15. first use formula 5 with a = 3 and b = 1 to find the ...homepage.smc.edu/wong_betty/math28/s/c11...

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


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


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


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


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


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)

15. first use formula 5 with a = 3 and b = 1 to find the ...homepage.smc.edu/wong_betty/math28/s/c11...

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