388
CHAPTER 6
6.1
Techniques of Integration
I N T E G R AT I O N B Y S U B S T I T U T I O N
■
■
■
■
Use
Use
Use
Use
the basic integration formulas to find indefinite integrals.
substitution to find indefinite integrals.
substitution to evaluate definite integrals.
integration to solve real-life problems.
Review of Basic Integration Formulas
Each of the basic integration rules you studied in Chapter 5 was derived from a
corresponding differentiation rule. It may surprise you to learn that, although you
now have all the necessary tools for differentiating algebraic, exponential, and
logarithmic functions, your set of tools for integrating these functions is by no
means complete. The primary objective of this chapter is to develop several
techniques that greatly expand the set of integrals to which the basic integration
formulas can be applied.
Basic Integration Formulas
1. Constant Rule:
2. Simple Power Rule n 1:
3. General Power Rule n 1:
k dx kx C
x n dx un
x n1
C
n1
du
dx dx
4. Simple Exponential Rule:
5. General Exponential Rule:
6. Simple Log Rule:
7. General Log Rule:
un du
u n1
C
n1
e x dx e x C
eu
du
dx dx
eu du
eu C
1
dx ln x C
x
dudx
dx u
1
du
u
ln u C
As you will see once you work a few integration problems, integration is not
nearly as straightforward as differentiation. A major part of any integration problem is determining which basic integration formula (or formulas) to use to solve
the problem. This requires remembering the basic formulas, familiarity with
various procedures for rewriting integrands in the basic forms, and lots of practice.
SECTION 6.1
Integration by Substitution
389
Integration by Substitution
STUDY
There are several techniques for rewriting an integral so that it fits one or more of
the basic formulas. One of the most powerful techniques is integration by
substitution. With this technique, you choose part of the integrand to be u and
then rewrite the entire integral in terms of u.
When you use integration by
substitution, you need to realize
that your integral should contain
just one variable. For instance,
the integrals
EXAMPLE 1
Integration by Substitution
Use the substitution u x 1 to find the indefinite integral.
and
x
dx
x 12
SOLUTION
From the substitution u x 1,
x u 1,
du
1,
dx
and
u1
du
u2
u
1
2 du
2
u
u
1
1
du
u u2
1
ln u C
u
1
ln x 1 C.
x1
x
dx
x 12
u1
du
u2
are in the correct form, but the
integral
dx du.
By replacing all instances of x and dx with the appropriate u-variable forms, you
obtain
x
dx x 12
TIP
x
dx
u2
is not.
Substitute for x and dx.
Write as separate fractions.
TRY
Simplify.
Find antiderivative.
Substitute for u.
The basic steps for integration by substitution are outlined in the guidelines
below.
Guidelines for Integration by Substitution
1. Let u be a function of x (usually part of the integrand).
2. Solve for x and dx in terms of u and du.
3. Convert the entire integral to u-variable form and try to fit it to one
or more of the basic integration formulas. If none fits, try a different
substitution.
4. After integrating, rewrite the antiderivative as a function of x.
IT
1
Use the substitution u x 2
to find the indefinite integral.
x
dx
x 22
390
CHAPTER 6
Techniques of Integration
D I S C O V E RY
Suppose you were asked to evaluate the integrals below. Which
one would you choose? Explain
your reasoning.
x 2 1 dx
x x 2 1 dx
EXAMPLE 2
Find
Integration by Substitution
xx2 1 dx.
Consider the substitution u x2 1, which produces du 2x dx.
To create 2x dx as part of the integral, multiply and divide by 2.
SOLUTION
or
1
2
1
2
xx 2 1 dx u12
du
x2 112 2x dx
Multiply and divide by 2.
u12 du
Substitute for x and dx.
1 u32
C
2 32
1
u32 C
3
Power Rule
Simplify.
1
x2 132 C
3
Substitute for u.
You can check this result by differentiating.
TRY
IT
Find
xx2 4 dx.
2
EXAMPLE 3
Find
Integration by Substitution
e 3x
dx.
1 e 3x
SOLUTION
To create
Consider the substitution u 1 e3x, which produces du 3e3x dx.
dx as part of the integral, multiply and divide by 3.
3e3x
1u
e3x
1
1
dx 3e3x dx
3x
1e
3 1 e3x
1 1
du
3 u
1
ln u C
3
1
ln1 e3x C
3
TRY
IT
Find
e2x
dx.
1 e2x
3
du
Multiply and divide by 3.
Substitute for x and dx.
Log Rule
Substitute for u.
Note that the absolute value is not necessary in the final answer because the quantity 1 e3x is positive for all values of x.
SECTION 6.1
EXAMPLE 4
Integration by Substitution
Find the indefinite integral.
xx 1 dx
SOLUTION
x u 1.
Consider the substitution u x 1, which produces du dx and
xx 1 dx u 1u12 du
Substitute for x and dx.
u32 u12 du
Multiply.
u52 u32
C
52
32
2
2
x 152 x 132 C
5
3
Power Rule
Substitute for u.
This form of the antiderivative can be further simplified.
2
2
6
10
x 152 x 132 C x 152 x 132 C
5
3
15
15
2
x 1323x 1 5
C
15
2
x 132 3x 2 C
15
You can check this answer by differentiating.
TRY
IT
4
Find the indefinite integral.
xx 2 dx
Example 4 demonstrates one of the characteristics of integration by substitution. That is, you can often simplify the form of the antiderivative as it exists
immediately after resubstitution into x-variable form. So, when working the exercises in this section, don’t assume that your answer is incorrect just because it
doesn’t look exactly like the answer given in the back of the text. You may be able
to reconcile the two answers by algebraic simplification.
T E C H N O L O G Y
If you have access to a symbolic integration utility, try using it to
find an antiderivative of f x x 2x 1 and check your answer
analytically using the substitution u x 1. You can also use the utility
to solve several of the exercises in this section.
Integration by Substitution
391
392
CHAPTER 6
Techniques of Integration
Substitution and Definite Integrals
The fourth step outlined in the guidelines for integration by substitution on page
389 suggests that you convert back to the variable x. To evaluate definite integrals,
however, it is often more convenient to determine the limits of integration for the
variable u. This is often easier than converting back to the variable x and evaluating the antiderivative at the original limits.
EXAMPLE 5
5
Evaluate
1
Using Substitution with a Definite Integral
x
dx.
2x 1
Use the substitution u 2x 1, which implies that u2 2x 1,
1, and dx u du. Before substituting, determine the new upper and
x
lower limits of integration.
SOLUTION
1 2
2 u
Lower limit: When x 1, u 21 1 1.
Upper limit: When x 5, u 25 1 3.
y
x
y
2x
5
Now, substitute and integrate, as shown.
1
Integration limits for x
Integration limits for u
4
3
5
2
( 5, )
5
3
(1, 1)
1
x
dx 2x 1
1
x
−1
1
2
3
4
f )u)
Region Before
u2
f (u)
3
1 u2 1
u du
u
2
1
2
1
Substitute for x, dx, and
limits of integration.
3
u2 1 du
1
u3
Simplify.
3
1
u
2 3
1
1
1
9 3 1
2
3
16
3
5
−1
FIGURE 6.1
Substitution
Find antiderivative.
Apply Fundamental Theorem.
Simplify.
1
2
5
(3, 5)
TRY
IT
5
2
4
Evaluate
3
x4x 1 dx.
0
2
(1, 1)
1
STUDY
u
−1
1
2
3
4
−1
FIGURE 6.2
Substitution
5
In Example 5, you can interpret the equation
5
Region After
TIP
1
x
dx 2x 1
3
1
1 u2 1
u du
u
2
graphically to mean that the two different regions shown in Figures 6.1 and 6.2
have the same area.
SECTION 6.1
393
Integration by Substitution
Application
Integration can be used to find the probability that an event will occur. In such an
application, the real-life situation is modeled by a probability density function f,
and the probability that x will lie between a and b is represented by
b
Pa ≤ x ≤ b f x dx.
a
Researchers such as psychologists
use definite integrals to represent
the probability that an event will
occur. For instance, a probability of
0.5 means that an event will occur
about 50% of the time.
The probability Pa ≤ x ≤ b must be a number between 0 and 1.
EXAMPLE 6
Finding a Probability
A psychologist finds that the probability that a participant in a memory experiment will recall between a and b percent (in decimal form) of the material is
b
Pa ≤ x ≤ b a
28 3
x1 x dx,
9
0 ≤ a ≤ b ≤ 1.
Find the probability that a randomly chosen participant will recall between 0%
and 87.5% of the material.
ALGEBRA
For help on the algebra in
Example 6, see Example 1(a) in
the Chapter 6 Algebra Review on
page 446.
3 1 x.
Let u Then u3 1 x, x 1 u3, and dx 3u2 du.
SOLUTION
3 1 0 1.
Lower limit: When x 0, u 3 1 0.875 0.5.
Upper limit: When x 0.875, u To find the probability, substitute and integrate, as shown.
0.875
0
REVIEW
28 3
x 1 x dx 9
12
289 1 u u3u du
3
3
28
9 x
y
1
x
1.5
2
283 u u du
u
28 u
3 7
4
1
y
1.0
12
6
3
1
7
1
0.865
So, the probability is about 86.5%, as indicated in Figure 6.3.
TRY
IT
6
Use Example 6 to find the probability that a participant will recall between
0% and 62.5% of the material.
TA K E
A N O T H E R
0.5
4 12
L O O K
Probability
In Example 6, explain how you could find a value of b such that P0 ≤ x ≤ b 0.5.
x
0.5
0.875
FIGURE 6.3
1.0
1.5
394
Techniques of Integration
CHAPTER 6
P R E R E Q U I S I T E
R E V I E W 6 . 1
The following warm-up exercises involve skills that were covered in earlier sections. You will
use these skills in the exercise set for this section.
In Exercises 1–8, evaluate the indefinite integral.
1.
3.
5.
7.
5 dx
2.
x32 dx
4.
2xx2 13 dx
6.
6e6x dx
8.
1
dx
3
x 23 dx
3x 2x 3 12 dx
2
dx
2x 1
In Exercises 9–12, simplify the expression.
9. 2 xx 12 xx 1
10. 6x x 43 3x2 x 42
11. 3x 712 2 x x 712
E X E R C I S E S
12. x 513 5x 523
6 . 1
In Exercises 1–38, find the indefinite integral.
1.
3.
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
x 24 dx
2
dt
t 92
2t 1
dt
t2 t 2
1 x dx
12x 2
dx
3x 2 x
1
dx
5x 13
2.
4.
6.
8.
10.
12.
1
dx
x 1
14.
e3x
dx
1 e3x
16.
2x
2 dx
e 3x
x2
dx
x1
xx 2 4 dx
e5x dx
18.
20.
22.
24.
25.
x 532 dx
4
dt
1 t 3
3
2y
dy
y4 1
27.
29.
31.
3 x52 dx
6x 2 2
dx
x3 x
1
dx
3x 12
1
dx
5x 1
4e 2x
dx
1 e 2x
33.
35.
37.
e
t
te t
2
1
x
dx
x 14
28.
x
dx
3x 12
30.
1
t 1
dt
2t 1
dt
t
x
32.
34.
dx
36.
t 21 t dt
38.
2x 1
ex
dx
1 ex
x2
dx
x 13
5x
dx
x 43
1
x 1
dx
6x x
dx
x
x2
x 1
dx
3
y 2
y 1 dy
41.
0
4
43.
40.
45.
4 x 1 dx
2
2
2
3xe x dx
x
dx
x 42
42.
x1 x3 dx
e2x dx
0
1
44.
0
0.5
dt
4
2 x 1 dx
0
1
0
dt
26.
4
39.
dx
2x
dx
x4
1 t 2
ex
dx
2
ex
In Exercises 39–46, evaluate the definite integral.
x1
x 1
x x 54 dx
0
0.5
46.
0
x 21 x3 dx
In Exercises 47–54, find the area of the region bounded by the
graphs of the equations. Then use a graphing utility to graph the
region and verify your answer.
395
Integration by Substitution
SECTION 6.1
63. Probability The probability of recall in an experiment is
modeled by
b
Pa ≤ x ≤ b 47. y xx 3, y 0, x 7
a
15
x 1 x dx
4
48. y x2x 1, y 0, x 4
where x is the percent of recall (see figure).
49. y x21 x, y 0, x 3
(a) What is the probability of recalling between 40% and
80%?
50. y x2x 2, y 0, x 7
51. y x2 1
, y 0, x 1, x 5
2 x 1
52. y 1
2x 1
, y 0, x , x 6
2
x 3
53. y 3
x
(b) What is the median percent recall? That is, for what
value of b is P0 ≤ x ≤ b 0.5?
y
15
4
y
y
x
x
1
3
x 1, y 0, x 0, x 7
2
54. y x x 2, y 0, x 2, x 10
P(a
x
y
1155 3
3 2 x (1
P(a
b)
3
x) 3 2
x
b)
2
1
In Exercises 55–58, find the area of the region bounded by the
graphs of the equations.
55. y xx 2, y 0
56. y y
3 1
x
x, y 0
a
b 0.5
x
1
0.5
y
1
x
2
−1
1.0
x
a
1155 3
x 1 x32 dx
32
(see figure). Find the probabilities that a sample will contain between (a) 0% and 25% and (b) 50% and 100% iron.
58. y 11 x ,
(Hint: Find the area of
the region bounded by
y x1 x 2 and y 0.
Then multiply by 4.)
65. Meteorology During a two-week period in March in a
small town near Lake Erie, the measurable snowfall S (in
inches) on the ground can be modeled by
y 0, x 0, x 4
y
St t14 t ,
3
0 ≤ t ≤ 14
2
where t represents the day.
1
(a) Use a graphing utility to graph the function.
x
1
2
3
4
(b) Find the average amount of snow on the ground during
the two-week period.
(c) Find the total snowfall over the two-week period.
In Exercises 59 and 60, find the volume of the solid generated by
revolving the region bounded by the graph(s) of the equation(s)
about the x-axis.
66. Revenue A company sells a seasonal product that generates a daily revenue R (in dollars per day) modeled by
R 0.06t 2365 t12 1250, 0 ≤ t ≤ 365
where t represents the day.
59. y x1 x 2
60. y x 1 x
2,
y0
In Exercises 61 and 62, find the average amount by which the
function f exceeds the function g on the interval.
61. f x b
57. y 2 x 2 1 x 2
a b 1
Figure for 64
Pa ≤ x ≤ b x
−2
0.5
64. Probability The probability of finding between a and b
percent iron in ore samples is modeled by
0.5
1
2
x
1
Figure for 63
y
2
1
1 ,
x
, 0, 1
gx x1
x 12
62. f x x4x 1, gx 2x3, 0, 2
(a) Find the average daily revenue over a period of 1 year.
(b) Describe a product whose seasonal sales pattern resembles the model. Explain your reasoning.
In Exercises 67 and 68, use a program similar to the Midpoint Rule
program on page 366 with n 10 to approximate the area of
the region bounded by the graph(s) of the equation(s).
3 67. y x 4 x,
68. y x 1 x 2
2
2
y0