EXERCISE 13-3 13-21
44.
2
1
x3ex2 dx
Let t = x2, then dt = 2x dx.
x3ex2 dx =
x2ex2 2
2x dx = 1
2
tet dt
46.
2
0
ln(4 – x)dx
Let u = ln(4 – x) and dv = dx. Then du = – 1
4
x
dx and v = x.
x
x
x
(44)
x
x
x
x
x
48.
xex+1 dx
50.
x ln(1 + x)dx
Let u = ln(1 + x) and dv = x dx. Then du = 1
1
x
dx and v = 1
2x2.
x
x
x
13-22 CHAPTER 13: ADDITIONAL INTEGRATION TOPICS
= 1
2x2 ln(1 + x) – 1
2
(1)(1)
1
xx
x
dx – 1
2 ln(1 + x)
= 1
2x2 ln(1 + x) – 1
2
(x – 1)dx – 1
2ln(1 + x)
2
1(1)
1
52
ln 1
x
x
dx
Let t =
x
= x1/2, then dt = 1
2x–1/2 dx.
ln 1
x
dx =
x
x
x
x
x
x
x
x
x
x
x
x
54.
x(ln x)2 dx
Let u = (ln x)2 and dv = x dx. Then du = 2 ln
x
x
dx and v = 1
2x2.
x(ln x)2 dx = 1
2x2(ln x)2 –
x(ln x)dx
To compute
x(ln x)dx we use integration-by-parts.
x
2
x
2x2(ln x)2 – 1
2x2 ln x + 1
4x2 + C =
2
2
x
2
2
x
2
4
x
EXERCISE 13-3 13-23
56.
x(ln x)3 dx
Let u = (ln x)3 and dv = x dx. Then du =
2
3(ln )
x
x
dx and v = 1
x
x
x
58.
1
e
ln(x4)dx =
1
e
4 ln x dx = 4
1
e
ln x dx
x
x
1
60.
2
1
ln(xex)dx =
2
1
(ln x + ln ex)dx =
2
1
(ln x + x)dx =
2
1
ln x dx +
2
1
x dx
2
x
2
2
1
62.
5
(ln )xdx
x
x
5
56 6
(ln ) 1 1 (ln ) .
xdx u du u C x C
66. 23
ln
x
xdx
13-24 CHAPTER 13: ADDITIONAL INTEGRATION TOPICS
Copyright © 2019 Pearson Education, Inc.
23
11
3
xxx C
68.
3.1 3
0.1 ln 20.47xxdx
70.
2
2
x
xedx
72. The total production over the first 12 months is given by the definite integral:
12
12
Thus,
12
10te–0.1t dt = 10(–10te–0.1t – 100e–0.1t)
12
= 10[(–120e-1.2 – 100e-1.2) – (–100)]
76. From page 395, Future Value = erT
0
T
f(t)e-rt dt. Now r =0.0415,
T = 4, f(t) = 1,000 – 250t. Thus,
13-26 CHAPTER 13: ADDITIONAL INTEGRATION TOPICS
82. S‘(t) = 350 ln(t + 1), S(0) = 0
S(t) =
350 ln(t + 1)dt = 350
ln(t + 1)dt
Let u = ln(t + 1) and dv = dt. Then du = 1
1tdt and v = t.
t
11
t
1
Monthly sales will reach 15,000 games after 20 months.
0
20,000
30
84. p = S(x) = 5 ln(x + 1); p = $26. To find
x
, solve
5 ln(
x
+ 1) = 26
ln(
x
+ 1) = 26
5 = 5.2
x
x
0
0
86.
The area bounded by the price-supply equation, p = 5 ln(x + 1), and the price
equation, y = p = 26, from x = 0 to
x
EXERCISE 13-3 13-27
88. R(t) = te-0.2t
Total amount =
10
0
R(t)dt =
10
0
te-0.2t dt
0.2
t
0.2
Thus,
10
10
90. N‘(t) = (t + 10)e-0.1t, 0 ≤ t ≤ 15; N(0) = 0
N(t) – N(0) =
0
t
N‘(x)dx =
0
t
(x + 10)e-0.1x dx
t
t
t
t
0.1
x
e
t
x
x
0
EXERCISE 13-4 13-29
8.
10 22
2
,() .
ln ln
x
x
dx f x
x
x
Partition [2,10] into four equal subintervals:
01234
2, 4, 6, 8, 10xxxxx
, 2.x
x
()
f
x
2 5.7708
6 20.0920
10 43.4294
By Simpson’s rule:
10. Use Formula 10 with a = b = 1:
2
1
(1 )
x
xdx = 1
x
+ ln 1
x
x
+ C
12. Use Formula 19 with a = 5, b = 2, c = 2, d = 1:
x
x
x
x
14. Use Formula 27 with a = 16 and b = 1:
1
x
x
16. Use Formula 31 with a = 3:
2
9
x
x
dx = 2
9
x
– 3 ln
2
39
x
x
+ C
18. Use Formula 45 with a = 4:
1
216
x
+ C
22. Use Formula 48 with a = 3, c = 5, d = 2:
1
x
– 1
13-30 CHAPTER 13: ADDITIONAL INTEGRATION TOPICS
24. Use Formula 6 with a = 6, b = 1:
6
2
2
(6 )
x
x
dx = 6
ln 6 6
x
x
6
2
26. Use Formula 16 with a = 3, b = 1, c = 1, d = 1:
x
7
x
7
28. Use Formula 40 with a = 4:
5
4
216xdx = 22
116 16 ln 16
2xx x x
5
4
30.
11
33
,() .
x
dx f x x
Partition [1,11] into five equal subintervals:
01 2 3 4 5
1, 3, 5, 7, 9, 11xxx x x x
x
()
f
x
1 1
7 343
9 729
5[(1) 2(3) 2(5) 2(7) 2(9) (11)](2/2)Tf f f f f f
Exact value:
3
1
1
11 1 3,660
444
x
xdx
EXERCISE 13-4 13-31
32.
5
44
,() .
x
dx f x x
Partition [1,5] into eight equal subintervals:
01 2 3 4 5 6 7 8
1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5xx x x x x x x x
x
()
f
x
1 1
2 16
3.5 150.0625
4 256
5 625
8[(1) 4(1.5) 2(2) 4(2.5) 2(3) 4(3.5) 2(4) 4(4.5) (5)](1/6)Sf f f f f f f f f
11
555
34. As explained in the text, if f is a linear function, then the trapezoidal rule gives the exact value of
f
b
36. () 3 2, 1 5.fx x x
Simpson’s rule with 1, 6 subintervals:x
1
(1)4(0)2(1)4(2)2(3)4(4) (5)3
Sf f f f f f f
EXERCISE 13-4 13-33
44.
2
64
x
xdx
Let u = x3, then du = 3x2 dx, and
xdx = 1
46.
x
Let u = x2, then du = 2x dx, and
44x
dx =
4
4x
2
24u
du
x
2
2
x
48.
2
(4 ) (2 )
x
x
x
e
ee
dx
Let u = ex, then du = ex dx and
x
x
e
dx =
1
du
x
x
x
x
x
50.
1
ln 4 ln
x
xxdx
x
13-34 CHAPTER 13: ADDITIONAL INTEGRATION TOPICS
Using Formula 27 with a = 4, b = 1, we have:
x
52.
x2e–4x dx
From Formula 47 with n = 2, a = –4, we have:
x
24
x
xe–4x dx =
x
4
– 1
4
e–4x dx = –1
4xe–4x – 1
16 e–4x + C1
54.
x3e2x dx
From Formula 47 with n = 3, a = 2, we have:
x3e2x dx =
32
x
x
e – 3
x2e2x dx
22
x
x
2
x
x
56.
(ln x)4 dx
EXERCISE 13-4 13-35
(ln x)2 dx = x(ln x)2 – 2
(ln x)dx
(ln x)dx = x(ln x) –
dx = x ln x – x + C
58.
3
x229xdx
From Formula 41 with a = 3, we have:
5
1
22 2
5
60.
4
2
22
(1)
x
xdx
22
(1)
xdx =
22
(1)x 2
2x dx =
2
u · 1
2du = 1
2
u–2 du = –1
2u–1 + C = –1
2(x2 – 1)–1 + C
Thus,
62.
2
(ln )
x
x
dx
Let u = ln x, then du = 1
x
dx, and
x
x
x
64.
21
x
xdx
21
xdx = 1
2
13-36 CHAPTER 13: ADDITIONAL INTEGRATION TOPICS
66.
1
32 32
1
();()ax bx cx d dx f x ax bx cx d
. Partition [1,1] into two equal subintervals:
012
1, 0, 1xxx
x
()
f
x
22
3
bd
68. f(x) = 2
1
x
, g(x) = 5x – x2
0.21
0.21
5
For the second integral we use Formula 32 with a = 1:
x
2
22
EXERCISE 13-4 13-37
70. f(x) = 4
x
x
3.59
4
xdx –
3.59
5
x
72. Find
x
, the supply when the price p = 20:
20 = 10
300
x
x
6,000 – 20
x
= 10
x
30
x
= 6,000
x
x
x
x
x