12 INTEGRATION
EXERCISE 12–1
2. 9
9
66.
x
x
4.
2
123
3
51 5.
xx
x
xx
x
3
1/3
3
44
x
x
10. 10 10dx x C
dxC
12. 2
14 7
x
dx x C
x
2
d
14. 23
15 5
x
dx x C
x
322
d
16. 89
1
9
x
dx x C
;
x
11
d
18. 43
1
3
x
dx x C
x
344
11
d
20. 1/3 4/3
86
x
dx x C
x
4/3 1/3 1/3
4
d
22. 77lndz z C
z
17
dzC
dz z z
24. 55
uu
edu e C
uuu
deC e e
28. () ln , ‘() 1 ln 1 ln ();yes.Fx x x x e F x x x f x
12-2 CHAPTER 12: INTEGRATION
30.
43
3
(3 2) 4(3 2) (3)
() , ‘() 3(3 2) ();no.
44
xx
Fx F x x f x
xx x x
36. True, since any antiderivative of k(x) = 0 is of the form K(x) = C and K(x) = 0 is an antiderivative of k(x) =
0.
38. False, since any antiderivative of g(x) = 5eπ is of the form G(x) = (5eπ)x + C which obviously is not equal
42. The graphs in this set could be graphs from a family of antiderivative functions since they appear to be
44.
x2(1 + x3)dx =
(x2 + x5)dx =
x2dx +
x5dx
46.
3
dt
t =
1/3
dt
t =
t–1/3 dt = 1
11
t(–1/3)+1 + C = 3
2t2/3 + C
3
‘ = 3
2
1
1
48.
2
6dm
m = 6
m–2 dm = 6(–m–1) + C = –6
m + C
6
50.
2
1
3
y
y
dy =
1
3ydy –
2
3
y
ydy = 1
3
1
ydy – 1
3
y dy = 1
3 ln|y| – 1
3 · 1
2y2 + C
Check: 2
ln
36
yyC
3 · 1
y – 1
6(2y) + 0 = 1
3y – 3
3
y
52.
2
t
etdt =
22
t
et
dt =
2
t
edt –
2
tdt =
2
1
24
tt
eC [using Indefinite Integral Formula]
2
1
24
tt
‘ = 1
2et – 2
4
t + 0 = 22 2
tt
etet
EXERCISE 12-1 12-3
54.
3
3
2
4x
x
dx = 4
x3 dx + 2
x–3 dx = 4
4
4
x
+ 2 2
1
2x
+ C = x4 – x–2 + C
2
x
56. R‘(x) = 500 – 0.4x
2
2
x
58. dR
50
1CCC
60. 12
‘( ) 2 1; (1) 5fx x x f
62. 32; (0) 2
t
dy ey
64. dy
x
x
Given y(1) = 3: 3 = 4(1)3 – 6(1)2 + C. Hence, C = 5 and y = 4x3 – 6x2 + 5.
66.
14
2
x
x
x
dx =
14
22
x
x
x
x
dx =
(x–3 – x2)dx =
x–3 dx –
x2 dx
x
x
12-4 CHAPTER 12: INTEGRATION
68.
4
2
13
x
x
dx =
4
22
13
x
x
x
dx =
(x–2 – 3x2)dx =
x–2 dx – 3
x2 dx
x
x
70.
1
x
x
e
x
x
dx =
1
x
x
e
dx =
(x–1 – ex)dx =
x–1 dx –
ex dx = ln|x| – ex + C
72. 22
(4 35) (83) 4 3
d
x
xdx xdxxxC
dx
27 27
tt
detdtet
dfxdx fx
78. d
dx (ln|x| + C) = d
dx (ln(–x) + C) since x < 0
x
x
80. By the properties of the indefinite integral (#5): [ () ()] () () .
f
x gx dx f xdx gxdx
82. For
f
t
= 0.002t + 0.03, we have
84. (A) R‘(x) > 0 for 0 < x < 500 and R‘(x) < 0 for 500 < x < 1,000. Therefore, the graph of R(x) is rising
from 0 to 500 and falling from 500 to 1,000. R‘(x) is decreasing, so R“(x) < 0 and hence the graph of R(x)
is concave downward on (0, 1,000). It has a local maximum at x = 500.
Rx
EXERCISE 12-1 12-5
Hence, C = 0 and R(x) = 300x – 0.3x2.
86. S‘(t) = 500 t1/4
S(t) =
S‘(t)dt =
500t1/4 dt = 500
t1/4 dt = 500 14
1
1
t1+(1/4) + C
88. S‘(t) = 500t1/4 + 300
90. L‘(x) = 2,000x–1/3
(1/3)1
x
92. dA
dt = –4t–3, 1 ≤ t ≤ 10
2
t
12-6 CHAPTER 12: INTEGRATION
94. V‘(t) = 15
t, 1 ≤ t ≤ 5
EXERCISE 12–2
2. 655
( ) (4 3) ; ‘( ) 6(4 3) (4) 24(4 3) .fx x f x x x
6. 33 2
() 6 ; ‘() 6 (3 ) 18 .
f
xefxex xe
8. 2
125
() ln( 5 4); ‘() (2 5) .
x
fx x x f x x
10.
(6x – 1)3(6)dx
Let u = 6x – 1, then du = 6 dx and
4
u + C [using Indefinite Integral Formulas]
12.
(x6 + 1)4(6x5)dx
Let u = x6 + 1, then du = 6x5dx and
(x6 + 1)4(6x5)dx =
u4 du =
5
5
u + C [using Indefinite Integral Formulas]
65
(1)
x + C
12-8 CHAPTER 12: INTEGRATION
24.
(5t + 1)3 dt
Let u = 5t + 1, then du = 5 dt and dt = 1
5du and
4
u + C [using Indefinite Integral Formulas]
26.
(t3 + 4)–2 t2 dt
Let u = t3 + 4, then du = 3t2 dt, t2 dt = 1
28.
e–0.01x dx
Let u = –0.01x, then du = –0.01 dx, dx = –100 du and
30.
2
1
x
x
dx
Let u = 1 + x2, then du = 2x dx, x dx = 1
x
x
x
EXERCISE 12-2 12-9
32.
3
2tdt
Let u = 2 – t, then du = –dt, dt = –du and
3
1
34.
2
35
(2)
t
tdt
Let u = t3 – 2, then du = 3t2 dt, and
2
t
(t3 – 2)–5 3
u–5 1
u–5 du = 1
12-10 CHAPTER 12: INTEGRATION
Copyright © 2019 Pearson Education, Inc.
3/2 1/2
2(5) 10(5)
3
40.
x(x + 6)8 dx
Let u = x + 6, then du = dx and x = u – 6.
10
u –
9
6
u + C
42. Let u = 1 – e-x, then du = –e-x(–1)dx = e-x dx.
-x dx =
5
u + C = 1
44. Let u = x3 – 3x + 7, then du = (3x2 – 3)dx = 3(x2 – 1)dx.
2
3
1
37
x
xx
dx =
(x3 – 3x + 7)–1 3
3(x2 – 1)dx =
u–1 1
3du = 1
3
u–1 du = 1
3 ln|u| + C
dx
3
3 · 3
37xx
3 ·
3
37
xx
3
37
xx
46. Let 47,ux then 7.du dx
74 7
x
dx
= udu
=
2
2
uC =
2
47
2
xC
= 2
49 28 8
2
x
xC
x
49 28
2
EXERCISE 12-2 12-11
48. Let 31,ux then 2
3.du x dx
2
363
=
63
2
50. Let 8,ux then 7
8.du x dx
3
78
8
x
xdx
= 3
udu
=
4
uC=
4
8
x
C
=
32
4
52. 1
11
() , ‘() () ln;no.Fx x F x f x x
12-12 CHAPTER 12: INTEGRATION
66. Let 43
3 1, then 12 .ux du xdx
31/2
34 1/2 4 1/2 3 4
12 1 1 1
xu
41
xdx =
= 1
u–1/2 du = 1
1/2
u + C = 1
68. Let u = 2ex 1. Then du = 2ex dx.
x
70. Let 1
ln( 5), Then .
5
ux du dx
x
22
ln(5) [ln(5)]
xux
72. dm
dn = 10n(n2 – 8)7
74. dy
dx =
2
34
5
(7)
x
x
12-14 CHAPTER 12: INTEGRATION
82. Sꞌ(t) = 20 – 20e–0.05t, 0 ≤ t ≤ 24
(A) S(t) =
(20 – 20e–0.05t)dt =
20 dt – 20
e–0.05t dt = 20t – 20 1
0.05
e–0.05t + C
(B) S(12) = 20(12) + 400e–0.05(12) – 400 ≈ 60
84. (A) R(t) = 2
120
1
t
t+ 3, 0 ≤ t ≤ 20
22
12
tt
2
120(1 )
t
Sign chart for R‘(t):
+ + + – – –
DecreasingIncreasing
R‘(t)
R(t)
x
021
0Test Numbers
72
5
‘( )
0 120( )
2()
tRt
Thus, the rate of production is greatest at t = 1.
(B) Q(t) =
R(t)dt =
2
120 3
1
t
t
dt = 120
21
t
tdt +
3 dt = 60 ln(t2 + 1) + 3t + C
(C) Q(t) = 250 thousands. Now, we need to solve: 250 = 60 ln(t2 + 1) + 3t for t
EXERCISE 12-3 12-17
40. dy
dt = –3y
1
dy
–3 dt
ydy =
–3 dt
42. dy
dx = 0.1y, y(0) = –2.5
1
dy
0.1 dx
y
dy =
0.1 dx
ln |y| = 0.1x + K (K an arbitrary constant)
44. dx
dt = 4t
x =
2
4
2
t + C = 2t2 + C (General solution)
46. dx
dt = 4x
1
x
dx
x
52. Figure (B). When y = –1, the slope dy
dx = –1 + 1 = 0 for any x.
54. y = Cex – 1
dy
12-18 CHAPTER 12: INTEGRATION
Cex = y + 1
56.
58.
60. Given y = 2
x
C = (x2 + C)1/2.
dy
x
x
x
x
62. Given y = C
x
= Cx–1,
dy
x
x
x
x
x
2
C
= –1
· C
= –1
· y = –y
x
x
x
64. Given y = 6
2
(1 )
t
Ce
= 2(1 + Ce–6t)–1
6
t
EXERCISE 12-3 12-19
Copyright © 2019 Pearson Education, Inc.
1 + Ce–6t = 2
y or Ce–6t = 2
y – 1
Thus dy
dt = 3Ce–6ty2 = 3 21
y
y2 = 3(2 – y)y.
To find the desired particular solution, we have to find the constant C from the following equation:
2
66. y = 5,250e0.12t
0 ≤ t ≤ 10, 0 ≤ y ≤ 20,000
68. p = 1,000e–0.08x
0 ≤ x ≤ 40, 0 ≤ p ≤ 1,000
70. N = 1,000(1 – e–0.07t)
72. N = 0.4
400
74. y = 1kMt
M
ce
= 2
M
. Thus,
78. dA
80. dA
dt = rA, A(0) = 5,000
12-20 CHAPTER 12: INTEGRATION
5r = ln 5, 282.70
82. (A) dp
dx = rp, p(0) = 20
23.47 = 20e40r
e40r = 1.174
(B) p(100) = 20e0.004(100) = 20e0.4 ≈ $29.84 per unit
84. dN
dt = k(L – N), N(0) = 0
0.33L = L(1 – e–8k)
1 – e–8k = 0.33
Therefore, N(t) = L(1 – e–0.05t)
(B) Solve L(1 – e–0.05t) = 0.66L:
1 – e–0.05t = 0.66
(D)