7. ∫xe-xdx
Let u = x and dv = e-xdx. Then du = dx and v = -e-x.
∫xe-xdx = x(-e-x) - ∫(-e-x)dx = -xe-x + ∫e-xdx = -xe-x - e-x + C
1 x2
1
e 2xdx = ∫eudu
∫
2
2
2
2
2
1
1
Let u = x ,
=
eu + C = ex + C
2
2
then du = 2x dx.
2
9. ∫xex dx =
11.
1
!0 (x
∫ex
22
xdx =
- 3)exdx
Let u = (x - 3) and dv = exdx. Then du = dx and v = ex.
∫(x - 3)exdx = (x - 3)ex - ∫exdx = (x - 3)ex - ex + C
= xex - 4ex + C.
Thus,
13.
3
!1 ln
1
!0 (x
1
- 3)exdx = (xex - 4ex) 0 = (e - 4e) - (-4)
= -3e + 4 ≈ -4.1548.
2xdx
dx
and v = x.
x
dx
∫ln 2xdx = (ln 2x)(x) - ∫x · x = x ln 2x - x + C
Let u = ln 2x and dv = dx.
Thus,
15.
17.
436
3
!1 ln
Then du =
3
2xdx = (x ln 2x - x) 1
= (3 ln 6 - 3) - (ln 2 - 1)
≈ 2.6821.
2x
1
dx = ∫ du = ln|u| + C = ln(x2 + 1) + C
+ 1
u
Substitution: u = x2 + 1
du = 2xdx
[Note: Absolute value not needed, since x2 + 1 ≥ 0.]
∫ x2
u2
(ln x)2
ln x
dx
=
udu
=
+
C
=
+ C
∫ x
∫
2
2
Substitution: u = ln x
1
du =
dx
x
CHAPTER 7
ADDITIONAL INTEGRATION TOPICS
19.
∫
x ln xdx = ∫x1/2ln xdx
dx
2 3/2
and v =
x .
x
3
dx
2
2
2
2
!
∫x1/2ln xdx = 3 x3/2 ln x - ∫ 3 x3/2 x = 3 x3/2 ln x - 3 ∫x1/2dx
4
2
= x3/2 ln x - x3/2 + C
9
3
6
21. ∫(x – 1) (x + 3)dx
Let u = ln x and dv = x1/2dx.
Then du =
Let u = x + 3 and dv = (x – 1)6!. Then du = dx and v =
(x " 1)7
7
(x " 1)7
(x " 1)7
- ∫
dx
7
7
(x + 3)(x " 1)7
(x " 1)8
!
=
+ C
7
56
!
!
23. ∫(x + 1)2(x – 1)2dx = ∫(x2 - 1)2dx = ∫(x4 – 2x2 + 1)dx
∫(x – 1)6(x + 3)dx = (x + 3)
!
!
=
2
1 5
x - x3 + x + C
3
5
25.
Since f!
(x) = (x - 3)e x < 0 on [0, 1], the
integral represents the negative of the area
between the graph of f and the x-axis from
x = 0 to x = 1.
A
27.
y = ln 2x
A
The integral represents the area between the
curve y = ln 2x and the x-axis from x = 1 to
x = 3.
29. ∫x2exdx
Let u = x2 and dv = exdx. Then du = 2x dx and v = ex.
∫x2exdx = x2ex - ∫ex(2x)dx = x2ex - 2∫xexdx
∫xexdx
can be computed by using integration-by-parts again.
EXERCISE 7-3
437
Let u = x and dv = exdx. Then du = dx and v = ex.
∫xexdx = xex - ∫exdx = xex - ex + C
and
∫x2exdx = x2ex - 2(xex - ex) + C = x2ex - 2xex + 2ex + C
= (x2 - 2x + 2)ex + C
31. ∫xeaxdx
Let u = x and dv = eaxdx.
∫xeaxdx = x ·
33.
e ln x
x2
!1
eax
a
Then du = dx and v =
eax
eax
xe ax
dx
=
+ C
∫ a
a2
a
eax
.
a
dx
dx
!1
and v =
.
x
x
ln x
dx
1 dx
ln x
ln x
1
" 1
∫ x 2 dx = (ln x) # ! x $% - ∫- x · x = - x + ∫ x 2 = - x - x + C
e ln x
1
ln e
1
1
" ln x
" ln 1
Thus, !
dx = # !
! $% e = !
! #!
! $%
2
1 x
x
x 1
e
e
1
1
2
= + 1 ≈ 0.2642.
e
[Note: ln e = 1.]
Let u = ln x and dv =
35.
2
!0 ln(x
dx
.
x2
Then du =
+ 4)dx
Let t = x + 4. Then dt = dx and
∫ln(x + 4)dx = ∫ln t dt.
Now, let u = ln t and dv = dt.
Then du =
dt
and v = t.
t
!1
t dt = t ln t - ∫t " #$ dt = t ln t - ∫dt = t ln t - t + C
t
Thus, ∫ln(x + 4)dx = (x + 4) ln(x + 4) - (x + 4) + C
∫ln
and
2
!0 ln(x
2
+ 4)dx = [(x + 4) ln(x + 4) - (x + 4)] 0
= 6 ln 6 - 6 - (4 ln 4 - 4) = 6 ln 6 - 4 ln 4 – 2
≈ 3.205.
37. ∫xex-2dx
Let u = x and dv = ex-2dx. Then du = dx and v = ex-2.
∫xex-2dx = xex-2 - ∫ex-2 dx = xex-2 - ex-2 + C
438
CHAPTER 7
ADDITIONAL INTEGRATION TOPICS
39. ∫x ln(1 + x2)dx
Let t = 1 + x2.
Then dt = 2x dx and
dt
1
=
∫ln t dt.
2
2
dt
Now, for ∫ln t dt, let u = ln t, dv = dt. Then du =
and v = t.
t
1
!1
ln t dt = t ln t - ∫t " #$ dt = t ln t - ∫dt = t ln t - t + C
∫
t
2
Therefore,
1
1
∫x ln(1 + x2)dx = 2 (1 + x2)ln(1 + x2) - 2 (1 + x2) + C.
∫x
ln(1 + x2)dx =
∫ln(1
+ x2)x dx =
∫ln
t
41. ∫ex ln(1 + ex)dx
Let t = 1 + ex. Then dt = exdx and
∫ex ln(1 + ex)dx = ∫ln t dt.
Now, as shown in Problems 31 and 35,
∫ln t dt = t ln t - t + C.
Thus,
∫ex ln(1
+ ex)dx = (1 + ex)ln(1 + ex) - (1 + ex) + C.
43. ∫(ln x)2dx
Let u = (ln x)2 and dv = dx.
∫(ln x)2dx =
∫ln x dx can
Then du =
2 ln x
dx and v = x.
x
2 ln x
dx = x(ln x)2 - 2∫ln x dx
x
be computed by using integration-by-parts again.
x(ln x)2 - ∫x ·
As shown in Problems 31 and 35,
∫ln
x dx = x ln x - x + C.
Thus,
∫(ln
x)2dx = x(ln x)2 - 2(x ln x - x) + C
= x(ln x)2 - 2x ln x + 2x + C.
45. ∫(ln x3)dx
1
dx and v = x.
x
1
∫(ln x3)dx = x(ln x)3 - ∫x · 3(ln x)2 · x dx = x(ln x)3 - 3∫(ln x3)dx
Now, using Problem 39,
∫(ln x2)dx = x(ln x)2 - 2x ln x + 2x + C.
Let u = (ln x)3 and dv = dx.
Therefore,
∫(ln
Then du = 3(ln x)2 .
x3)dx = x(ln x)3 - 3[x(ln x)2 - 2x ln x + 2x] + C
= x(ln x)3 - 3x(ln x)2 + 6x ln x - 6x + C.
EXERCISE 7-3
439
47.
e
"1
ln(x2)dx =
e
e
"1
2 ln x dx = 2 "1 ln x dx
By Example 4, ∫ln x dx = x ln x – x + C. Therefore
e
2 "1 ln x dx = 2(x ln x – x) e = 2[e ln e – e] – 2[ln 1 – 1] = 2
1
!
!
!
49.
!
1
"0 ln(ex
2
)dx =
1
"0 x2
dx =
1 3 1
1
x
=
0
3
3
2
!
(Note: ln(ex ) = x2 ln e = x2.)
!
!
!
!
2
!
0
4
51. y = x - 2 - ln x, 1 ≤ x ≤ 4
y = 0 at x ≈ 3.146
-2
A =
3.146
4
=
[-(x - 2 - ln x)]dx + !
!1
3.146
3.146
4
(ln x + 2 - x)dx + !
(x
!1
3.146
(x - 2 - ln x)dx
- 2 - ln x)dx
∫ln
x dx is found using integration-by-parts. Let u = ln x and
1
dv = dx. Then du =
dx and v = x.
x
!1
∫ln x dx = x ln x - ∫x " x #$ dx = x ln x - ∫dx = x ln x - x + C
Thus,
1
"
"1
A = # x ln x ! x + 2x ! x 2$% 3.146 + # x 2 ! 2x ! x ln x + x $% 4
1
3.146
2
2
1 2$ 3.146
"
"1 2
= # x ln x + x !
x
+ # x ! x ! x ln x $% 4
3.146
2 % 1
2
≈ (1.803 - 0.5) + (-1.545 + 1.803) = 1.561
Now,
10
53. y = 5 - xe x , 0 ≤ x ≤ 3
y = 0 at x ≈ 1.327
A =
1.327
(5 - xex )dx +
3
0
3
[-(5 - xex )]dx
!0
!1.327
1.327
3
= !
(5 - xex )dx + !
(xex - 5)dx
0
1.327
Now, ∫xex dx is found using integration-by-parts.
-50
Let u = x and
dv = e x dx. Then, du = dx and v = e x .
∫xex dx = xe x - ∫e x dx = xe x - e x + C
Thus,
A = (5x - [xe x - e x ]) 1.327 + (xe x - e x - 5x) 3
0
≈ (5.402 - 1) + (25.171 - [-5.402]) ≈ 34.98
440
CHAPTER 7
ADDITIONAL INTEGRATION TOPICS
1.327
55. Marginal profit: P'(t) = 2t - te-t.
The total profit over the first 5 years is given by the definite
integral:
5
!0
(2t - te-t)dt =
5
!0
2tdt -
5
-tdt
!0 te
We calculate the second integral using integration-by-parts.
Let u = t and dv = e-t dt. Then du = dt and v = -e-t
∫te-tdt = -te-t - ∫-e-tdt = -te-t - e-t + C = -e-t [t + 1] + C
Thus,
5
5
Total profit = t 2 0 + (e-t [t + 1]) 0
≈ 25 + (0.040 - 1) = 24.040
To the nearest million, the total profit is $24 million.
P'(t)
57.
P'(t) = 2t - te - t
10
The total profit for the first five years
(in millions of dollars) is the same as the
area under the marginal profit function,
P'(t) = 2t - te-t, from t = 0 to t = 5.
5
Total
Profit
0
5
t
T
59. From Section 7-2, Future Value = erT ! f(t)e-rtdt.
0
T = 5, f(t) = 1000 - 200t.
Now r = 0.08,
Thus,
5
FV = e(0.08)5 ! (1000 - 200t)e-0.08tdt
=
0
0.4 5 -0.08t
1000e
e
dt
0
!
5
- 200e0.4 ! te-0.08tdt.
0
We calculate the second integral using integration-by-parts.
e!0.08t
.
!0.08
te !0.08t
e!0.08t
e!0.08t
-0.08t
-0.08t
te
dt
=
dt
=
-12.5
te
+ C
∫
∫ !0.08
!0.08
0.0064
= -12.5te-0.08t - 156.25e-0.08t + C
Thus, we have:
Let u = t, dv = e-0.08tdt.
Then du = dt and v =
e!0.08t 5
- 200e0.4[-12.5te-0.08t - 156.25e-0.08t] 5
0
!0.08 0
0.4
0.4
-0.4
-0.4
= -12,500 + 12,500e
- 200e [-62.5e
- 156.25e
+ 156.25]
0.4
0.4
= -12,500 + 12,500e
+ 43,750 - 31,250e
0.4
= 31,250 - 18,750e
≈ 3,278 or $3,278
FV = 1000e0.4
EXERCISE 7-3
441
1
61. Gini Index = 2 ! (x - xex-1)dx
0
1
1
0
0
= 2 ! xdx - 2 ! xex-1dx
We calculate the second integral using integration-by-parts.
Let u = x, dv = ex-1dx. Then du = dx, v = ex-1.
∫xex-1dx = xex-1 - ∫ex-1dx = xex-1 - ex-1 + C
1
1
Therefore, 2 ! xdx - 2 ! xex-1dx = x2 1 - 2[xex-1 - ex-1] 1
0
0
0
0
-1
y
63.
1
y=x
y = xe(x-1)
0.5
0
0.5
1
x
= 1 - 2[1 - 1 + (e )]
= 1 - 2e-1 ≈ 0.264.
The area bounded by y = x and the Lorenz curve
y = xe(x-1) divided by the area under the curve
y = x from x = 0 to x = 1 is the index of income
concentration, in this case 0.264. It is a
measure of the concentration of income—the
closer to zero, the closer to all the income
being equally distributed; the closer to one, the
closer to all the income being concentrated in a
few hands.
65. S'(t) = -4te0.1t, S(0) = 2,000
S(t) = ∫-4te0.1t dt = -4∫te0.1t dt
e0.1t
= 10e0.1t
0.1
∫te0.1t dt = 10te0.1t - ∫10e0.1t dt = 10te0.1t - 100e0.1t + C
Let u = t and dv = e0.1t dt. Then du = dt and v =
2000
S(t) = -40te0.1t + 400e0.1t + C
Now,
Since S(0) = 2,000, we have
2,000 = 400 + C, C = 1,600
0
30
Thus,
S(t) = 1,600 + 400e0.1t - 40te0.1t
0
To find how long the company will continue to manufacture this
computer, solve S(t) = 800 for t.
The company will manufacture the computer for 15 months.
67. p = D(x) = 9 - ln(x
9 - ln( x + 4)
ln( x + 4)
x + 4
x
Now,
CS =
1,000
!0
+
=
=
=
≈
4); p = $2.089. To find x , solve
2.089
6.911
e6.911 (take the exponential of both sides)
1,000
(D(x) - p )dx =
=
442
CHAPTER 7
1,000
!0
1,000
!0
[9 - ln(x + 4) - 2.089]dx
6.911dx -
ADDITIONAL INTEGRATION TOPICS
1,000
!0
ln(x + 4)dx
To calculate the second integral, we first let z = x + 4 and dz = dx
to get
∫ln(x + 4)dx = ∫ln zdz
Then we use integration-by-parts. Let u = ln z and dv = dz.
1
Then du =
dz and v = z.
z
1
∫ln zdz = z ln z - ∫z · z dz = z ln z - z + C
Therefore,
∫ln(x + 4)dx = (x + 4)ln(x + 4) - (x + 4) + C
and
CS = 6.911x 1,000 - [(x + 4)ln(x + 4) - (x + 4)] 1,000
0
0
≈ 6911 - (5935.39 - 1.55) ≈ $977
69.
The area bounded by
! the price-demand
equation,
p = 9 - ln(x + 4), and the price equation,
y = p = 2.089, from x = 0 to x = x =
1,000, represents the consumers' surplus.
This is the amount saved by consumers who
are willing to pay more than $2.089.
!
p = D(x)
p = 2.089
CS
x = 1000
71. Average concentration: =
5 20 ln(t + 1)
5 ln(t + 1)
1
dt = 4 !
2
!
0 (t + 1)2
5 ! 0 0 (t + 1)
ln(t + 1)
dt is found using integration-by-parts.
+ 1)2
∫ (t
Let u = ln(t + 1) and dv = (t + 1)-2dt.
1
Then du =
dt = (t + 1)-1dt and v = -(t + 1)-1.
t + 1
ln(t + 1)
ln(t + 1)
dt = 2
+ 1)
t + 1
ln(t + 1)
= +
t + 1
∫ (t
∫
- (t + 1)-1(t + 1)-1dt
∫(t
+ 1)-2dt = -
1
ln(t + 1)
- t + 1
t + 1
+ C
Therefore, the average concentration is:
1  5
 ln(t + 1)
1 5 20 ln(t + 1)
dt
=
4
t + 1
t + 1 0
5 !0 (t + 1)2
1
" ln 6
= 4# !
! $% - 4(-ln 1 - 1)
6
6
= 4 -
1
2
2
ln 6 =
(10 - 2 ln 6) ≈ 2.1388 ppm
3
3
3
EXERCISE 7-3
443
73. N'(t) = (t + 6)e-0.25t, 0 ≤ t ≤ 15; N(0) = 40
N(t) - N(0) =
N(t) = 40 +
t
t
!0 N'(x)dx;
!0 (x
t
+ 6)e-0.25xdx = 40 + 6 ! e-0.25xdx +
0
t
= 40 + 6(-4e-0.25x) 0
= 64 - 24e-0.25t +
t
+
t
!0 xe
-0.25x
!0 xe
-0.25x
t
!0 xe
-0.25x
dx
dx
dx
Let u = x and dv = e-0.25xdx. Then du = dx and v = -4e-0.25x;
∫xe-0.25xdx = -4xe-0.25x - ∫ -4e-0.25xdx = -4xe-0.25x - 16e-0.25x + C
Now,
t
!0 xe
-0.25x
t
dx = (-4xe-0.25x - 16e-0.25x) 0
= -4te-0.25t - 16e-0.25t + 16
and
N(t) = 80 - 40e-0.25t - 4te-0.25t
To find how long it will take a student to
achieve the 70 words per minute level, solve
N(t) = 70:
100
0
15
It will take 8 weeks.
By the end of the course, a student should be
0
able to type
N(15) = 80 - 40e-0.25(15) - 60e-0.25(15) ≈ 78 words per minute.
1 5
(20 + 4t - 5te-0.1t)dt
5 !0
5
1 5
=
(20
+
4
t
)
dt
te-0.1tdt
!
!
0
0
5
-0.1t
dt is found using integration-by-parts.
∫te
75. Average number of voters =
e!0.1t
= -10e-0.1t.
!0.1
∫te-0.1tdt = -10te-0.1t - ∫-10e-0.1tdt = -10te-0.1t + 10∫e-0.1tdt
Let u = t and dv = e-0.1tdt.
Then du = dt and v =
10e !0.1t
= -10te
+
+ C = -10te-0.1t - 100e-0.1t + C
!0.1
Therefore, the average number of voters is:
5
1 5
(20 + 4t)dt - ! te-0.1tdt
!
0
5 0
5
5
1
= 5 (20t + 2t2) 0 - (-10te-0.1t - 100e-0.1t) 0
-0.1t
1
= 5 (100 + 50)
= 30 + (50e-0.5 + 100e-0.5) - 100
= 150e-0.5 - 70
≈ 20.98 (thousands) or 20,980
444
CHAPTER 7
5
+ (10te-0.1t + 100e-0.1t) 0
ADDITIONAL INTEGRATION TOPICS
EXERCISE 7-4
1. Use Formula 9 with a = b = 1.
x
x
1
1
∫ x(1 + x)dx = 1 ln 1 + x + C = ln x + 1
+ C
3. Use Formula 18 with a = 3, b = 1, c = 5, d = 2:
1
2
5 + 2x
1
1
∫(3 + x)2(5 + 2x)dx = 3 ! 2 " 5 ! 1 · 3 + x + (3 ! 2 " 5 ! 1)2 ln 3 + x
5 + 2x
1
=
+ 2 ln
+ C
3 + x
3+ x
+ C
5. Use Formula 25 with a = 16 and b = 1:
2(x " 2 # 16)
x
2(x " 32)
16 + x + C =
dx =
16 + x + C
2
3
16 + x
3#1
∫
7. Use Formula 29 with a = 1:
!
∫
1 !
x 1 " x2
dx = -
1
1+
ln
1
= -ln
!
1+
1 " x2
x
1 " x2
x
!
+ C
+ C
!
9. Use Formula 37 with a = 2 (a2 = 4):
!
1
dx
=
ln
∫ 2
2
x x + 4
2 +
1
x
x2 + 4
+ C
11. Use Formula 51 with n = 2:
!
2
∫x
2+ 1
x 2 +1
x3
x3
! x
ln x dx =
ln x + C =
ln x + C
(2 + 1)2
2+ 1
3
9
13. First let u = ex. Then du = exdx and dx =
1
1
dx = ∫
du
x
+ e
u(1 + u)
Now use Formula 9 with a = b = 1:
u
1
1
∫ u(1 + u)du = 1 ln 1 + u + C
ex
= ln
+ C
1 + ex
Thus,
1
1
du = du .
x
e
u
∫1
= ln| e x | - ln|1 + ex| + C
= x - ln|1 + ex| + C
EXERCISE 7-4
445