5.3244.
2
Finally, from Formula (4.5) we get:
.
h
y0 + y8 + 4(y1 + y3 + y5 + y7 ) + 2(y2 + y4 + y6 ) = 5.3389.
3
√
Since the antiderivative of the integrand x 3 − 1 is not elementary we cannot apply the Fundamental
Theorem of Calculus to find the exact value I for comparison. Using the Maple program, for example,
.
we get I = 5.356 314 216. Therefore, both Simpson’s rule and also the trapezoidal method gave quite
good result although the number of nodes was relatively small.
N
I8S =
Finally, we look at the question of assessing the accuracy of our results. As we mentioned, for
sufficiently smooth integrands there exist errors formulae for all three of the methods we presented. They
contain an unknown number ξ being somewhere in the integration domain. So we do not know the exact
value of error. These formulae can be only used to estimate the highest absolute value of error as we
explained in the section on the rectangle method (page 336). The result is often rather pessimistic and
the true degree of accuracy is much better then the estimate.
In practical problems, namely in connection with the use of computers, the following approach is
used. A small number ε > 0 is set. Then using a composite formula an approximate value of the integral
is found for some number n of subintervals and for the double number 2n of subintervals. We denote
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these values In and I2n . If |In − I2n | < ε , the value I2n is declared to be the approximate value of the
integral. Otherwise, the value I4n is found corresponding to the quadruple number of subinterval 4n. If
|I2n − I4n | < ε, the value I24 is declared to be the approximate value. Otherwise, we again double the
number of subintervals etc. We continue until the distance of two consecutive values is smaller then ε .
In general it would be sufficient to increase the number of subintervals anyhow. But this doubling is
appropriate because the values from the previous steps can be shared (such rules are called nested).
Z 4.5
12
Example 4.2 Using Simpson’s rule evaluate the value of
sin
d x . Start with the number of
x
0
subintervals n = 30 and double this number until the difference of the last two results is in absolute
value smaller then ε = 0.01.
Solution. The integrand is graphed in Fig. 2.5 b). As this function oscillates extremely quickly near
zero it can be expected that the number of nodes needed to access the given accuracy will be high. The
integrand g is not defined at zero. We set, for example, g(0) = 0. It has no impact on the exact value of
the integral but the numeric result can be influenced if the number of nodes is small. In similar cases it
is reasonable to complete the missing value so the function is continuous. This is impossible in this case
because the limit of integrand for x → 0+ does not exist.
Evidently, this problem would be difficult to solve without a programmable calculator or a computer
with appropriate software. Using Formula (4.5) we evaluate approximate values of the integral for n =
= 30, 60, 120, . . . . Calculating with six significant figures we obtain:
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S
I30
S
I60
S
I120
.
= −0.637 770
.
= −0.672 977
.
= −0.705 837
S
I240
S
I480
S
I960
.
= −0.714 130
.
= −0.793 987
.
= −0.697 963
S
I1920
S
I3840
S
I7680
.
= −0.720 183
.
= −0.727 127
.
= −0.715 383
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S
S We have I120
− I240
= 0.008
287 < ε = 0.01. But the differences of the following pairs of values
S
S increase. Only I1920 − I3840 = 0.006 953 < ε = 0.01 again. Then a slight increase occurs. There
are troubles with the numerical evaluation of this integral. Let us have a look at the result provided by
.
Maple: I = −0.716 973.
Because the problems are clearly caused by the behaviour of the integrand in the right neighbourhood
of zero it is more appropriate to split the integration domain into two parts—the smaller left interval
containing zero, where more nodes are used, and the larger right interval, where a smaller number of
nodes is sufficient. Denoting InS (α; β) the value obtained using Simpson’s rule for n + 1 nodes of the
.
S
S
interval hα, βi we get I120
(0; 0.5) + I60
(0.5; 4.5) = −0.716 513. Comparing it with the result of
S
. This clearly proves that
Maple we can conclude that this result is much better than for example I7680
when solving a similar problem it is not possible to apply blindly and without thinking formulae found
in some handbook.
N
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4.4 Exercises to Chapter 4
1. Evaluate the following definite integrals using the rectangle, trapezoidal, and Simpson’s method with
n + 1 nodes:
Z 1p
Z 1p
4
a)
x + 1 d x, n = 6,
b)
x 3 + 1 d x, n = 6,
−1
2
Z
c)
0
p
x 1 + x 4 d x,
n = 6,
d)
0
Z
π
e)
0
Z
5
sin x
d x,
x
2
e−x d x,
g)
n = 6,
n = 4,
f)
h)
0
Z
i)
2
3
ex
d x, n = 4,
1 x
Z π
sin x
√ d x, n = 6,
x
0
Z 2
2
x 2 e−x d x, n = 8,
Z
1
3
1
d x,
ln x
Z
n = 4,
j)
2
2
(x + 1) e−x d x,
n = 4.
1
Hint: In e) and f) define the function value at x = 0 so the integrand f is continuous.
2. Using Simpson’s rule find approximate values of the given integrals for n and 2n nodes and decide if
their difference is smaller, in absolute value, then the given ε :
Z 2p
Z 3π
cos x
3
4
a)
x + 1 d x, n = 4, ε = 0.005,
b)
d x, n = 4, ε = 0.05,
x
0
π/2
Z 4
Z 2
3x − 1
2
c)
d x, n = 2, ε = 0.02,
d)
x 2 e−2x d x, n = 4, ε = 0.05.
1/3 ln 3x
1
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Hint: In c) define the function value at x = 1/3 so the integrand f is continuous.
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Numerical Methods of Solving Definite Integrals
3. Evaluate the following definite integrals and compare the results with the results obtained using the
rectangle, trapezoidal, and Simpson’s method with n + 1 nodes:
Z π/2
Z 3
a)
cos y dy, n = 4,
b)
(x 3 − 3x 2 + 1) d x, n = 6,
−π/2
π/3
−1
3
Z
Z
5 sin 4x d x,
c)
n = 6,
Z
n = 4,
0
3
e)
0
Z
ex/3 d x,
d)
0
12
d x,
2
x +9
n = 6,
f)
−1
Z 4
0
2 sin 2x d x,
g)
n = 4,
h)
0
−π/2
Z 1
n = 4,
j)
Z
2 cos y sin y dy,
k)
n = 6,
x−1
d x,
x+1
n = 8,
1
d x,
+x
n = 6,
1
4y arctan 2y dy,
l)
−π
n = 6,
x2
2
−1
Z π/3
1
d x,
x−4
3
Z
ln(x + 2) d x,
i)
1
Z
n = 4.
−1
4. Using an appropriate Computer Algebra System (Maple, Mathematica, Matlab, Mathcad, etc.) evaluate the following integrals:
Z 3
Z 3
a)
sin x 2 d x,
b)
cos x 2 d x,
2
Z
2
8π
cos
c)
0
Z
e)
0
5π
1
d x,
x
sin x 2
√ d x,
x
Z
8
(x − 3)
d)
p
5
x 3 − 7 d x,
2
Z
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cos(x ln x) d x.
f)
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350
Numerical Methods of Solving Definite Integrals
Answers to Exercises
Numerical calculations were made using Maple with the accuracy set to five significant figures (in all
calculations). So using a different program or calculator the obtained results can somewhat differ.
1. In the sequel, IR , IT , and IS denotes the value of the integral obtained using the rectangle, trapezoidal, and
Simpson’s method, respectively.
a) IR = 2.2051,
IT = 2.2051,
IS = 2.1786,
b)
IR = 1.0794,
IT = 1.1139,
IS = 1.1114,
c)
IR = 3.3732,
IT = 4.7476,
IS = 4.6473,
d)
IR = 7.1368,
IT = 8.1310,
IS = 8.0409,
e)
IR = 2.1065,
IT = 1.8520,
IS = 1.8520,
IT = 1.9596,
IS = 1.9340,
f (0) = 1,
f)
IR = 2.2215,
f (0) = 0,
g)
IR = 1.5144,
IT = 0.88942,
IS = 0.76764,
h)
IR = 0.25138,
IT = 0.23296,
IS = 0.23325,
i)
IR = 1.1890,
IT = 1.1224,
IS = 1.1185,
j)
IR = 0.39988,
IT = 0.31478,
IS = 0.30989.
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351
Numerical Methods of Solving Definite Integrals
2. In is the result obtained using n + 1 nodes.
a) I4 = 2.9111,
I8 = 2.9109,
c)
I2 = 10.499,
I4 = 10.529,
yes,
b)
I4 = −0.45116,
I8 = −0.45898,
no,
no ,
d)
I4 = 0.040673,
I8 = 0.040780,
yes.
f (1/3) = 1,
2,
IR = 1.8961,
IT = 1.8961,
IS = 2.0046,
b)
−4,
IR = −5.3333,
IT = −4,
IS = −4,
c)
1.8750,
IR = 2.1761,
IT = 1.7982,
IS = 1.8778,
d)
5.1549,
IR = 4.5373,
IT = 5.1817,
IS = 5.1549,
e)
3.1416,
IR = 3.3036,
IT = 3.1370,
IS = 3.1416,
f)
g)
−0.5108,
−2,
IR = −0.48926,
IR = −1.8961,
IT = −0.51148,
IT = −1.8961,
IS = −0.51083,
IS = −2.0046,
h)
0.7812,
IR = 0.34206,
IT = 0.74206,
IS = 0.77831,
i)
1.2958,
IR = 1.0075,
IT = 1.2821,
IS = 1.2953,
j)
0.1178,
IR = 0.12494,
IT = 0.11799,
IS = 0.11778,
k)
0.75,
IR = 0.32166,
IT = 0.62395,
IS = 0.77037,
l)
3.5355,
IR = 3.7850,
IT = 3.7850,
IS = 3.5705.
3. a)
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4. a)
d)
−0.03130,
b)
0.14967,
c)
23.582,
36.718,
e)
0.67990,
f)
0.38210.
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352
Bibliography
[1] Bronstein, M. Symbolic Integration: toward Practical Algorithms. Computer Algebra and Differential Equations, E. Tournier (Ed.), Academic Press, 1989, p. 59–85.
[2] Bronstein, M. Integration of Elementary functions. J. Symbolic Computation (1990), p. 117–173.
[3] Děmidovič, B. P. – Maron, I. A. Základy numerické matematiky. Praha: SNTL, 1966. 721 s.
[4] Došlá, Z. – Kuben, J. Diferenciální počet funkcí jedné proměnné. 2. dotisk prvního vydání. Skriptum. Brno: Masarykova univerzita v Brně, 2008. vi, 209 s. ISBN 80-210-3121-2.
[5] Fichtengoľc, G. M. Kurs differenciaľnogo i integraľnogo isčislenija, díl II. 7. vydání. Moskva:
Nauka, 1969. 800 s.
[6] Hildebrandt, T. H. Definitions of Riemann-Stieltjes Integral. Amer. Math. Monthly, vol. 45 (1938),
p. 265–278.
[7] Hošková, Š. – Kuben, J. Integrální počet funkcí jedné proměnné. Skriptum. 1. vydání. Brno: Vojenská akademie v Brně, 2004. vi, 197 s. ISBN 80-85960-75-3.
[8] Hošková, Š. – Kuben, J. – Račková, P. Integrální počet funkcí jedné proměnné. Studijní opora.
Součást projektu Operační program Rozvoje lidských zdrojů CZ.04.1.03/3.2.15.1/0016 Studijní opory s převažujícími distančními prvky pro předměty teoretického základu studia. Ostrava:
VŠB–TU, 2006. vi, 219 s. ISBN 80-248-1191-X.
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[9] Jarník, V. Integrální počet (I). 5. vydání. Praha: Academia, 1974. 243 s.
[10] Jarník, V. Integrální počet (II). 2. vydání. Praha: Academia, 1976. 764 s.
[11] Kropáč, J. – Kuben, J. Funkce gama a beta, transformace Laplaceova, Z a Fourierova. Skriptum,
3. vydání. Brno: Vojenská akademie v Brně, 2002. vi, 136 s.
[12] Kropáč, J. – Kuben, Skalární a vektorové pole, křivkový a plošný integrál. Skriptum. Brno: Vojenská akademie v Brně, 1999. v, 118 s.
[13] Kuben, J. Integral calculus for functions of a single variable. Student text. 1st edition. Brno: University of Defence, 2011. x, 227 pp. ISBN 978-80-7231-798-1.
[14] Kuben, J. – Šarmanová, P. Diferenciální počet funkcí jedné proměnné. Studijní opora. Součást projektu Operační program Rozvoje lidských zdrojů CZ.04.1.03/3.2.15.1/0016 Studijní opory s převažujícími distančními prvky pro předměty teoretického základu studia. Ostrava: VŠB–TU, 2006.
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[15] Kurzweil, J. Henstock-Kurzweil Integration: Its Relation to Topological Vector Spaces. World Scientific, Singapore 2000. 144 pp. ISBN 978-981-02-4207-7.
[16] Kurzweil, J. Integration between the Lebesgue Integral and the Henstock-Kurzweil Integral. Its
Relation to Local Convex Vector Spaces. World Scientific, Singapore 2002. 140 pp.
ISBN 978-981-23-8046-3.
[17] Lazard, D. Primitives des fonctions élémentaires. Séminaire BOURBAKI. 36 ème année, 1983–84,
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[18] Nagy, J. – Nováková, E. – Vacek, M. Integrální počet. Matematika pro vysoké školy technické, sešit
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vsb.cz/sites/mi21.vsb.cz/files/symbolicke_integrovani.pdf
[cit. 2011-06-29].
[20] Novák, V. Integrální počet funkcí jedné proměnné. Skriptum. Brno: UJEP Brno, 1980. 89 s.
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[23] Ralston, A. Základy numerické matematiky. Praha: Academia, 1973. 635 s.
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Index
A
additivity of definite integral
with respect to domain of integration, 190
with respect to integrand, 189
antiderivative, 12
generalized, 213
astroid, 252
C
cardioid, 252
choice of representatives of partition, 179
comparison test, 318
limit version, 320
constant rule
for definite integral, 189
for indefinite integral, 17
convergence of improper integral
absolute, 324
conditional, 324
relative, 324
curve
plane
coordinates of centre of mass, 261, 262
length, 231, 233
mass, 261, 262
parametric equations, 233
space
coordinates of centre of mass, 265
length, 235
mass, 265
parametric equations, 235
cycloid, 252
D
Dirichlet test, 325
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356
Index
F
function
elementary, 18
integrable, 181
primitive, 12
transcendental
higher, 153
lower, 153
Fundamental Theorem of Calculus, 195
generalized, 213
G
Gaussian quadrature formulae, 344
H
helix, 253
homogeneity of the definite integral, 189
I
integral
definite, 181
as function of lower limit, 216
as function of upper limit, 215
Henstock-Kurzweil, 182
Lebesgue, 182
Newton, 217
Riemann, 182
improper
absolutely convergent, 324
conditionally convergent, 324
convergent, 277, 286, 296
divergent, 277, 286, 296
principal value, 306
relatively convergent, 324
indefinite, 12
integral sum, 180
integrand, 13, 181
integration by parts
for definite integral, 202, 216
for indefinite integral, 39
integration constant, 14
integration domain, 181
L
limit
lower, 181
upper, 181
limits of integration, 181
M
mean value of function, 193
Mean Value Theorem of Integral Calculus, 193
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Index
N
Newton-Cotes formulae, 343
Newton-Leibniz formula, 195
generalized, 213
nodal value, 334
node, 334
norm of the partition, 179
P
partial fraction, 79
first type, 79
second type, 79
partition of interval, 178
equidistant, 179, 334
plane figure
coordinates of centre of mass, 266
mass, 266
Q
quadrature formula, 334
R
rectangle
curvilinear, 225
area, 226
rectangle formula, 335
composite, 335
Romberg’s method, 344
S
semicubical parabola, 252
Simpson’s rule, 340
composite, 341
singular point, 285, 289, 295
solid of revolution, 236
area of lateral surface, 237, 245
lateral surface, 236
volume, 236, 245
subgraph of function, 223
area, 224, 245
substitution, 57
linear, 63
substitution method
for definite integral, 205
for indefinite integral
the first, 56
the second, 67
sum rule
for definite integral, 189
for indefinite integral, 17
system moment
of curve, 261, 265
of figure, 266
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358
T
torus, 253
trapezoid
curvilinear, 225
area, 226
trapezoidal formula, 337
composite, 338
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