Section 6-1
Antiderivatives and Indefinite Integrals
➤
➤
➤
Antiderivatives
Indefinite Integrals: Formulas and Properties
Applications
Many operations in mathematics have reverses—compare addition and subtraction, multiplication and division, and powers and roots. We now know how
to find the derivatives of many functions. The reverse operation, antidifferentiation (the reconstruction of a function from its derivative) will receive our attention in this and the next two sections. A function F is an antiderivative of a
function f if F¿(x) = f(x). We develop special antiderivative formulas in much
the same way as we developed derivative formulas.
369
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370 Chapter 6
Integration
➤ Antiderivatives
1
(A) Find three antiderivatives of 2x.
(B) How many antiderivatives of 2x exist, and how are they related to
each other?
(C) What notation would you use to represent all antiderivatives
of 2x?
x3
The function F(x) =
is an antiderivative of the function f(x) = x2
3
because
d x3
a b = x2
dx 3
However, F(x) is not the only antiderivative of x2. Note also that
d x3
a
+ 2 b = x2
dx 3
d x3
a
- b = x2
dx 3
d x3
a
+ 25 b = x2
dx 3
Therefore,
x3
+2
3
x3
-
3
x3
+ 25
3
are also antiderivatives of x2, because each has x2 as a derivative. In fact, it appears that
x3
+C
3
for any real number C
is an antiderivative of x2, because
d x3
a
+ C b = x2
dx 3
Antidifferentiation of a given function does not lead to a unique function, but
to an entire family of functions.
Does the expression
x3
+C
3
with C any real number
include all antiderivatives of x2? Theorem 1 (which we state without proof) indicates that the answer is yes.
THEOREM 1
On Antiderivatives
If the derivatives of two functions are equal on an open interval (a, b), then
the functions differ by at most a constant. Symbolically: If F and G are differentiable functions on the interval (a, b) and F¿(x) = G¿(x) for all x in (a, b),
then F(x) = G(x) + k for some constant k.
Insight
Suppose that F(x) is an antiderivative of f(x). If G(x) is any other antiderivative of f(x), then, by Theorem 1, the graph of G(x) is a vertical translation of
the graph of F(x) (see Section 1-2).
●
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Antiderivatives and Indefinite Integrals
Section 6-1
371
E XAMPLE 1 A Family of Antiderivatives Note that
d x2
a b =x
dx 2
(A) Find all antiderivatives of f(x) = x.
(B) Graph the antiderivative of f(x) = x that passes through the point (0, 0);
through the point (0, 1); through the point (0, 2).
(C) How are the graphs of the three antiderivatives in part (B) related?
Solution
y
(A) By Theorem 1, any antiderivative of f(x) has the form
F2(x)
F(x) =
F1(x)
F0(x)
5
where k is a real number.
(B) Because F(0) = (022) + k = k, the functions
F0(x) =
3
x
0
3
FIGURE 1
✓1
Matched
Problem
x2
+k
2
x2
,
2
F1(x) =
x2
+ 1,
2
and F2(x) =
x2
+2
2
pass through the points (0, 0), (0, 1), and (0, 2), respectively (see Fig. 1).
(C) The graphs of the three antiderivatives are vertical translations of each
other.
■
Note that
d 3
(x ) = 3x2
dx
(A) Find all antiderivatives of f(x) = 3x2.
(B) Graph the antiderivative of f(x) = 3x2 that passes through the point
(0, 0); through the point (0, 1); through the point (0, 2).
(C) How are the graphs of the three antiderivatives in part (B) related?
➤ Indefinite Integrals: Formulas and Properties
Theorem 1 states that if the derivatives of two functions are equal, then the
functions differ by at most a constant. We use the symbol
3
f(x) dx
called the indefinite integral, to represent the family of all antiderivatives of
f(x), and write
3
f(x) dx = F(x) + C
if
F¿(x) = f(x)
The symbol 1 is called an integral sign, and the function f(x) is called the
integrand. The symbol dx indicates that the antidifferentiation is performed
with respect to the variable x. (We will have more to say about the symbols 1
and dx later in this chapter.) The arbitrary constant C is called the constant of
integration. Referring to the preceding discussion, we can write
3
x2 dx =
x3
+C
3
since
d x3
a
+ C b = x2
dx 3
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372 Chapter 6
Integration
Of course, variables other than x can be used in indefinite integrals. For
example,
t2 dt =
t3
+C
3
since
d t3
a + C b = t2
dt 3
u2 du =
u3
+C
3
since
d u3
a
+ C b = u2
du 3
3
or
3
The fact that indefinite integration and differentiation are reverse operations, except for the addition of the constant of integration, can be expressed
symbolically as
d
c f(x) dx d = f(x) The derivative of the indefinite integral of f(x) is f(x).
dx 3
and
3
F¿(x) dx = F(x) + C
The indefinite integral of the derivative of F(x)
is F(x) + C.
We can develop formulas for the indefinite integrals of certain basic functions
from the formulas for derivatives that were established in Chapters 3 and 5.
FORMULAS
Indefinite Integrals of Basic Functions
For C a constant:
1.
3
2.
3
3.
xn dx =
xn +1
+ C,
n +1
n Z -1
ex dx = ex + C
1
dx = ln∑ x∑ + C,
3x
x Z0
Each formula can be justified by showing that the derivative of the righthand side is the integrand of the left-hand side (see Problems 93–96 in Exercise 6-1). Note that formula 1 does not give the antiderivative of x-1 (because
xn +1(n + 1) is undefined when n = -1), but formula 3 does.
2
Formulas 1, 2, and 3 do not provide a formula for the indefinite integral of
the function ln x. Show, however, assuming x 7 0, that
3
ln x dx = x ln x - x + C
by differentiating the right-hand side. [We will discuss a technique for deriving such indefinite integral formulas in Section 7-3.]
We can obtain properties of the indefinite integral from properties of the
derivative that were established in Chapter 3.
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Antiderivatives and Indefinite Integrals
Section 6-1
PROPERTIES
373
Of Indefinite Integrals
For k a constant:
4.
3
kf(x) dx = k
5.
3
[f(x) ; g(x)] dx =
3
f(x) dx
3
f(x) dx ;
3
g(x) dx
Property 4 states that
The indefinite integral of a constant times a function is the constant times the indefinite integral of the function.
Property 5 states that
The indefinite integral of the sum of two functions is the sum of
the indefinite integrals, and the indefinite integral of the difference of two functions is the difference of the indefinite integrals.
To establish property 4, let F be a function such that F¿(x) = f(x). Then
k
3
f(x) dx = k
3
F¿(x) dx = k[F(x) + C1] = kF(x) + kC1
and since [kF(x)]¿ = kF¿(x) = kf(x), we have
3
kf(x) dx =
3
kF¿(x) dx = kF(x) + C2
But kF(x) + kC1 and kF(x) + C2 describe the same set of functions, since C1
and C2 are arbitrary real numbers. Thus, property 4 is established. Property 5
can be established in a similar manner (see Problems 97–98 in Exercise 6-1).
Caution
It is important to remember that property 4 states that a constant factor can
be moved across an integral sign. A variable factor cannot be moved across an
integral sign:
CONSTANT FACTOR
3
5x12 dx = 5
3
VARIABLE FACTOR
x12 dx
3
xx12 dx Z x
3
x12 dx
Indefinite integral formulas and properties can be used together to find indefinite integrals for many frequently encountered functions. For example,
formula 1, for n = 0, gives
3
dx = x + C
Therefore, by property 4,
k dx = k(x + C) = kx + kC
3
Because kC is a constant, it is customary to replace it by a single symbol that
denotes an arbitrary constant (usually C), so we write
3
k dx = kx + C
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374 Chapter 6
Integration
In words,
The indefinite integral of a constant function with value k is
kx C.
Similarly, using property 5, and then formulas 2 and 3,
3
a ex +
1
1
b dx =
ex dx +
dx
x
3
3x
= ex + C1 + ln∑ x∑ + C2
Because C1 + C2 is a constant, it is usually replaced by the symbol C, and we
write
1
a ex + b dx = ex + ln∑x∑ + C
x
3
E XAMPLE 2 Using Indefinite Integral Properties and Formulas
(A)
3
(B)
3
(C)
(D)
3
3
5 dx = 5x + C
9ex dx = 9
5t7 dt = 5
3
3
ex dx = 9ex + C
t7 dt = 5
t8
5
+ C = t8 + C
8
8
(4x3 + 2x - 1) dx =
3
4x3 dx +
3
2x dx -
3
dx
x3 dx + 2 x dx dx
3
3
3
4x4
2x2
=
+
-x +C
4
2
=4
Property 4 can
be extended to
the sum and
difference of
an arbitrary
number of
functions.
= x4 + x2 - x + C
(E)
3
a 2ex +
1
3
b dx = 2 ex dx + 3
dx
x
3
3x
= 2ex + 3 ln∑ x + C
■
To check any of the results in Example 2, we differentiate the final result to
obtain the integrand in the original indefinite integral. When you evaluate an
indefinite integral, do not forget to include the arbitrary constant C.
✓2
Matched
Problem
Find each indefinite integral:
(A)
(C)
3
3
2 dx
(B)
3x4 dx
(D)
3
16et dt
3
(2x5 - 3x2 + 1) dx
5
a - 4ex b dx
(E)
3 x
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Antiderivatives and Indefinite Integrals
Section 6-1
375
E XAMPLE 3 Using Indefinite Integral Properties and Formulas
(A)
(B)
4x-3 +1
4
dx =
4x-3 dx =
+ C = -2x-2 + C
3
-3 + 1
3
3x
3
3 2
52
u du = 5
3
u23 du = 5
=5
(C)
u(23) +1
+C
2
3 + 1
u53
5
3
+ C = 3u53 + C
x3 - 3
x3
3
dx
=
a
- 2 b dx
2
2
x
3 x
3 x
=
3
(x - 3x-2) dx
x dx - 3 x-2 dx
3
3
1 +1
x
x-2 +1
=
-3
+C
1 +1
-2 + 1
=
= 12 x2 + 3x-1 + C
(D)
a
3
2
3
2x
- 62x b dx =
3
(2x-13 - 6x12) dx
x-13 dx - 6 x12 dx
3
3
x(-13) +1
x(12) +1
=2 1
-6 1
+C
-3 + 1
2 + 1
=2
=2
x23
2
3
-6
x32
3
2
+C
= 3x23 - 4x32 + C
(E)
✓3
Matched
Problem
3
x(x2 + 2) dx =
3
(x3 + 2x) dx =
x4
+ x2 + C
4
Find each indefinite integral:
(A)
3
(B)
3
a 2x23 -
3
b dx
x4
5
42
w3 dw
(C)
x4 - 8x3
dx
x2
3
(D)
3
a 82
x -
(E)
3
3
6
2x
b dx
(x2 - 2)(x + 3) dx
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■
376 Chapter 6
Integration
Caution
1. Note from Example 3(E) that
x(x2 + 2) dx Z
x2 x3
a
+ 2x b + C
2 3
3
In general, the indefinite integral of a product is not the product of the indefinite integrals (this is to be expected, because the derivative of a product
is not the product of the derivatives).
2.
3
ex dx Z
ex +1
+C
x +1
The power rule applies only to power functions of the form xn, where the
exponent n is a real constant not equal to -1 and the base x is the variable. The function ex is an exponential function with variable exponent x
and constant base e. The correct form is
ex dx = ex + C
3
3. Not all elementary functions have elementary antiderivatives. It is impos2
sible, for example, to give a formula for the antiderivative of f(x) = ex in
terms of elementary functions. Nevertheless, finding such a formula when
it exists can markedly simplify the solution of certain problems.
➤ Applications
Let us now consider some applications of the indefinite integral to see why we
are interested in finding antiderivatives of functions.
E XAMPLE 4 Curves Find the equation of the curve that passes through (2, 5) if its slope
is given by dydx = 2x at any point x.
Solution
We are interested in finding a function y = f(x) such that
dy
= 2x
dx
(1)
and
y =5
y
when
x =2
(2)
If dydx = 2x, then
y x2 3
y =
yx 1
2
5
2x dx
3
= x2 + C
(2, 5)
yx 1
2
(3)
Since y = 5 when x = 2, we determine the particular value of C so that
5 = 22 + C
3
0
x
3
FIGURE 2 y = x2 + C
✓4
Matched
Problem
Thus, C = 1, and
y = x2 + 1
is the particular antiderivative out of all those possible from equation (3) that
satisfies both equations (1) and (2) (see Fig. 2).
■
Find the equation of the curve that passes through (2, 6) if its slope is given by
dydx = 3x2 at any point x.
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Section 6-1
Antiderivatives and Indefinite Integrals
377
In certain situations, it is easier to determine the rate at which something
happens than how much of it has happened in a given length of time (for example, population growth rates, business growth rates, rate of healing of a
wound, rates of learning or forgetting). If a rate function (derivative) is given
and we know the value of the dependent variable for a given value of the independent variable, then—if the rate function is not too complicated—we can
often find the original function by integration.
E XAMPLE 5
Cost Function If the marginal cost of producing x units is given by
C¿(x) = 0.3x2 + 2x
and the fixed cost is $2,000, find the cost function C(x) and the cost of producing 20 units.
Solution
Recall that marginal cost is the derivative of the cost function and that fixed
cost is cost at a 0 production level. Thus, the mathematical problem is to find
C(x) given
C¿(x) = 0.3x2 + 2x
C(0) = 2,000
We now find the indefinite integral of 0.3x2 + 2x and determine the arbitrary
integration constant using C(0) = 2,000:
C¿(x) = 0.3x2 + 2x
(0.3x2 + 2x) dx
3
= 0.1x3 + x2 + K
Since C represents the cost, we
use K for the constant of integration.
C(x) =
y
But
$4,000
C(0) = (0.1)03 + 02 + K = 2,000
$3,200
Thus, K = 2,000, and the particular cost function is
$2,000
y C(x)
C(x) = 0.1x3 + x2 + 2,000
We now find C(20), the cost of producing 20 units:
0
10
20
x
Units
C(20) = (0.1)203 + 202 + 2,000
= $3,200
See Figure 3 for a geometric representation.
FIGURE 3
✓5
Matched
Problem
■
Find the revenue function R(x) when the marginal revenue is
R¿(x) = 400 - 0.4x
and no revenue results at a 0 production level. What is the revenue at a production level of 1,000 units?
E XAMPLE 6 Advertising An FM radio station is launching an aggressive advertising
campaign in order to increase the number of daily listeners. The station currently has 27,000 daily listeners, and management expects the number of daily
listeners, S(t), to grow at the rate of
S¿(t) = 60t12
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378 Chapter 6
Integration
listeners per day, where t is the number of days since the campaign began.
How long should the campaign last if the station wants the number of daily listeners to grow to 41,000?
Solution
We must solve the equation S(t) = 41,000 for t, given that
S¿(t) = 60t12
and
S(0) = 27,000
First, we use integration to find S(t):
S(t) = 60t12 dt
t32
= 60 3 + C
2
32
= 40t
+C
Since
S(0) = 40(0)32 + C = 27,000
we have C = 27,000, and
S(t) = 40t32 + 27,000
Now we solve the equation S(t) = 41,000 for t:
40t32 + 27,000
40t32
t32
t
=
=
=
=
=
41,000
14,000
350
35023
49.664 419 p
Use a calculator.
Thus, the advertising campaign should last approximately 50 days.
✓6
Matched
Problem
■
The current monthly circulation of the magazine Computing News is 640,000
copies. Due to competition from a new magazine in the same field, the
monthly circulation of Computing News, C(t), is expected to decrease at the
rate of
C¿(t) = -6,000t13
copies per month, where t is the time in months since the new magazine began
publication. How long will it take for the circulation of Computing News to
decrease to 460,000 copies per month?
Answers to Matched Problems
1. (A) x3 + C
(B)
y
5
x
5
5
5
(C) The graphs are vertical translations of each other.
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Antiderivatives and Indefinite Integrals
Section 6-1
(B) 16et + C
2. (A) 2x + C
(D)
3. (A)
1 6
3x
3
-x +x +C
6 53
5x
43
(D) 6x
+ x-3 + C
12
- 12x
+C
(C)
3 5
5x
+C
(E) 5 ln ∑ x∑ - 4ex + C
(B)
5 85
2w
(E)
1 4
4x
+C
(C)
3
1 3
3x
- 4x2 + C
2
+ x - x - 6x + C
3
4. y = x - 2
5. R(x) = 400x - 0.2x2; R(1,000) = $200,000
6. t = (40)34 L 16 mo
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379