12-34 CHAPTER 12: INTEGRATION
94. The average number of children in the city over the six year time period is given by:
6
6
2
6
CHAPTER 12 REVIEW
x
2
x
2.
20
20
10 10
55dx x
= 5(20) – 5(10) = 50 (12-5)
3.
92
(4 )tdt
=
9
4dt
–
92
tdt
= 9
0
4t –
9
3
0
t= 36 – 243 = –207 (12-5)
6.
0
x
x
edx
Let u = –2x2 . Then du = –4x dx.
x
22
41
xx u
x
F
x
8. 22
() ln 2ln, ‘() ();
F
xx xFx fx
x
yes. (12-1)
2
12ln
x
F
212ln
x
12-38 CHAPTER 12: INTEGRATION
43.
1
3
1
2dx
x
= –
1
3
1()
2dx
x
= –
11/ 2
5
udu
=
51/2
1
udu
= 5
1/2
1
2u
= 2 5 – 2 ≈ 2.472
44. Let u = 1 + x2. Then du = 2x dx.
x
x
x
3
45. Let u = 1 + x2. Then du = 2x dx.
3
xdx
3
3
2 ·
0
1
=
2
0
2(1 )x
= – 11 9
20 2 20
46. 34 5
(2 5)
x
xdx
Let u = 2x4 + 5. Then du = 8x3dx.
x
6
=
48
(12-2)
47. 11 1
x
x
xx
edx e dx du
= –ln|u| + C = –ln|e–x + 3| + C = –ln(e–x + 3) + C
48. 22
2(2)
(2)
x
xx
x
edx e e dx u du
e
dx =
1
1
u
+ C = –(ex + 2)–1 + C = 1
(2)
x
e
+ C
49. dy
dx = 3x–1 – x–2
x
x
12-40 CHAPTER 12: INTEGRATION
Copyright © 2019 Pearson Education, Inc.
7
7
52 32
2(16 ) 32(16 )
xx
52 32 52 32
2 9 32 9 2 16 32 16
56.
19
1
(1) 0xdx
. (12-4)
57.
dy
dx = 9x2ex3
, f(0) = 2
Let u = x3. Then du = 3x2dx.
22
xx u
58. dN
dt = 0.06N, N(0) = 800, N > 0
59. N = 50(1 – e–0.07t),
60. p = 500e–0.03x,
12-42 CHAPTER 12: INTEGRATION
66. a = 200, b = 600, n = 2, ∆x = 600 200
2
= 200
L2 = C‘(200)∆x + C‘(400)∆x
200
67. The graph of C‘(x) is a straight line with y-intercept = 600 and slope = 300 600
600 0
= – 1
2
Thus, C‘(x) = – 1
2x + 600
x
200
=
200
68. The total change in profit for a production change from 10 units per week to 40 units per week is given by:
40
2
2
2
69. P‘(x) = 100 – 0.02x
x
2
x
70. The required definite integral is:
15 15
71. Average inventory from t = 3 to t = 6:
66
12-44 CHAPTER 12: INTEGRATION
Copyright © 2019 Pearson Education, Inc.
Now A(1) = 5
1 + C = 5. Therefore, C = 0 and
A(t) = 5
t
76. The total amount of seepage during the first four years is given by:
4
0
4
2
0
1, 000
(1 ) dt
t
= 1,000
42
0
4
1
(1 )
1
t
77. (A) The exponential growth law applies and we have:
dP
dt = 0.0107P, P(0) = 116 (million)
(B) Time to double:
116e0.0107t = 232
e0.0107t = 2
78. Let Q = Q(t) be the amount of carbon-14 present in the bone at time t. Then,
dQ
CHAPTER 12 REVIEW 12-45
79. N‘(t) = 7e–0.1t and N(0) = 25.
N(t) = 0.1 0.1 0.1
7
77 (0.1)
0.1
tt t
edt edt e dt
= –70e–0.1t + C, 0 ≤ t ≤ 15