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446 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS
15. First use Formula 5 with a = 3 and b = 1 to find the indefinite integral.
∫
x2
3 + xdx =
(3 + x)2
2 ! 13"2 ! 3(3 + x)
13+32
13ln|3 + x| + C
=
(3 + x)2
2 - 6(3 + x) + 9 ln|3 + x| + C
Thus, 1
3
!
x2
3 + xdx =
!
(3 + x)2
2" 6(3 + x) + 9 ln3 + x
#
$ % %
&
' ( ( 13
=
(3 + 3)2
2 - 6(3 + 3) + 9 ln|3 + 3|
-
!
(3 + 1)2
2" 6(3 + 1) + 9 ln3 + 1
#
$ % %
&
' ( (
= 9 ln
3
2 - 2 ≈ 1.6492.
17. First use Formula 15 with a = 3, b = c = d = 1 to find the indefinite integral.
∫
1
(3 + x)(1 + x)dx =
1
3 ! 1 " 1 ! 1ln
1 + x
3 + x + C =
1
2 ln
1 + x
3 + x + C
Thus, 0
7
!
1
(3 + x)(1 + x)dx =
1
2ln
1 + x
3 + x 0
7 =
1
2 ln
1 + 7
3 + 7 -
1
2 ln
1
3
=
1
2 ln
4
5 -
1
2ln
1
3 =
1
2 ln
12
5 ≈ 0.4377.
19. First use Formula 36 with a = 3 (a2 = 9) to find the indefinite integral:
∫
!
1
x2 + 9
dx = ln
!
x + x2 + 9 + C
Thus, 0
4
!
!
1
x2 + 9
dx = ln
!
x + x2 + 9 0
4 = ln
!
4 + 16 + 9 - ln
!
9
= ln 9 - ln 3 = ln 3 ≈ 1.0986.
21. Consider Formula 35. Let u = 2x. Then u2 = 4x2, x =
u
2, and dx =
du
2.
∫
!
4x2 + 1
x2dx = ∫
!
u2 + 1
u2
4
du
2 = 2∫
!
u2 + 1
u2du
= 2
!
"u2 + 1
u+ lnu + u2 + 1
#
$
% %
&
'
( ( + C
= 2
!
"4x2 + 1
2x+ ln2x + 4x2 + 1
#
$
% %
&
'
( ( + C
= -
!
4x2 + 1
x + 2 ln
!
2x + 4x2 + 1 + C
EXERCISE 7-4 447
23. Let u = x2. Then du = 2x dx.
∫
!
x
x4 " 16
dx =
1
2∫
!
1
u2 " 16
du
Now use Formula 43 with a = 4 (a2 = 16):
1
2∫
!
1
u2 " 16
du =
1
2 ln
!
u + u2 " 16 + C =
!
1
2ln x2 + x4 " 16 + C
25. Let u = x3. Then du = 3x2dx.
∫x2
!
x6 + 4dx =
1
3∫
!
u2 + 4du
Now use Formula 32 with a = 2 (a2 = 4):
1
3∫
!
u2 + 4du =
1
3 ·
!
1
2u u2 + 4 + 4 lnu + u2 + 4"
# $ %
& ' + C
=
!
1
6x3 x6 + 4 + 4 lnx3 + x6 + 4"
# $ %
& ' + C
27. ∫
!
1
x3 4 " x4dx = ∫
!
x
x4 4 " x4dx
Let u = x2. Then du = 2x dx.
∫
!
x
x4 4 " x4dx =
1
2∫
!
1
u2 4 " u2du
Now use Formula 30 with a = 2 (a2 = 4):
1
2∫
!
1
u2 4 " u2du = -
1
2 ·
!
4 " u2
4u + C =
!
" 4 " x4
8x2 + C
29. ∫
ex
(2 + ex)(3 + 4ex)dx = ∫
1
(2 + u)(3 + 4u)du
Substitution: u = ex, du = ex dx. Now use Formula 15 with a = 2, b = 1, c = 3, d = 4:
∫
1
(2 + u)(3 + 4u)du =
1
2 ! 4 " 3 ! 1 ln
3 + 4u
2 + u + C =
1
5 ln
3 + 4ex
2 + ex + C
31. ∫
!
ln x
x 4 + ln xdx = ∫
!
u
4 + udu
Substitution: u = ln x, du =
1
xdx.
Use Formula 25 with a = 4, b = 1:
∫
!
u
4 + udu =
!
2(u " 2 # 4)
3 # 124 + u + C =
!
2(u " 8)
34 + u + C
=
!
2(ln x " 8)
34 + ln x + C
448 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS
33. Use Formula 47 with n = 2 and a = 5:
∫x2e5xdx =
x2e5x
5 -
2
5∫xe5xdx
To find ∫xe5xdx, use Formula 47 with n = 1, a = 5:
∫xe5xdx =
xe5x
5 -
1
5∫e5xdx =
xe5x
5 -
1
5 ·
e5x
5
Thus, ∫x2e5xdx =
x2e5x
5 -
!
2
5
xe5x
5"
1
25e5x
#
$ % %
&
' ( ( + C
=
x2e5x
5 -
2xe5x
25 +
2e5x
125 + C.
35. Use Formula 47 with n = 3 and a = -1.
∫x3e-xdx =
x3e!x
!1 -
3
!1∫x2e-xdx = -x3e-x + 3∫x2e-xdx
Now ∫x2e-xdx =
x2e!x
!1 -
2
!1∫xe-xdx = -x2e-x + 2∫xe-xdx
and ∫xe-xdx =
xe!x
!1 -
1
!1∫e-xdx = -xe-x - e-x, using Formula 47.
Thus, ∫x3e-xdx = -x3e-x + 3[-x2e-x + 2(-xe-x - e-x)] + C = -x3e-x - 3x2e-x - 6xe-x - 6e-x + C.
37. Use Formula 52 with n = 3: ∫(ln x)3dx = x(ln x)3 - 3∫(ln x)2dx Now ∫(ln x)2dx = x(ln x)2 - 2∫ln xdx using Formula 52 again, and ∫ln xdx = x ln x - x by Formula 49. Thus, ∫(ln x)3dx = x(ln x)3 - 3[x(ln x)2 - 2(x ln x - x)] + C = x(ln x)3 - 3x(ln x)2 + 6x ln x - 6x + C.
39. 3
5
! x
!
x2 " 9dx. First consider the indefinite integral.
Let u = x2 - 9. Then du = 2x dx or x dx =
1
2du. Thus,
∫x
!
x2 " 9dx =
1
2∫u1/2du =
1
2!u3 2
3 2 + C =
1
3(x
2! 9)3/2 + C.
Now, 3
5
! x
!
x2 " 9dx =
1
3(x
2! 9)3/2
3
5 =
1
3 · 163/2 =
64
3.
EXERCISE 7-4 449
41. 2
4
!
1
x2 ! 1dx. Consider the indefinite integral:
∫
1
x2 ! 1dx =
1
2 ! 1 ln
x ! 1
x + 1 + C, using Formula 13 with a = 1.
Thus,
2
4
!
1
x2 ! 1dx =
1
2 ln
x ! 1
x + 1 2
4 =
1
2 ln
3
5 -
1
2 ln
1
3 =
1
2ln
9
5 ≈ 0.2939.
43. ∫
ln x
x2dx = ∫x-2ln xdx
=
x !1
!1ln x -
x !1
(!1)2 + C [Formula 51 with n = -2]
= -
1
xln x -
1
x + C =
!1 ! ln x
x + C
45. ∫
!
x
x2 " 1
dx = ∫
!
1
x2 " 1
!
2
2
"
# $ %
& ' xdx =
1
2∫u-1/2du
Let u = x2 - 1 = u1/2 + C
Then du = 2xdx =
!
x2 " 1 + C
47. f(x) =
!
10
x2 + 1
, g(x) = x2 + 3x
The graphs of f and g are shown at the right. The x-coordinates of the points of intersection are: x1 ≈ -3.70, x2 ≈ 1.36
A = !3.70
1.36
"
!
1
x2 + 1
" (x2 + 3x)#
$ % %
&
' ( ( dx
= 10 !3.70
1.36
"
!
1
x2 + 1
dx - !3.70
1.36
" (x2 + 3x)dx
5-5
10
-5
For the first integral, use Formula 36 with a = 1:
A = (10 ln |x +
!
x2 + 1|) !3.70
1.36 -
!
1
3x3 +
3
2x2
"
# $
%
& ' !3.70
1.36
≈ [11.15 - (-20.19)] - [3.61 - (3.65)] = 31.38
49. f(x) = x
!
x + 4, g(x) = 1 + x
The graphs of f and g are shown at the right. The x-coordinates of the points of intersection are: x1 ≈ -3.49, x2 ≈ 0.83
2-4
3
-3 A =
!3.49
0.83
" [1 + x - x
!
x + 4]dx = !3.49
0.83
" (1 + x)dx - !3.49
0.83
" x
!
x + 4dx
450 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS
For the second integral, use Formula 22 with a = 4 and b = 1:
A =
!
x +1
2x2
"
# $
%
& ' !3.49
0.83 -
!
2[3x " 8]
15(x + 4)3
#
$ %
&
' ( !3.49
0.83
≈ (1.17445 - 2.60005) - (-7.79850 + 0.89693) ≈ 5.48
51. Find x , the demand when the price p = 15:
15 =
7500 ! 30x
300 ! x
4500 - 15 x = 7500 - 30 x 15 x = 3000 x = 200
Consumers' surplus:
CS = 0
x
! [D(x) - p]dx = 0200
!
!
7500 " 30x
300 " x" 15
#
$ % &
' ( dx =
0
200
!
!
3000 " 15x
300 " x
#
$ % &
' ( dx
Use Formula 20 with a = 3000, b = -15, c = 300, d = -1:
CS =
!
"15x
"1+3000("1) " ("15)(300)
("1)2ln300 " x
#
$ %
&
' ( 0
200
= [15x + 1500 ln|300 - x|] 0
200
= 3000 + 1500 ln(100) - 1500 ln(300)
= 3000 + 1500 ln
!
1
3
"
# $ %
& ' ≈ 1352
Thus, the consumers' surplus is $1352.
53.
CS
p = D(x)
x = 200
p = 15
The shaded region represents the consumers' surplus.
55. C'(x) =
250 + 10x
1 + 0.05x, C(0) = 25,000
C(x) = ∫
250 + 10x
1 + 0.05xdx = 250∫
1
1 + 0.05xdx + 10∫
x
1 + 0.05xdx
= 250
!
1
0.05ln1 + 0.05x
"
# $
%
& ' + 10
!
x
0.05"
1
(0.05)2ln1 + 0.05x
#
$ %
&
' ( + K
(Formulas 3 and 4)
= 5,000 ln |1 + 0.05x| + 200x - 4,000 ln |1 + 0.05x| + K = 1,000 ln |1 + 0.05x| + 200x + K
Since C(0) = 25,000, K = 25,000 and C(x) = 1,000 ln(1 + 0.05x) + 200x + 25,000, x ≥ 0
EXERCISE 7-4 451
To find the production level that produces a cost of $150,000, solve C(x) = 150,000 for x: The production level is x = 608 pairs of skis.
At a production level of 850 pairs of skis,
300,000
1,000
0
0
C(850) = 1,000 ln(1 + 0.05[850]) + 200(850) + 25,000 ≈ $198,773.
57. FV = erT 0
T
! f(t)e-rtdt
Now, r = 0.1, T = 10, f(t) = 50t2.
FV = e(0.1)10 0
10
! 50t2e-0.1tdt = 50e 0
10
! t2e-0.1tdt
To evaluate the integral, use Formula 47 with n = 2 and a = -0.1:
∫t2e-0.1tdt =
t2e!0.1t
!0.1 -
2
!0.1∫te-0.1tdt = -10t2e-0.1t + 20∫te-0.1tdt
Now, using Formula 47 again:
∫te-0.1tdt =
te!0.1t
!0.1 -
1
!0.1∫e-0.1tdt = -10te-0.1t + 10
e-0.1t-0.1
= -10te-0.1t - 100e-0.1t Thus, ∫t2e-0.1tdt = -10t2e-0.1t - 200te-0.1t - 2000e-0.1t + C.
FV = 50e[-10t2e-0.1t - 200te-0.1t - 2000e-0.1t] 0
10
= 50e[-1000e-1 - 2000e-1 - 2000e-1 + 2000] = 100,000e - 250,000 ≈ 21,828 or $21,828
59. Gini Index:
2 0
1
! [x - f(x)]dx = 2 0
1
!
!
x "1
2x 1 + 3x
#
$ % &
' ( dx =
0
1
! [2x - x
!
1 + 3x ]dx
= 0
1
! 2xdx - 0
1
! x
!
1 + 3x dx
For the second integral, use Formula 22 with a = 1 and b = 3:
= x2 0
1 -
!
2(3 " 3x # 2 " 1)
15(3)2(1 + 3x)3
0
1
= 1 -
!
2(9x " 2)
135(1 + 3x)3
0
1
= 1 -
!
14
13543 -
!
4
13513
= 1 -
112
135 -
4
135 =
19
135 ≈ 0.1407
452 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS
61.
As the area bounded by the two curves gets smaller, the Lorenz curve approaches y = x and the distribution of income approaches perfect equality — all individuals share equally in the income.
63. S'(t) =
t2
(1 + t)2; S(t) = ∫
t2
(1 + t)2dt
Use Formula 7 with a = 1 and b = 1:
S(t) =
1 + t
13!
12
13(1 + t)!2(1)
13ln|1 + t| + C
= 1 + t -
1
1 + t - 2 ln|1 + t| + C
Since S(0) = 0, we have 0 = 1 - 1 - 2 ln 1 + C and C = 0. Thus, S(t) = 1 + t -
1
1 + t - 2 ln|1 + t|.
Now, the total sales during the first two years (= 24 months) is given by: S(24) = 1 + 24 -
1
1 + 24 - 2 ln|1 + 24| = 24.96 - 2 ln 25 ≈ 18.5
Thus, total sales during the first two years is approximately $18.5 million.
65.
12 24
0.5
1
0t
S'(t)
y = S'(t)
Total Sales
(millions of dollars)
Mo
nth
ly R
ate
of
Sal
es
(mil
lio
ns
of
do
llar
s p
er m
on
th)
Months
The total sales, in millions of dollars, over the first two years (24 months) is the area under the curve y = S'(t) from t = 0 to t = 24.
67. P'(x) = x
!
2 + 3x , P(1) = -$2,000
P(x) = ∫x
!
2 + 3x dx =
!
2(9x " 4)
135(2 + 3x)3/2 + C
(Formula 22)
P(1) =
2(5)
13553/2 + C = -2,000
C = -2,000 -
2
2753/2 ≈ -2,000.83
Thus, P(x) =
!
2(9x " 4)
135(2 + 3x)3/2 - 2,000.83.
26,000
100
0
0
The number of cars that must be sold to have a profit of $13,000: 54 Profit if 80 cars are sold per week:
P(80) =
!
2(716)
135(242)3/2 - 2,000.83 ≈ $37,932.20
CHAPTER 7 REVIEW 453
69. dRdt =
!
100
t2 + 9
. Therefore,
R = ∫
!
100
t2 + 9
dt = 100∫
!
1
t2 + 9
dt
Using Formula 36 with a = 3 (a2 = 9), we have:
R = 100 ln
!
t + t2 + 9 + C
Now R(0) = 0, so 0 = 100 ln|3| + C or C = -100 ln 3. Thus,
R(t) = 100 ln
!
t + t2 + 9 - 100 ln 3
and
R(4) = 100 ln(4 +
!
42
+ 9) - 100 ln 3 = 100 ln 9 - 100 ln 3 = 100 ln 3 ≈ 110 feet
71. N'(t) =
!
60
t2 + 25
The number of items learned in the first twelve hours of study is given by:
N = 0
12
!
!
60
t2 + 25
dt = 60 0
12
!
!
1
t2 + 25
dt
= 60
!
lnt + t2 + 25"
# $
%
& ' 0
12, using Formula 36
= 60
!
ln12 + 122
+ 25 " ln 25#
$ % &
' (
= 60(ln 25 - ln 5) = 60 ln 5 ≈ 96.57 or 97 items
73.
6 12
6
12
0t
y = N'(t)
y
Total Number of
Items Learned
Hours of Study
Rat
e of
Lea
rnin
g
The area under the rate of learning curve, y = N'(t), from t = 0 to t = 12 represents the total number of items learned in that time interval.
CHAPTER 7 REVIEW
1. A = a
b
! f(x)dx (7-1) 2. A = b
c
! [-f(x)]dx (7-1)
3. A = a
b
! f(x)dx + b
c
! [-f(x)]dx (7-1)