454 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

4. A = 0.5

1

! [-ln x]dx + 1

e

! ln xdx

We evaluate the integral using integration-by-parts. Let u = ln x, dv = dx.

Then du =

1

xdx, v = x, and ∫ln xdx =

x ln x - ∫x

1

x

! "

# $ dx = x ln x - x + C

Thus, A = -

0.5

1

! ln xdx + 1

e

! ln xdx

= (-x ln x + x) 0.5

1 + (x ln x - x) 1

e

≈ (1 - 0.847) + (1) = 1.153 (7-1)

y

y = ln x

x0.5

e

1

- 1

0 1

(7-1)

5. ∫xe4xdx. Use integration-by-parts:

Let u = x and dv = e4xdx. Then du = dx and v =

e4x

4.

∫xe4xdx =

xe4x

4 - ∫

e4x

4dx =

xe4x

4 -

e4x

16 + C (7-3, 7-4)

6. ∫x ln x dx. Use integration-by-parts:

Let u = ln x and dv = x dx. Then du =

1

xdx and v =

x2

2.

∫x ln x dx =

x2 ln x

2 - ∫

1

x ·

x2

2dx =

x2 ln x

2 -

1

2∫x dx

=

x2 ln x

2 -

x2

4 + C (7-3, 7-4)

7. ∫

ln x

xdx

Let u = ln x. Then du =

1

xdx and

∫

ln x

xdx = ∫u du =

1

2u2 + C =

1

2[ln x]2 + C (6-2)

8. ∫

x

1 + x2dx

Let u = 1 + x2. Then du = 2x dx and

∫

x

1 + x2dx = ∫

1 2 du

u =

1

2∫

1

udu =

1

2 ln|u| + C =

1

2 ln(1 + x2) + C (7-2)

9. Use Formula 11 with a = 1 and b = 1.

∫

1

x(1 + x)2dx =

1

1(1 + x)+

1

12ln

x

1 + x + C =

1

1 + x + ln

x

1 + x + C

(7-4)

CHAPTER 7 REVIEW 455

10. Use Formula 28 with a = 1 and b = 1.

∫

!

1

x2 1 + xdx = -

!

1 + x

1 " x#

1

2 " 1 1ln

1 + x # 1

1 + x + 1 + C

= -

!

1 + x

x -

1

2 ln

!

1 + x " 1

1 + x + 1 + C (7-4)

11. y = 5 – 2x – 6x2; y = 0 on [1, 2] A = -

!

1

2" (5 – 2x – 6x2)dx

=

!

1

2" (6x2 + 2x – 5)dx = (2x3 + x2 – 5x)

!

1

2

= 10 – (-2) = 12

(7-1)

12. y = 5x + 7; y = 12 on [-3, 1] A =

!

"3

1# [12 – (5x + 7)]dx

=

!

"3

1# (5 – 5x)dx =

!

5x "5

2x2

#

$ %

&

' (

!

"3

1

=

!

5 "5

2

#

$ %

&

' ( " "15 "

45

2

#

$ %

&

' ( = 40

(7-1)

13. y = -x + 2; y = x2 + 3 on [-1, 4] A =

!

"1

4# [(x2 + 3) – (-x + 2)]dx

=

!

"1

4# (x2 + x + 1)dx

=

!

1

3x3 +

1

2x2 + x

"

# $

%

& '

!

"1

4

=

!

64

3 + 8 + 4 -

!

"1

3+1

2" 1

#

$ %

&

' (

=

!

205

6 ≈ 34.167

(7-1)

14. y =

!

1

x; y = -e-x on [1, 2]

A =

!

1

2"

!

1

x" e"x

#

$ %

&

' ( dx =

!

1

2"

!

1

x+ e"x

#

$ %

&

' ( dx

= (ln x – e-x)

!

1

2

= (ln 2 – e-2) – (-e-1) = ln 2 – e-2 + e-1 ≈ 0.926

(7-1)

2 –2

–25

x

y 1

–3

y = 12

y = 5x + 7

x

y

4 x –1

12

y

1 2

1

–1

x

y

456 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

15. y = x; y = -x3 on [-2, 2] A =

!

"2

0# (-x3 – x)dx +

!

0

2" x – (-x3)dx

= -

!

"2

0# (x3 + x)dx +

!

0

2" (x + x3)dx

= -

!

1

4x4 +

1

2x2

"

# $

%

& '

!

"2

0 +

!

1

2x2 +

1

4x4

"

# $

%

& '

!

0

2

= 0 + 6 + 6 – 0 = 12

(7-1)

16. y = x2; y = -x4 on [-2, 2] A =

!

"2

2# [x2 – (-x4)]dx

=

!

"2

2# (x2 + x4)dx =

!

1

3x3 +

1

5x5

"

# $

%

& '

!

"2

2

=

!

8

3 +

!

32

5 -

!

"8

3"32

5

#

$ %

&

' ( =

!

16

3 +

!

64

5 ≈ 18.133

(7-1)

17. A = a

b

! [f(x) - g(x)]dx (7-1) 18. A = b

c

! [g(x) - f(x)]dx (7-1)

19. A = b

c

! [g(x) - f(x)]dx + c

d

! [f(x) - g(x)]dx (7-1)

20. A = a

b

! [f(x) - g(x)]dx + b

c

! [g(x) - f(x)]dx + c

d

! [f(x) - g(x)]dx (7-1)

21. A = 0

5

! [(9 - x) - (x2 - 6x + 9)]dx

= 0

5

! (5x - x2)dx

=

5

2x2 !

1

3x3"

# $ % 05

=

125

2!125

3=125

6 ≈ 20.833 (7-1)

5

5

10

y

0x

y = x 2 – 6x + 9

y = 9 – x

(0, 9)

(5, 4)

(7-1) 22.

0

1

! xexdx. Use integration-by-parts.

Let u = x and dv = exdx. Then du = dx and v = ex. ∫xexdx = xex - ∫exdx = xex - ex + C

Therefore, 0

1

! xexdx = (xex - ex) 0

1 = 1·e - e - (0·1 - 1) = 1 (7-3, 7-4)

23. Use Formula 38 with a = 4

0

3

!

!

x2

x2 + 16

dx =

!

1

2x x2 + 16 " 16 ln x + x2 + 16#

$ % &

' ( 03

=

!

1

23 25 " 16 ln(3 + 25)[ ] -

!

1

2("16 ln 16)

=

1

2[15 - 16 ln 8] + 8 ln 4

=

15

2 - 8 ln 8 + 8 ln 4 ≈ 1.955 (7-4)

2 –2

4

–4

x

y

2 –2

4

–8

x

y

CHAPTER 7 REVIEW 457

24. Let u = 3x, then du = 3 dx. Now, use Formula 40 with a = 7.

∫

!

9x2 " 49dx =

1

3∫

!

u2 " 49du

=

1

3 ·

!

1

2u u2 " 49 " 49 lnu + u2 " 49#

$ %

&

' ( + C

=

!

1

63x 9x2 " 49 " 49 ln3x + 9x2 " 49#

$ %

&

' ( + C (7-4)

25. ∫te-0.5t dt. Use integration-by-parts.

Let u = t and dv = e-0.5tdt. Then du = dt and v = e-0.5t-0.5 .

∫te-0.5t dt =

!te!0.5t

0.5 + ∫

e!0.5t

0.5dt =

!te!0.5t

0.5 +

e!0.5t

!0.25 + C

= -2te-0.5t - 4e-0.5t + C (7-3, 7-4)

26. ∫x2 ln x dx. Use integration-by-parts.

Let u = ln x and dv = x2dx. Then du =

1

xdx and v =

x3

3.

∫x2 ln x dx =

x3 ln x

3 - ∫

1

x ·

x3

3dx =

x3 ln x

3 -

1

3∫x2dx

=

x3 ln x

3 -

x3

9 + C (7-3, 7-4)

27. Use Formula 48 with a = 1, c = 1, and d = 2.

∫

1

1 + 2exdx =

x

1!

1

1 " 1ln|1 + 2ex| + C = x - ln|1 + 2ex| + C (7-4)

28. (A) (4, 4)

(2,

2)

0

y = x

y = x3 – 6x2 + 9x

A = 0

2

! [(x3 - 6x2 + 9x) - x]dx + 2

4

! [x - (x3 - 6x2 + 9x)]dx

= 0

2

! (x3 - 6x2 + 8x)dx + 2

4

! (-x3 + 6x2 - 8x)dx

=

1

4x4 ! 2x

3+ 4x

2" #

$ % 02 +

!1

4x4

+ 2x3 ! 4x

2" #

$ % 24

= 4 + 4 = 8

458 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

(B)

0

y = x3 – 6x2 + 9x

(1.7

5,

2.7

5)

(4.11, 5.11)

y = x + 1

(0.14, 1.14)

The x-coordinates of the points of intersection are: x1 ≈ 0.14,

x2 ≈ 1.75, x3 ≈ 4.11.

A = 0.14

1.75

! [(x3 - 6x2 + 9x) - (x + 1)]dx

+ 1.75

4.11

! [(x + 1) - (x3 - 6x2 + 9x)]dx

= 0.14

1.75

! (x3 - 6x2 + 8x - 1)dx + 1.75

4.11

! (1 - x3 + 6x2 - 8x)dx

=

1

4x4 ! 2x

3+ 4x

2 ! x" #

$ % 0.141.75 +

x !

1

4x4

+ 2x3 ! 4x

2" #

$ % 1.754.11

= [2.126 - (-0.066)] + [4.059 - (-2.126)] ≈ 8.38 (7-1)

29. ∫

(ln x)2

xdx = ∫u2du =

u3

3 + C =

(ln x)3

3 + C Substitution: u = ln x

du =

1

xdx (6-2)

30. ∫x(ln x)2dx. Use integration-by parts.

Let u = (ln x)2 and dv = xdx. Then du = 2(ln x)

1

xdx and v =

x2

2.

∫x(ln x)2dx =

x2(ln x)2

2 - ∫2(ln x)

1

x·

x2

2dx =

x2(ln x)2

2 - ∫x ln xdx

Let u = ln x and dv = xdx. Then du =

1

xdx and v =

x2

2.

∫x ln xdx

Thus, ∫x(ln x)2dx =

x2(ln x)2

2 -

!

x2 ln x

2"x2

4

#

$ % %

&

' ( ( + C

=

x2(ln x)2

2 -

x2 ln x

2 +

x2

4 + C. (7-3, 7-4)

31. Let u = x2 - 36. Then du = 2xdx.

∫

!

x

x2 " 36

dx = ∫

x

(x2 ! 36)1 2dx =

1

2∫

1

u1 2du =

1

2∫u-1/2du

=

1

2 ·

u1 2

1 2 + C = u1/2 + C =

!

x2 " 36 + C (6-2)

CHAPTER 7 REVIEW 459

32. Let u = x2, du = 2xdx. Then use Formula 43 with a = 6.

∫

!

x

x4 " 36

dx =

1

2∫

!

du

u2 " 36

=

1

2ln

!

u + u2 " 36 + C

=

1

2ln

!

x2 + x4 " 36 + C (7-4)

33. 0

4

! x ln(10 - x)dx Consider ∫x ln(10 - x)dx = ∫(10 - t)ln t(-dt) = ∫t ln tdt - 10∫ln tdt.

Substitution: t = 10 - x dt = -dx x = 10 - t

Now use integration-by-parts on the two integrals.

Let u = ln t, dv = tdt. Then du =

1

tdt, v =

t2

2.

∫t ln tdt =

t2

2ln t - ∫

t2

2 ·

1

tdt =

t2 ln t

2 -

t2

4 + C

Let u = ln t, dv = dt. Then du =

1

tdt, v = t.

∫t ln tdt = t ln t - ∫t ·

1

tdt = t ln t - t + C

Thus, 0

4

! x ln(10 - x)dx =

!

(10 " x)2 ln(10 " x)

2"(10 " x)2

4

#

$ % %

!

"10(10 " x)ln(10 " x) + 10(10 " x)#

$ % 04

=

36 ln 6

2!36

4 - 10(6) ln 6 + 10(6)

-

!

100 ln 10

2"100

4" 10(10)ln 10 + 10(10)

#

$ % &

' (

= 18 ln 6 - 9 - 60 ln 6 + 60 - 50 ln 10 + 25 + 100 ln 10 - 100

= 50 ln 10 - 42 ln 6 - 24 ≈ 15.875. (7-3, 7-4)

34. Use Formula 52 with n = 2.

∫(ln x)2dx = x(ln x)2 - 2∫ln x dx Now use integration-by-parts to calculate ∫ln x dx. Let u = ln x, dv = dx. Then du =

1

xdx, v = x.

∫ln x dx = x ln x - ∫x ·

1

xdx = x ln x - x + C

Therefore, ∫(ln x)2dx = x(ln x)2 - 2[x ln x - x] + C = x(ln x)2 - 2x ln x + 2x + C. (7-3, 7-4)

460 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

35. ∫xe-2x2 dx

Let u = -2x2. Then du = -4x dx.

∫xe-2x2 dx = -

1

4eudu = -

1

4eu + C

= -

1

4e-2x

2 + C (6-2)

36. ∫x2e-2x dx. Use integration-by-parts. Let u = x2 and dv = e-2xdx.

Then du = 2x dx and v = -

1

2e-2x.

∫x2e-2x dx = -

1

2x2e-2x + xe-2x dx

Now use integration-by-parts again. Let u = x and dv = e-2xdx. Then du = dx and v = -

1

2e-2x.

∫xe-2x dx = -

1

2xe-2x +

1

2∫e-2x dx

= -

1

2xe-2x -

1

4e-2x + C

Thus,

∫x2e-2x dx = -

1

2x2e-2x +

!

"1

2xe"2x "

1

4e"2x

#

$ % &

' ( + C

= -

1

2x2e-2x -

1

2xe-2x -

1

4e-2x + C (7-3, 7-4)

37. First graph the two functions to find the points of intersection.

The curves intersect at the points where x = 1.448 and x = 6.915.

Area A = 1.448

6.915

!

!

6

2 + 5e"x" [0.2x + 1.6]

#

$ %

&

' ( dx

≈ 1.703

5

0

0 10

(7-1)

38. (A) Probability (0 ≤ t ≤ 1) = 0

1

! 0.21e-0.21t dt

= -e-0.21t

0

1

= -e-0.21 + 1 ≈ 0.189

(B) Probability (1 ≤ t ≤ 2) = 1

2

! 0.21e-0.21t dt

= -e-0.21t

1

2

= e-0.21 - e-0.42 ≈ 0.154 (7-2)

CHAPTER 7 REVIEW 461

39.

1 2 3

0.1

0.2

y

y = f(t)

t0 (7-2)

The probability that the product will fail during the second year of warranty is the area under the probability density function y = f(t) from t = 1 to t = 2. (7-2)

40. R'(x) = 65 - 6 ln(x + 1), R(0) = 0 R(x) = ∫[65 - 6 ln(x + 1)]dx = 65x - 6∫ln(x + 1)dx Let z = x + 1. Then dz = dx and ∫ln(x + 1)dx = ∫ln z dz. Now, let u = ln z and dv = dz. Then du =

1

zdz and v = z:

∫ln z dz = z ln z - ∫z

1

z

! " # $ dz = z ln z - ∫dz = z ln z - z + C

Therefore, ∫ln(x + 1)dx = (x + 1)ln(x + 1) - (x + 1) + C and R(x) = 65x - 6[(x + 1)ln(x + 1) - (x + 1)] + C Since R(0) = 0, C = -6. Thus, R(x) = 65x - 6[(x + 1)ln(x + 1) - x]

To find the production level for a revenue of $20,000 per week, solve R(x) = 20,000 for x.

The production level should be 618 hair dryers per week.

At a production level of 1,000 hair dryers per week, revenue

40,000

1,000

0

0

R(1,000) = 65,000 - 6[(1,001)ln(1,001) - 1,000] ≈ $29,506 (7-3)

41. (A)

5

3000

1 4

y

y = f(t)

0t

Total

Income

(B) Total income =

1

4

! 2,500e0.05t dt

= 50,000e0.05t

1

4

= 50,000[e0.2 - e0.05] ≈ $8,507 (7-2)

462 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

42. f(t) = 2,500e0.05t, r = 0.15, T = 5

(A) FV = e(0.15)5 0

5

! 2,500e0.05t e-0.15t dt = 2,500e0.75 0

5

! e-0.1t dt

= -25,000e0.75 e-0.1t

0

5

= 25,000[e0.75 - e0.25] ≈ $20,824

(B) Total income = 0

5

! 2,500e0.05t dt = 50,000e0.05t

0

5

= 50,000[e0.25 - 1] ≈ $14,201

Interest = FV - Total income = $20,824 - $14,201 = $6,623 (7-2)

43. (A)

0.5 1

0.5

1

y

0x

y = f(x)

y = x

CURRENT

y

0x

y = g(x)

y = x

0.5 1

0.5

1 PROJECTED

(B) The income will be more equally distributed 10 years from now since the area between y = x and the projected Lorenz curve is less than the area between y = x and the current Lorenz curve.

(C) Current: Gini Index = 2

0

1

! [x - (0.1x + 0.9x2)]dx

= 2 0

1

! (0.9x - 0.9x2)dx = 2(0.45x2 - 0.3x3) 0

1 = 0.30

Projected: Gini Index = 2

0

1

! (x - x1.5)dx

= 2 0

1

! (x - x3/2)dx = 2

1

2x2 !

2

5x5 2"

# $ % 01 = 2

1

10

! "

# $ = 0.2

Thus, income will be more equally distributed 10 years from now, as indicated in part (B). (7-1)

44. (A) p = D(x) = 70 - 0.2x, p = S(x) = 13 + 0.0012x2 Equilibrium price: D(x) = S(x)

70 - 0.2x = 13 + 0.0012x2 0.0012x2 + 0.2x - 57 = 0

x =

!

"0.2 ± 0.04 + 0.2736

0.0024 =

!0.2 ± 0.56

0.0024

Therefore, x =

!0.2 + 0.56

0.0024 = 150, and p = 70 - 0.2(150) = 40.

CHAPTER 7 REVIEW 463

CS = 0

150

! (70 - 0.2x - 40)dx = 0

150

! (30 - 0.2x)dx

= (30x - 0.1x2) 0

150

= $2,250

PS = 0

150

! [40 - (13 + 0.012x2)]dx = 0

150

! (27 - 0.012x2)dx

C S

P S

p = D(x)

p = S(x)

p =40

x = 150

= (27x - 0.0004x3) 0

150

= $2,700

(B) p = D(x) = 70 - 0.2x, p = S(x) = 13e0.006x Equilibrium price: D(x) = S(x)

70 - 0.2x = 13e0.006x

Using a graphing utility to solve for x, we get x ≈ 170 and p = 70 - 0.2(170) ≈ 36.

CS = 0

170

! (70 - 0.2x - 36)dx = 0

170

! (34 - 0.2x)dx

= (34x - 0.1x2) 0

170

= $2,890

PS = 0

170

! (36 - 13e0.006x)dx = (36x - 2,166.67e0.006x) 0

170

p = 36

C S

P S

p = D(x)

p = S(x)

x = 170

≈ $2,278

(7-2)

464 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

45. (A)

Graph the quadratic regression model and the line p = 52.50 to find the point of intersection.

75

25

0 50

The demand at a price of 52.50 cents per pound is 25,403 lbs.

(B) Let S(x) be the quadratic regression model found in part (A). Then the producers' surplus at the price level of 52.5 cents per pound is given by PS =

0

25.403

! [52.5 - S(x)]dx ≈ $1,216 (7-2)

46. R(t) =

60t

(t + 1)2(t + 2)

The amount of the drug eliminated during the first hour is given by

A = 0

1

!

60t

(t + 1)2(t + 2)dt

We will use the Table of Integration Formulas to calculate this integral. First, let u = t + 2. Then t = u - 2, t + 1 = u - 1, du = dt and

∫

60t

(t + 1)2(t + 2)dt = 60∫

u ! 2

(u ! 1)2 " udu

= 60∫

1

(u ! 1)2du - 120∫

1

u(u ! 1)2du

In the first integral, let v = u - 1, dv = du. Then

60∫

1

(u ! 1)2du = 60∫v-2dv = -60v-1 =

!60

u ! 1

For the second integral, use Formula 11 with a = -1, b = 1:

-120∫

1

u(u ! 1)2du = -120

!1

u ! 1+ ln

u

u ! 1

"

# $ %

& '

Combining these results and replacing u by t + 2, we have:

∫

60t

(t + 1)2(t + 2)dt =

!60

t + 1+

120

t + 1! 120 ln

t + 2

t + 1 + C

=

60

t + 1! 120 ln

t + 2

t + 1 + C

CHAPTER 7 REVIEW 465

Now,

A = 0

1

!

60t

(t + 1)2(t + 2)dt =

60

t + 1! 120 ln

t + 2

t + 1

" #

$ %

&

' ( )

* + 01

= 30 - 120 ln

3

2

! " # $ - 60 + 120 ln 2

≈ 4.522 milliliters The amount of drug eliminated during the 4th hour is given by:

A = 3

4

!

60t

(t + 1)2(t + 2)dt =

60

t + 1! 120 ln

t + 2

t + 1

" #

$ %

&

' ( )

* + 34

= 12 - 120 ln

6

5

! " # $ - 15 + 120 ln

5

4

! " # $

≈ 1.899 milliliters (6-5, 7-4)

47.

5

8

y = R(t)

y

t0 1 3 4 (6-5, 7-1)

48. f(t) =

!

4 3

(t + 1)20 " t " 3

0 otherwise

#

$ %

& %

(A) Probability (0 ≤ t ≤ 1) = 0

1

!

4 3

(t + 1)2dt

To calculate the integral, let u = t + 1, du = dt. Then,

∫

4 3

(t + 1)2dt =

4

3∫u-2du =

4

3

u!1

!1 = -

4

3u + C =

!

"4

3(t + 1) + C

Thus,

0

1

!

4 3

(t + 1)2dt =

!4

3(t + 1) 0

1 = -

2

3 +

4

3 =

2

3 ≈ 0.667

(B) Probability (t ≥ 1) = 1

3

!

4 3

(t + 1)2dt

=

!

"4

3(t + 1) 1

3

= -

1

3 +

2

3 =

1

3 ≈ 0.333 (7-2)

466 CHAPTER 7 ADDITIONAL INTEGRATION TOPICS

49.

1 2 3 4

0.5

1

1.5

0

y = f(t)

t

y

(7-2)

The probability that the doctor will spend more than an hour with a randomly selected patient is the area under the probability density function y = f(t) from t = 1 to t = 3. (7-2)

50. N'(t) =

100t

(1 + t2)2. To find N(t), we calculate

∫

100t

(1 + t2)2dt

Let u = 1 + t2. Then du = 2t dt, and

N(t) = ∫

100t

(1 + t2)2dt = 50∫

1

u2du = 50∫u-2du

= -50

1

u + C

=

!50

1 + t2 + C

At t = 0, we have N(0) = -50 + C

Therefore, C = N(0) + 50 and

N(t) =

!50

1 + t2 + 50 + N(0)

Now, N(3) =

!5

1 + 32 + 50 + N(0) = 45 + N(0)

Thus, the population will increase by 45 thousand during the next 3 years. (6-5, 7-1)

51. We want to find Probability (t ≥ 2) = 2

!

" f(t)dt Since

!"

"

# f(t)dt = !"

2

# f(t)dt + 2

!

" f(t)dt = 1,

2

!

" f(t)dt = 1 - !"

2

# f(t)dt = 1 - 0

2

! f(t)dt (since f(t) = 0 for t ≤ 0) = 1 - Probability (0 ≤ t ≤ 2)

Now, Probability (0 ≤ t ≤ 2) = 0

2

! 0.5e-0.5t dt

= -e-0.5t

0

2

= -e-1 + 1 ≈ 0.632

Therefore, Probability (t ≥ 2) = 1 - 0.632 = 0.368 (7-2)