12-22 CHAPTER 12: INTEGRATION
94. dS
dR = k
R
1
96. Initially, the graph of x is concave up which implies that x’ is increasing. There is a point of inflection at
EXERCISE 12–4
4. 5[3(2 4 6 8 10)] 5[3(30)] 5(90) 450 sq.ft.
8. A and D.
12. G and H
16.
18. For Figure (C):
L3 = u(1)·1 + u(2)·1 + u(3)·1 = 1 + 2 + 4 = 7
20. L3 ≤
4
1
()uxdx
≤ R3, R3 ≤
4
1
()vxdx
≤ L3; since u(x) is increasing on [1, 4], L3 underestimates the area
12-24 CHAPTER 12: INTEGRATION
40.
a
c
f(x)dx = –
c
a
f(x)dx = –{(Area A) – (Area B) + (Area C)}= –{2.817 – 4.951 + 10.667} = – 8.553
b
c
44.
4
3x2dx = 3
4
x2dx = 3(21) = 63
48.
1
(4x2 – 9x)dx =
1
4x2dx –
1
9xdx = 4
1
x2dx – 9
1
xdx = 4(21) – 9(7.5) = 16.5
5
4
5
4
5
58. True. Take x0 = 0, x1 = 1, x2 = 2, …, x10 = 10 and c1 = 1
2, c2 = 3
2, …, c10 = 19
2, i.e. c1, c2, … c10 are
the midpoints of the intervals (0, 1), (1, 2), …, (9, 10) and of course ∆x = 1. Then
60. False. Let f(x) = –2x on [–10, 0]. The exact area under the graph of f from x = –10 to x = 0 is 100 (see
problem 58 above). For n = 10,
62. h(x) is an increasing function; ∆x = 100
R
EXERCISE 12-4 12-25
Error bound for R10:
(500) 1000
≤ 1000 or n ≥ 500.
12-26 CHAPTER 12: INTEGRATION
72.
4
1
xx dx; f(x) = xx
0.5 ≈ 1,530
74. L4 = [N(20) + N(40) + N(60) + N(80)]∆t = [51 + 68 + 76 + 81](20) = 5,520 units
76. L5 = [A‘(5) + A‘(6) + A‘(7) + A‘(8) + A‘(9)]∆t, ∆t = 1
78. [0, 6], ∆x = 2
L3 = [N‘(0) + N‘(2) + N‘(4)]∆x = [29 + 26 + 23](2) = 156
EXERCISES 12-5
4. ( ) 2 on [ 3, 1].fx x The region bounded by the graph of f and the x-axis is a trapezoid with
2
6. ( ) 10 on [ 100,50].fx x The region bounded by the graph of f and the x-axis consists of two right
triangles: 1
T with vertices 2
(0,0), ( 100,1000), ( 100,0) and with verticesT
EXERCISES 12-5 12-27
8. 2
() 25 on[5,5].fx x The region bounded by the graph of f and the x-axis is a semi-circle of
1(5 ) 12.5 39.27.
2
10. F (x) = 9x + 120
12. F(x) = x2 + 30x + 210
14.
44
222
0
0
2(4)(0)16xdx x
16.
33
0
0
5 5 5(3) 5(0) 15dx x
18.
6
333
62
33
63 72 9 63
333
x
xdx
3
33
2
1112
11
28.
15
51
21 21 20xdx xdx
44
5
45 5
4
a
12-28 CHAPTER 12: INTEGRATION
Copyright © 2019 Pearson Education, Inc.
2
2
36.
4
x
dx =
4
2x–1/2 dx = 4x1/2 25
4
= 4(25)1/2 – 4(4)1/2 = 20 – 8 = 12
1
40.
2
1
x
dx
Let u = x + 1, then du = dx.
1
x
x
1
42.
10
e–0.01x dx
Let u = –0.01x, then du = –0.01dx.
e–0.01x dx =
e–0.01 0.01
eu du = –100eu + C = –100e–0.01x + C
44.
e
tdt
Let u = ln t, then du = 1
tdt. Thus,
2
e
EXERCISES 12-5 12-29
46.
1
0
xex2dx
Let u = x2, then du = 2x dx.
xex2dx =
ex2 2
2x dx = 1
2
eu du = 1
2eu + C = 1
2ex2 + C
1
0
48. 2
1
0
x
edx
Note:
0
a
fxdx
50. g(x) = 2x + 7 on [0, 5]
5
52. g(t) = 4t – 3t2 on [–2, 2]
2
54. g(x) = 1
x
on [3, 8]
(A) Ave g(x) = 1
8
1
x
dx (B)
8
12-30 CHAPTER 12: INTEGRATION
56. f(x) = 64e0.08x on [0, 10]
10
= 64
10
0.08
e
0
58.
1
0
x2
32xdx =
1
0
x(3x2 + 2)1/2 dx
Let u = 3x2 + 2, then du = 6x dx.
x(3x2 + 2)1/2 dx =
(3x2 + 2)1/2 6
6x dx =
1
6
u1/2 du = 1
6·
3/2
32
u + C
=
3/2
9
u + C =
23/2
(3 2)
9
x + C
1
23/2
(3 2)
x
1
0
3/2
(3 2)
–
3/2
(2)
3/2 3/2
(5 2 )
= 1
60.
2
1
2
1
244
x
xx
dx =
2
1
(2x2 + 4x + 4)–1(x + 1)dx
Let u = 2x2 + 4x + 4, then du = (4x + 4)dx = 4(x + 1)dx.
1
1
62.
7
6
ln( 5)
5
t
t
dt
Let u = ln(t – 5), then du = 1
5tdt.
ln( 5)
5
t
t
dt =
u du =
2
2
u + C =
2
(ln( 5))
2
t + C
2
7
6
2
2
2
EXERCISES 12-5 12-31
64.
1
1
ex2 dx ≈ 2.925
66.
3
0
2
9
x
dx ≈ 7.069
68. 110 21 1
1
( ) ( ) [ ( ) ( )] [ ( ) ( )] [F(x ) ( )] ( ) ( ).
n
kk nn
k
F
xFx FxFx FxFx Fx FbFa
70. C‘(x) = 500 – 3
x
on [0, 600]
The increase in cost from a production level of 0 bikes per month to a production level of 600 bikes per
x
x
72. Total maintenance costs from the end of the second year to the end of the seventh year:
7
7
74. (A)
(B) Let q(t) be the quadratic regression model found in part (A). The number of
76. To obtain the useful life, set C‘(t) = R‘(t) and solve for t.
3 = 15e–0.1t
12-32 CHAPTER 12: INTEGRATION
The total profit accumulated during the useful life is:
16
16
15 3
t
16
78. C(x) = 20,000 + 10x
(A) Average cost per unit:
(B) Ave C(x) = 1
1,000
(20,000 + 10x)dx
EXERCISES 12-5 12-33
88. Rate of production: R(t) = 2
120
1
t
t + 3, 0 ≤ t ≤ 20
Total production from year N to year M is given by:
P =
M
N
R(t)dt =
M
N
2
120 3
1
t
t
dt
M
t
M
26
26
90. A‘(t) = –0.9e–0.1t
The change during the first five days is given by:
5
0.1
t
e
5
0
5
0
The change during the second five days, i.e., from the 5th day to the 10th day, is given by:
10
10
5
92. C(t) = 2
0.14
1
t
t
Average concentration during the first hour after injection is given by:
1
0.14
t
1
0
Average concentration during the first two hours after the injection is given by:
2
0.14
t
2
0