418
CHAPTER 5
Integrals
In the first integral on the far right side we make the substitution u − 2x. Then
du − 2dx and when x − 2a, u − a. Therefore
2y
2a
0
y
f sxd dx − 2y f s2ud s2dud − y f s2ud du
y
a
2a
a
0
a
x
a
_a
f sxd dx − y f s2ud du 1 y f sxd dx
a
a
0
0
(a) If f is even, then f s2ud − f sud so Equation 9 gives
(a) ƒ even, j ƒ dx=2 j ƒ dx
y
0
a
2a
f sxd dx − y f sud du 1 y f sxd dx − 2 y f sxd dx
a
0
a
a
0
0
(b) If f is odd, then f s2ud − 2f sud and so Equation 9 gives
y
_a
a
0
and so Equation 8 becomes
9
_a
a
0
y
0
a
2a
a
a
(b) ƒ odd, j ƒ dx=0
_a
FIGURE 3
x
f sxd dx − 2y f sud du 1 y f sxd dx − 0
a
a
0
Q
0
Theorem 7 is illustrated by Figure 3. For the case where f is positive and even, part
(a) says that the area under y − f sxd from 2a to a is twice the area from 0 to a because
of symmetry. Recall that an integral yab f sxd dx can be expressed as the area above the
x-axis and below y − f sxd minus the area below the axis and above the curve. Thus
part (b) says the integral is 0 because the areas cancel.
EXAMPLE 10 Since f sxd − x 6 1 1 satisfies f s2xd − f sxd, it is even and so
y
2
22
sx 6 1 1d dx − 2 y sx 6 1 1d dx
2
0
f
g
− 2 17 x 7 1 x
2
0
284
− 2(128
7 1 2) − 7
Q
EXAMPLE 11 Since f sxd − stan xdys1 1 x 2 1 x 4 d satisfies f s2xd − 2f sxd, it is odd
and so
y
1
21
1–6 Evaluate the integral by making the given substitution.
1.
y cos 2x dx,
2.
y xe
3.
y
4.
y sin
5.
y
2x 2
dx, u − 2x 2
cos d, u − sin x3
dx , u − x 4 2 5
x 25
4
y s2t 1 1 dt,
Q
u − 2t 1 1
u − 2x
x 2 sx 3 1 1 dx, u − x 3 1 1
2
6.
tan x
dx − 0
1 1 x2 1 x4
7–48 Evaluate the indefinite integral.
7.
y x s1 2 x
9.
y s1 2 2xd
11.
2
dx
8.
yx
2
3
e x dx
dx
10.
y sin t s1 1 cost
y cossty2d dt
12.
y sec
9
2
dt
2 d
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
419
SECTION 5.5 The Substitution Rule
13.
15.
17.
19.
dx
5 2 3x
y
14.
y cos sin d
3
eu
du
s1 2 e u d2
a 1 bx 2
y
y
21.
y
23.
s3ax 1 bx 3
16.
18.
dx
sln xd2
dx
x
20.
yy
2
s4 2 y 3d 2y3 dy
ye
25r
dr
dx
59.
61.
63.
y sec tan d
24.
y x sx 1 2 dx
65.
25.
ye
s1 1 e x dx
26.
y
27.
y sx
2
28.
ye
29.
y5
sins5 t d dt
30.
y
31.
y
sarctan xd2
dx
x2 1 1
32.
y
33.
y cos s1 1 5td dt
34.
y
35.
y scot x csc x dx
36.
y
x
t
1 1dsx 3 1 3xd 4 dx
2
dx
sa ± 0d
ax 1 b
cos t
sin t dt
sec 2 x
dx
tan 2 x
x
dx
x2 1 4
cossyxd
dx
x2
2t
dt
2t 1 3
dt
cos2 t s1 1 tan t
37.
y sinh x cosh x dx
38.
y
39.
y
sin 2x
dx
1 1 cos2x
40.
y
41.
y cot x dx
42.
y
43.
y
44.
y
46.
yx
48.
y x sx
2
dx
s1 2 x sin x
21
2
11x
dx
1 1 x2
45.
y
47.
y xs2x 1 5d
8
dx
sin x
dx
1 1 cos2x
cossln td
dt
t
x
dx
1 1 x4
2
s2 1 x dx
3
2
y6
y
2
y
y4
y
13
y
a
y
2
y
e4
y
1
56.
sin t
dt
cos2 t
58.
1 1 dx
0
0
67.
1
69.
e
71.
0
73.
1
60.
dx
ss1 1 2xd
62.
64.
2
51.
ye
2
cos x
2 1d3 dx
50.
y tan sec d
sin x dx
52.
y sin x cos x dx
2
2
4
y
1
y
y2
y
a
y
y3
y
4
y
2
y
Ty2
0
x sx 2 1 a 2 dx sa . 0d
66.
x sx 2 1 dx
68.
csc 2 ( 12 t) dt
2
xe2x dx
cos x sinssin xd dx
x sa 2 2 x 2 dx
x 4 sin x dx
2y3
0
dx
x sln x
ez 1 1
dz
ez 1 z
dx
70.
0
72.
x
s1 1 2x
dx
2
sx 2 1de sx21d dx
0
sins2 tyT 2 d dt
4
3
74. Verify that f sxd − sin s
x is an odd function and use that fact
to show that
3
0 < y sin s
x dx < 1
3
22
; 75–76 Use a graph to give a rough estimate of the area of the
region that lies under the given curve. Then find the exact area.
75. y − s2x 1 1, 0 < x < 1
76. y − 2 sin x 2 sin 2x, 0 < x < 2
77. Evaluate y22
sx 1 3ds4 2 x 2 dx by writing it as a sum of
two integrals and interpreting one of those integrals in terms
of an area.
78. Evaluate y01 x s1 2 x 4 dx by making a substitution and
interpreting the resulting integral in terms of an area.
79. Which of the following areas are equal? Why?
y
y=2x´
; 49–52 Evaluate the indefinite integral. Illustrate and check that
your answer is reasonable by graphing both the function and its
antiderivative (take C − 0).
y xsx
2y3
0
y
49.
y
0
3
0
3
y3
sx 3 1 x 4 tan xd dx
y (1 1 sx )
dx
5x 1 1
y
0
e 1yx
dx
x2
2y4
y sin x sinscos xd dx
3
y
3
1 1 7x dx
s
1
22.
2
1
0
sx
z2
dz
3
z 11
y
y
0
57.
sin sx
y
55.
y=eœ„x
0
0
1 x
1 x
y
y=e sin x sin 2x
53–73 Evaluate the definite integral.
53.
y
1
0
coss ty2d dt
54.
y
1
0
s3t 2 1d50 dt
0
1
π x
2
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
420
CHAPTER 5
Integrals
80. A model for the basal metabolism rate, in kcalyh, of a young
man is Rstd − 85 2 0.18 coss ty12d, where t is the time in
hours measured from 5:00 am. What is the total basal metabolism of this man, y024 Rstd dt, over a 24-hour time period?
81. An oil storage tank ruptures at time t − 0 and oil leaks from
the tank at a rate of rstd − 100e20.01t liters per minute. How
much oil leaks out during the first hour?
82. A bacteria population starts with 400 bacteria and grows at a
rate of rstd − s450.268de1.12567t bacteria per hour. How many
bacteria will there be after three hours?
83. Breathing is cyclic and a full respiratory cycle from the
beginning of inhalation to the end of exhalation takes
about 5 s. The maximum rate of air flow into the lungs
is about 0.5 Lys. This explains, in part, why the function
f std − 12 sins2 ty5d has often been used to model the rate of
air flow into the lungs. Use this model to find the volume of
inhaled air in the lungs at time t.
84. The rate of growth of a fish population was modeled by the
equation
60,000e20.6 t
Gstd −
s1 1 5e20.6 t d2
where t is measured in years and G in kilograms per year.
If the biomass was 25,000 kg in the year 2000, what is the
predicted biomass for the year 2020?
85. Dialysis treatment removes urea and other waste products
from a patient’s blood by diverting some of the bloodflow
externally through a machine called a dialyzer. The rate at
which urea is removed from the blood (in mgymin) is often
well described by the equation
r
ustd − C0 e2rtyV
V
where r is the rate of flow of blood through the dialyzer (in
mLymin), V is the volume of the patient’s blood (in mL), and
C 0 is the amount of urea in the blood (in mg) at time t − 0.
Evaluate the integral y030 ustd dt and interpret it.
86. Alabama Instruments Company has set up a production line
to manufacture a new calculator. The rate of production of
these calculators after t weeks is
S
100
dx
− 5000 1 2
dt
st 1 10d2
D
calculatorsyweek
(Notice that production approaches 5000 per week as time
goes on, but the initial production is lower because of the
workers’ unfamiliarity with the new techniques.) Find the
number of calculators produced from the beginning of the
third week to the end of the fourth week.
87. If f is continuous and y f sxd dx − 10, find y f s2xd dx.
4
2
0
0
88. If f is continuous and y f sxd dx − 4, find y x f sx 2 d dx.
9
3
0
0
89. If f is continuous on R, prove that
y
b
a
f s2xd dx −
y
2a
2b
f sxd dx
For the case where f sxd > 0 and 0 , a , b, draw a diagram
to interpret this equation geometrically as an equality of
areas.
90. If f is continuous on R, prove that
y
b
a
f sx 1 cd dx − y
b1c
a1c
f sxd dx
For the case where f sxd > 0, draw a diagram to interpret this
equation geometrically as an equality of areas.
91. If a and b are positive numbers, show that
y
x a s1 2 xd b dx − y x b s1 2 xd a dx
1
1
0
0
92. If f is continuous on f0, g, use the substitution u − 2 x to
show that
y
0
y
2
x f ssin xd dx −
0
f ssin xd dx
93. Use Exercise 92 to evaluate the integral
y
0
x sin x
dx
1 1 cos2x
94. (a) If f is continuous, prove that
y
y2
0
f scos xd dx − y
y2
0
(b) Use part (a) to evaluate y0
y2
f ssin xd dx
cos 2 x dx and y0
y2
sin 2 x dx.
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.