SECTION 4.5
Section 4.5
Integration by Substitution
295
Integration by Substitution
•
•
•
•
•
Use pattern recognition to find an indefinite integral.
Use a change of variables to find an indefinite integral.
Use the General Power Rule for Integration to find an indefinite integral.
Use a change of variables to evaluate a definite integral.
Evaluate a definite integral involving an even or odd function.
Pattern Recognition
In this section you will study techniques for integrating composite functions. The
discussion is split into two parts—pattern recognition and change of variables. Both
techniques involve a u-substitution. With pattern recognition you perform the
substitution mentally, and with change of variables you write the substitution steps.
The role of substitution in integration is comparable to the role of the Chain Rule
in differentiation. Recall that for differentiable functions given by y Fu and
u gx, the Chain Rule states that
d
Fgx Fgxgx.
dx
From the definition of an antiderivative, it follows that
Fgxgx dx Fgx C
Fu C.
These results are summarized in the following theorem.
THEOREM 4.12
NOTE The statement of Theorem 4.12
doesn’t tell how to distinguish between
f gx and gx in the integrand. As you
become more experienced at integration,
your skill in doing this will increase. Of
course, part of the key is familiarity with
derivatives.
Antidifferentiation of a Composite Function
Let g be a function whose range is an interval I, and let f be a function that is
continuous on I. If g is differentiable on its domain and F is an antiderivative
of f on I, then
f gxgx dx Fgx C.
If u gx, then du gx dx and
f u du Fu C.
E X P L O R AT I O N
STUDY TIP There are several techniques for applying substitution, each
differing slightly from the others.
However, you should remember that the
goal is the same with every technique—
you are trying to find an antiderivative
of the integrand.
Recognizing Patterns The integrand in each of the following integrals fits the
pattern f gxgx. Identify the pattern and use the result to evaluate the integral.
a.
2xx 2 14 dx
b.
3x 2x3 1 dx
c.
sec2 xtan x 3 dx
The next three integrals are similar to the first three. Show how you can multiply
and divide by a constant to evaluate these integrals.
d.
xx 2 14 dx
e.
x 2x3 1 dx
f.
2 sec2 x(tan x 3 dx
296
CHAPTER 4
Integration
Examples 1 and 2 show how to apply Theorem 4.12 directly, by recognizing the
presence of f gx and gx. Note that the composite function in the integrand has an
outside function f and an inside function g. Moreover, the derivative gx is present
as a factor of the integrand.
Outside function
f gxgx dx Fgx C
Inside function
EXAMPLE 1
Find
Derivative of
inside function
Recognizing the f g xgx Pattern
x 2 122x dx.
Solution Letting gx x 2 1, you obtain
gx 2x
and
f gx f x 2 1 x 2 12.
TECHNOLOGY Try using
a computer algebra system, such
as Maple, Derive, Mathematica,
Mathcad, or the TI-89, to solve the
integrals given in Examples 1 and 2.
Do you obtain the same antiderivatives that are listed in the examples?
.
From this, you can recognize that the integrand follows the f gxgx pattern. Using
the Power Rule for Integration and Theorem 4.12, you can write
f gx
g x
x 2 122x dx 1 2
x 13 C.
3
Try using the Chain Rule to check that the derivative of 13x 2 1)3 C is the
integrand of the original integral.
Try It
EXAMPLE 2
Exploration A
Technology
Video
Recognizing the f g xgx Pattern
Find 5 cos 5x dx.
Solution Letting gx 5x, you obtain
gx 5
and
f gx f 5x cos 5x.
From this, you can recognize that the integrand follows the f gxgx pattern. Using
the Cosine Rule for Integration and Theorem 4.12, you can write
f gx g x
cos 5x5 dx sin 5x C.
.
You can check this by differentiating sin 5x C to obtain the original integrand.
Try It
Exploration A
SECTION 4.5
Integration by Substitution
297
The integrands in Examples 1 and 2 fit the f gxgx pattern exactly—you only
had to recognize the pattern. You can extend this technique considerably with the
Constant Multiple Rule
kf x dx k f x dx.
Many integrands contain the essential part (the variable part) of gx but are missing
a constant multiple. In such cases, you can multiply and divide by the necessary
constant multiple, as shown in Example 3.
EXAMPLE 3
Find
Multiplying and Dividing by a Constant
xx 2 12 dx.
Solution This is similar to the integral given in Example 1, except that the integrand
is missing a factor of 2. Recognizing that 2x is the derivative of x 2 1, you can let
gx x 2 1 and supply the 2x as follows.
xx 2 12 dx x 2 12
122x dx
gx
f gx
1
x 2 12 2x dx
2
1 x 2 13
C
2
3
1
x 2 13 C
6
.
Try It
Multiply and divide by 2.
Exploration A
Constant Multiple Rule
Integrate.
Simplify.
Exploration B
In practice, most people would not write as many steps as are shown in Example
3. For instance, you could evaluate the integral by simply writing
1
x 2 12 2x dx
2
1 x 2 13
C
2
3
1
x 2 13 C.
6
xx 2 12 dx NOTE Be sure you see that the Constant Multiple Rule applies only to constants. You cannot
multiply and divide by a variable and then move the variable outside the integral sign. For
instance,
x 2 12 dx 1
2x
x 2 12 2x dx.
After all, if it were legitimate to move variable quantities outside the integral sign, you could
move the entire integrand out and simplify the whole process. But the result would be incorrect.
298
CHAPTER 4
Integration
Change of Variables
With a formal change of variables, you completely rewrite the integral in terms of u
and du (or any other convenient variable). Although this procedure can involve more
written steps than the pattern recognition illustrated in Examples 1 to 3, it is useful for
complicated integrands. The change of variable technique uses the Leibniz notation
for the differential. That is, if u gx, then du gx dx, and the integral in
Theorem 4.12 takes the form
f gxgx dx EXAMPLE 4
Find
f u du Fu C.
Change of Variables
2x 1 dx.
Solution First, let u be the inner function, u 2x 1. Then calculate the differential
du to be du 2 dx. Now, using 2x 1 u and dx du2, substitute to obtain
2x 1 dx u
du2 Integral in terms of u
1
u12 du
2
1 u 32
C
2 32
1
u32 C
3
1
2x 132 C.
3
Constant Multiple Rule
STUDY TIP Because integration is
usually more difficult than differentiation,
you should always check your answer to
an integration problem by differentiating.
For instance, in Example 4 you should
.
differentiate 132x 132 C to verify
that you obtain the original integrand.
Try It
EXAMPLE 5
Find
Antiderivative in terms of u
Simplify.
Antiderivative in terms of x
Exploration A
Change of Variables
x2x 1 dx.
Solution As in the previous example, let u 2x 1 and obtain dx du2.
Because the integrand contains a factor of x, you must also solve for x in terms of u,
as shown.
x u 12
u 2x 1
Solve for x in terms of u.
Now, using substitution, you obtain
x2x 1 dx u 1 12 du
u
2
2
1
u32 u12 du
4
1 u52 u32
C
4 52
32
1
1
2x 152 2x 132 C.
10
6
.
Try It
Exploration A
Exploration B
Open Exploration
SECTION 4.5
Integration by Substitution
299
To complete the change of variables in Example 5, you solved for x in terms of
u. Sometimes this is very difficult. Fortunately it is not always necessary, as shown in
the next example.
EXAMPLE 6
Find
Change of Variables
sin2 3x cos 3x dx.
Solution Because sin2 3x sin 3x2, you can let u sin 3x. Then
du cos 3x3 dx.
Now, because cos 3x dx is part of the original integral, you can write
du
cos 3x dx.
3
STUDY TIP When making a change
of variables, be sure that your answer is
written using the same variables as in
the original integrand. For instance, in
Example 6, you should not leave your
answer as
1 3
9u
Substituting u and du3 in the original integral yields
sin2 3x cos 3x dx u2
du
3
1 2
u du
3
1 u3
C
3 3
1
sin3 3x C.
9
C
but rather, replace u by sin 3x.
You can check this by differentiating.
d 1 3
1
sin 3x 3sin 3x2cos 3x3
dx 9
9
sin2 3x cos 3x
.
Because differentiation produces the original integrand, you know that you have
obtained the correct antiderivative.
Try It
Exploration A
The steps used for integration by substitution are summarized in the following
guidelines.
Guidelines for Making a Change of Variables
1. Choose a substitution u gx. Usually, it is best to choose the inner part of
a composite function, such as a quantity raised to a power.
2. Compute du gx dx.
3. Rewrite the integral in terms of the variable u.
4. Find the resulting integral in terms of u.
5. Replace u by gx to obtain an antiderivative in terms of x.
6. Check your answer by differentiating.
300
CHAPTER 4
Integration
The General Power Rule for Integration
One of the most common u-substitutions involves quantities in the integrand that are
raised to a power. Because of the importance of this type of substitution, it is given a
special name—the General Power Rule for Integration. A proof of this rule follows
directly from the (simple) Power Rule for Integration, together with Theorem 4.12.
THEOREM 4.13
The General Power Rule for Integration
If g is a differentiable function of x, then
gx n1
C,
n1
gxn gx dx n 1.
Equivalently, if u gx, then
un du EXAMPLE 7
a.
un1
C,
n1
n 1.
Substitution and the General Power Rule
33x 14 dx u4
u55
du
3x 143 dx u1
b.
2x 1x 2 x dx 3x 2x3 2 dx x 2 x1 2x 1 dx Suppose you were asked to find one
of the following integrals. Which one
would
you choose? Explain your
.
reasoning.
a.
x3 1 dx
or
x 2x3 1 dx
b.
tan3x sec 2 3x dx
tan 3x dx
d.
4x
dx 1 2x 22
x3 212 3x 2 dx u11
1 2x 22 4x dx u2
e.
x3 232
2
C x 3 232 C
32
3
du
Try It
Exploration A
x 2 x2
C
2
u3232
du
u2
E X P L O R AT I O N
u22
du
u12
c.
3x 15
C
5
1 2x 21
1
C
C
1
1 2x2
u33
du
cos2 x sin x dx cos x2 sin x dx cos x3
C
3
Exploration B
Some integrals whose integrands involve quantities raised to powers cannot be
found by the General Power Rule. Consider the two integrals
or
xx2 12 dx
and
x 2 12 dx.
The substitution u x 2 1 works in the first integral but not in the second. In the
second, the substitution fails because the integrand lacks the factor x needed for du.
Fortunately, for this particular integral, you can expand the integrand as
x 2 12 x 4 2x 2 1 and use the (simple) Power Rule to integrate each term.
SECTION 4.5
Integration by Substitution
301
Change of Variables for Definite Integrals
When using u-substitution with a definite integral, it is often convenient to determine
the limits of integration for the variable u rather than to convert the antiderivative back
to the variable x and evaluate at the original limits. This change of variables is stated
explicitly in the next theorem. The proof follows from Theorem 4.12 combined with
the Fundamental Theorem of Calculus.
THEOREM 4.14
Change of Variables for Definite Integrals
If the function u gx has a continuous derivative on the closed interval a, b
and f is continuous on the range of g, then
b
f gxgx dx a
EXAMPLE 8
gb
ga
f u du.
Change of Variables
1
Evaluate
xx 2 13 dx.
0
Solution To evaluate this integral, let u x 2 1. Then, you obtain
u x 2 1 ⇒ du 2x dx.
Before substituting, determine the new upper and lower limits of integration.
Lower Limit
Upper Limit
When x 0, u 02
1 1.
When x 1, u 12 1 2.
Now, you can substitute to obtain
1
xx 2 13 dx 0
1
2
x 2 132x dx
1
2
u3 du
1
Integration limits for x
0
2
Integration limits for u
1
1 u4 2
2 4 1
1
1
4
2
4
15 .
8
Try rewriting the antiderivative 12u44 in terms of the variable x and evaluate the
definite integral at the original limits of integration, as shown.
1 x 2 14 1
2
4
1
0
1
1
15
4
2
4
8
2
1 u4
2 4
.
Notice that you obtain the same result.
Try It
Exploration A
Video
302
CHAPTER 4
Integration
Change of Variables
EXAMPLE 9
5
Evaluate A 1
x
dx.
2x 1
Solution To evaluate this integral, let u 2x 1. Then, you obtain
u2 2x 1
u2 1 2x
u2 1
x
2
u du dx.
Differentiate each side.
Before substituting, determine the new upper and lower limits of integration.
Lower Limit
Upper Limit
When x 1, u 2 1 1.
When x 5, u 10 1 3.
Now, substitute to obtain
y
5
5
x
dx 2x 1
1
4
3
1 u2 1
u du
u
2
1
3
y=
3
1
u2 1 du
2 1
3
1 u3
u
2 3
1
1
1
93 1
2
3
16 .
3
x
2x − 1
(5, 53 )
2
(1, 1)
1
x
−1
1
2
3
4
5
.
The region before substitution has an area
16
of 3 .
Figure 4.37
Try It
Exploration A
Exploration B
Video
Geometrically, you can interpret the equation
f(u)
5
5
2
f(u) = u + 1
2
(3, 5)
1
3
2
(1, 1)
u
−1
1
2
3
4
5
−1
The region after substitution has an area
16
of 3 .
Figure 4.38
3
1
u2 1
du
2
to mean that the two different regions shown in Figures 4.37 and 4.38 have the same
area.
When evaluating definite integrals by substitution, it is possible for the upper
limit of integration of the u-variable form to be smaller than the lower limit. If this
happens, don’t rearrange the limits. Simply evaluate as usual. For example, after
substituting u 1 x in the integral
4
1
x
dx 2x 1
1
x21 x12 dx
0
you obtain u 1 1 0 when x 1, and u 1 0 1 when x 0. So,
the correct u-variable form of this integral is
0
2
1
1 u22u2 du.
SECTION 4.5
Integration by Substitution
303
Integration of Even and Odd Functions
y
Even with a change of variables, integration can be difficult. Occasionally, you can
simplify the evaluation of a definite integral (over an interval that is symmetric about
the y-axis or about the origin) by recognizing the integrand to be an even or odd
function (see Figure 4.39).
THEOREM 4.15
x
−a
Integration of Even and Odd Functions
Let f be integrable on the closed interval a, a.
a
a
Even function
a
1. If f is an even function, then
a
f x dx 2
f x dx.
0
a
2. If f is an odd function, then
y
a
Proof Because f is even, you know that f x f x. Using Theorem 4.12 with
the substitution u x produces
x
−a
f x dx 0.
0
a
a
0
f x dx 0
f udu a
a
f u du a
a
f u du 0
f x dx.
0
Finally, using Theorem 4.6, you obtain
a
Odd function
a
Figure 4.39
0
f x dx a
a
0
a
f x dx f x dx
0
a
f x dx a
f x dx 2
0
f x dx.
0
This proves the first property. The proof of the second property is left to you (see
Exercise 133).
f(x) = sin3 x cos x + sin x cos x
EXAMPLE 10
y
2
Evaluate
2
1
Integration of an Odd Function
sin3 x cos x sin x cos x dx.
Solution Letting f x sin3 x cos x sin x cos x produces
−π
4
π
4
−1
f x sin3x cosx sinx cosx
sin3 x cos x sin x cos x f x.
So, f is an odd function, and because f is symmetric about the origin over
2, 2, you can apply Theorem 4.15 to conclude that
2
.
Because f is an odd function,
.
π
2
x
2
2
f x dx 0.
2
sin3 x cos x sin x cos x dx 0.
Try It
Exploration A
Figure 4.40
Editable Graph
NOTE From Figure 4.40 you can see that the two regions on either side of the y-axis have the
same area. However, because one lies below the x-axis and one lies above it, integration
produces a cancellation effect. (More will be said about this in Section 7.1.)
304
CHAPTER 4
Integration
Exercises for Section 4.5
The symbol
indicates an exercise in which you are instructed to use graphing technology or a symbolic computer algebra system.
Click on
to view the complete solution of the exercise.
Click on
to print an enlarged copy of the graph.
In Exercises 1– 6, complete the table by identifying u and du for
the integral.
f gxgx dx
1.
2.
3.
4.
5.
6.
u gx
du gx dx
x 2x3 1 dx
x
38.
sec 2x tan 2x dx
tan2 x sec2 x dx
cos x
dx
sin2 x
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
31.
33.
1 2x42 dx
9 x 2 2x dx
8.
10.
3 dx
12.
x 2x3 14 dx
14.
tt 2 2 dt
16.
x3
5x
x4
3 1
2
dy
x4
dx x 2 8x 1
dx
In Exercises 7–34, find the indefinite integral and check the
result by differentiation.
7.
dy
4x
4x dx
16 x 2
dy
10x 2
36.
dx 1 x3
dy
x1
37.
dx x 2 2x 32
35.
5x 2 1210x dx
x 2 1
In Exercises 35–38, solve the differential equation.
x2
dx
x
dx
1 x 23
18.
20.
2
x
dx
1 x32
x
dx
1 x 2
1 3 1
1
dt
t
t2
1
dx
2x
x 2 3x 7
dx
x
2
t2 t dt
t
9 yy dy
22.
24.
26.
28.
30.
32.
34.
x 2 932x dx
Slope Fields In Exercises 39–42, a differential equation, a
point, and a slope field are given. A slope field consists of line
segments with slopes given by the differential equation. These line
segments give a visual perspective of the directions of the
solutions of the differential equation. (a) Sketch two approximate
solutions of the differential equation on the slope field, one of
which passes through the given point. (To print an enlarged copy
of the graph, select the MathGraph button.) (b) Use integration
to find the particular solution of the differential equation and use
a graphing utility to graph the solution. Compare the result with
the sketches in part (a).
39.
dy
x4 x2
dx
40.
2, 2
3 1 2x 24x dx
dy
x2x3 12
dx
1, 0
y
x2
x3
5 dx
y
4
3
2
x4x 2 33 dx
t 3t 4 5 dt
u2u3
2 du
x3
dx
1 x 42
2
−2
−1
41.
2
x
−2
2
x
dx
16 x32
x3
dx
1 x 4
1
x2 dx
3x2
1
dx
2x
t 2t2
dt
t
t3
1
2 dt
3
4t
x
−2
dy
x cos x 2
dx
0, 1
42.
dy
2 sec2x tan2x
dx
0, 1
y
y
4
3
2 y8 y32 dy
x
−4
4
−4
x
−3
3
−3
SECTION 4.5
43.
45.
47.
49.
50.
51.
53.
55.
sin x dx
sin 2x dx
44.
46.
1
1
cos d
2
48.
75.
0
9
4x 3 sin x 4 dx
77.
1
2
cos 6x dx
79.
81.
82.
3
tan4 x sec2 x dx
52.
csc2 x
dx
cot 3 x
54.
56.
tan x sec2 x dx
sin x
dx
cos3 x
x
csc2
dx
2
57. fx cos
83.
0, 3
3
,
4
4
2 , 2
59. fx sin 4x
60. fx sec22x
61. fx 2x4x2 102
In Exercises 63–70, find the indefinite integral by the method
shown in Example 5.
65.
67.
69.
xx 2 dx
64.
x 21 x dx
66.
x2 1
dx
2x 1
x
dx
x 1) x 1
68.
70.
1
72.
2x 2x 3 1 dx
74.
1
2
73.
1
1
x
dx
2x 1
2x
cos
dx
3
x cos x dx
84.
48
dy
dx 3x 53
7
6
5
4
y
f
(0, 4)
(−1, 3)
2
1
x
−4 −3 −2
85.
6
5
4
f
x
−6 −5 −4 −3 −2 −1
1 2
−2
1 2 3 4
dy
2x
dx 2x2 1
86.
dy
9x2
4x 3
dx
3x 132
y
y
f
8
6
4
2
7
6
5
4
3
(5, 4)
x
−8 −6 −4
4 6 8
f
(0, 2)
−4
−6
−8
x
−3 −2 −1
1 2 3 4 5
x 12 x dx
In Exercises 87–92, find the area of the region. Use a graphing
utility to verify your result.
2x 1
dx
x 4
87.
7
2
1
0
88.
2
x1 x 2 dx
3 x 2 dx
x2 y
y
16
80
12
60
8
40
4
x 2x 3 82 dx
6
3 x 1 dx
x
0
3 t 4 dt
t
4
xx 2 13 dx
3 4 x 2 dx
x
0
5
80.
dx
x2x 1 dx
In Exercises 71–82, evaluate the definite integral. Use a graphing
utility to verify your result.
71.
78.
y
2, 10
2, 7
62. fx 2x8 x2
dx
13, 1
58. fx sec x tan x
63.
2
dy
18x22x3 12
dx
Point
x
2
0
2
x
1 2x 2
Differential Equations In Exercises 83–86, the graph of a
function f is shown. Use the differential equation and the given
point to find an equation of the function.
In Exercises 57–62, find an equation for the function f that has
the given derivative and whose graph passes through the given
point.
Derivative
76.
x 12 x dx
0
2
sec1 x tan1 x dx
1
x 1 x
2
1
dx
2x 1
1
2
x sin x 2 dx
sin 2x cos 2x dx
cot2 x dx
4
In Exercises 43–56, find the indefinite integral.
305
Integration by Substitution
20
x
2
4
6
8
−2
x
2
4
6
306
CHAPTER 4
Integration
89. y 2 sin x sin 2x
90. y sin x cos 2x
y
106. Use the symmetry of the graphs of the sine and cosine
functions as an aid in evaluating each definite integral.
y
2
(a)
3
1
(c)
π
4
π
2
23
sec2
2
3π
4
π
x
π
2
π
2x dx
92.
csc 2x cot 2x dx
4
3
2
2
π
2
3π
4
π
0
π
16
π
8
3π
16
π
4
x
xx 3 dx
96.
x3x
2 dx
x 2x 1 dx
1
3
cos
0
d
6
2
98.
sin 2x dx
0
Writing In Exercises 99 and 100, find the indefinite integral in
two ways. Explain any difference in the forms of the answers.
99.
100.
2x 12 dx
sin x cos x dx
In Exercises 101–104, evaluate the integral using the properties
of even and odd functions as an aid.
2
101.
x 2x 2 1 dx
102.
2
2
103.
2
2
xx 2 13 dx
2
sin2 x cos x dx
104.
2
sin x cos x dx
2
8
105. Use 02 x 2 dx 3 to evaluate each definite integral without
using the Fundamental Theorem of Calculus.
0
(a)
0
2
x 2 dx
(b)
2
2
(c)
sin 3x cos 3x dx
x5 x 23 dx u3 du
where u 5 x 2.
110. Without integrating, explain why
2
2
xx 2 12 dx 0.
5
3
97.
108.
0
7
95.
2
94.
x3 6x 2 2x 3 dx
109. Describe why
x
x
dx
2x 1
sin x cos x dx
2
Writing About Concepts
In Exercises 93–98, use a graphing utility to evaluate the
integral. Graph the region whose area is given by the definite
integral.
4
4
1
π
4
(d)
4
107.
y
3
2
cos x dx
In Exercises 107 and 108, write the integral as the sum of the
integral of an odd function and the integral of an even function.
Use this simplification to evaluate the integral.
12
4
cos x dx
4
2
x
4
y
93.
(b)
2
1
4
sin x dx
4
2
91.
4
4
x 2 dx
x 2 dx
2
0
(d)
2
3x 2 dx
111. Cash Flow The rate of disbursement dQdt of a 2 million
dollar federal grant is proportional to the square of 100 t.
Time t is measured in days 0 ≤ t ≤ 100, and Q is the
amount that remains to be disbursed. Find the amount that
remains to be disbursed after 50 days. Assume that all the
money will be disbursed in 100 days.
112. Depreciation The rate of depreciation dVdt of a machine is
inversely proportional to the square of t 1, where V is the
value of the machine t years after it was purchased. The initial
value of the machine was $500,000, and its value decreased
$100,000 in the first year. Estimate its value after
4 years.
113. Rainfall The normal monthly rainfall at the Seattle-Tacoma
airport can be approximated by the model
R 3.121 2.399 sin0.524t 1.377
where R is measured in inches and t is the time in months,
with t 1 corresponding to January. (Source: U.S. National
Oceanic and Atmospheric Administration)
(a) Determine the extrema of the function over a one-year
period.
(b) Use integration to approximate the normal annual rainfall.
(Hint: Integrate over the interval 0, 12.)
(c) Approximate the average monthly rainfall during the
months of October, November, and December.
SECTION 4.5
114. Sales The sales S (in thousands of units) of a seasonal
product are given by the model
S 74.50 43.75 sin
307
Integration by Substitution
(b) What is the median percent recall? That is, for what value
of b is it true that the probability of recalling 0 to b is 0.5?
y
t
6
1.5
where t is the time in months, with t 1 corresponding to
January. Find the average sales for each time period.
Pa, b
1.0
(a) The first quarter 0 ≤ t ≤ 3
(b) The second quarter 3 ≤ t ≤ 6
0.5
(c) The entire year 0 ≤ t ≤ 12
115. Water Supply A model for the flow rate of water at a pumping station on a given day is
Rt 53 7 sin
t
t
3.6 9 cos
8.9
6
12
where 0 ≤ t ≤ 24. R is the flow rate in thousands of gallons
per hour, and t is the time in hours.
(a) Use a graphing utility to graph the rate function and
approximate the maximum flow rate at the pumping
station.
(b) Approximate the total volume of water pumped in 1 day.
116. Electricity The oscillating current in an electrical circuit is
I 2 sin60 t cos120 t
where I is measured in amperes and t is measured in seconds.
Find the average current for each time interval.
(a) 0 ≤ t ≤
(b) 0 ≤ t ≤
(c) 0 ≤ t ≤
1
60
1
240
1
30
x
a b 0.5
1.5
1.0
Figure for 117
118. The probability that ore samples taken from a region contain
between a% and b % iron is
b
Pa, b a
1155 3
x 1 x32 dx
32
where x represents the percent of iron. (See figure.) What is
the probability that a sample will contain between
(a) 0% and 25% iron?
(b) 50% and 100% iron?
y
2
Pa, b
1
Probability In Exercises 117 and 118, the function
f x kx n1 xm,
where n > 0, m > 0, and k is a constant, can be used to represent various probability distributions. If k is chosen such that
f x dx 1
0
the probability that x will fall between a and b 0 ≤ a ≤ b ≤ 1
is
b
2
119. Temperature The temperature in degrees Fahrenheit in a
house is
t12 8
f x dx.
where t is time in hours, with t 0 representing midnight.
The hourly cost of cooling a house is $0.10 per degree.
(a) Find the cost C of cooling the house if its thermostat is set
at 72F by evaluating the integral
20
a
C 0.1
117. The probability that a person will remember between a% and
b % of material learned in an experiment is
a
15
x1 x dx
4
where x represents the percent remembered. (See figure.)
(a) For a randomly chosen individual, what is the probability
that he or she will recall between 50% and 75% of the
material?
72 12 sin
8
t 8
72 dt. (See figure.)
12
T
Temperature (in °F)
b
Pa, b x
b1
T 72 12 sin
1
Pa, b a
0≤ x≤ 1
84
78
72
66
60
Thermostat setting: 72°
t
2
4
6
8
10 12 14 16 18 20 22 24
Time (in hours)
308
CHAPTER 4
Integration
(b) Find the savings from resetting the thermostat to 78F by
evaluating the integral
18
72 12 sin
C 0.1
10
t 8
78 dt.
12
True or False? In Exercises 125–130, determine whether the
statement is true or false. If it is false, explain why or give an
example that shows it is false.
125.
(See figure.)
126.
Temperature (in °F)
T
84
2x 12 dx 132x 13 C
x x 2 1 dx 12x 2 13x3 x C
10
10
b
72
66
128.
60
Thermostat setting: 78°
6
4
8
10 12 14 16 18 20 22 24
F 100,000 1 sin
2 t 60
365
Consider the functions f and g, where
121. Graphical Analysis
t
f x 6 sin x cos2 x
and
gt 130.
f x dx.
0
(a) Use a graphing utility to graph f and g in the same
viewing window.
(b) Explain why g is nonnegative.
(e) Consider the function
f x dx.
Use a graphing utility to graph h. What is the relationship
between g and h? Verify your conjecture.
sinin
by evaluating an appropriate
n
definite integral over the interval 0, 1.
n
lim
n→ i1
123. (a) Show that 0 x21 x5 dx 0 x51 x2 dx.
1
1
(b) Show that 0 xa1 xb dx 0 xb1 xa dx.
1
2
124. (a) Show that 0
1
2
sin2 x dx 0
2
0
(b) Show that
positive integer.
sinn
x dx b
ca
f cx dx.
a
132. (a) Verify that sin u u cos u C u sin u du.
2
(b) Use part (a) to show that 0 sinx dx 2.
133. Complete the proof of Theorem 4.15.
134. Show that if f is continuous on the entire real number line,
then
b
bh
f x h dx f x dx.
ah
Putnam Exam Challenge
135. If a0, a1, . . ., an are real numbers satisfying
an
a0 a1 . . .
0
1
2
n1
t
2
f x dx c
a
(d) Does each of the zeros of f correspond to an extremum of
g? Explain.
sin x dx
a
131. Assume that f is continuous everywhere and that c is a
constant. Show that
(c) Identify the points on the graph of g that correspond to the
extrema of f.
ht b2
sin2 2x cos 2x dx 13 sin3 2x C
cb
where F is measured in pounds and t represents the time in
days, with t 1 corresponding to January 1. The manufacturer
wants to set up a schedule to produce a uniform amount of
fertilizer each day. What should this amount be?
bx 2 d dx
0
129. 4 sin x cos x dx cos 2x C
Time (in hours)
120. Manufacturing A manufacturer of fertilizer finds that
national sales of fertilizer follow the seasonal pattern
ax3 bx 2 cx d dx 2
sin x dx a
t
2
122. Find
10
127.
78
cos2 x dx.
2
0
cosn
x dx, where n is a
show that the equation a0 a1 x a 2 x 2 . . . an x n 0
has at least one real zero.
136. Find all the continuous positive functions f x, for 0 ≤ x ≤ 1,
such that
1
f x dx 1
0
1
f xx dx 0
1
f xx2 dx 2
0
where is a real number.
These problems were composed by the Committee on the Putnam Prize Competition.
© The Mathematical Association of America. All rights reserved.