Chapter 6 Differential Equations and Mathematical Modeling

264 Section 6.1

Chapter 6Differential Equations andMathematical Modeling

Section 6.1 Slope Fields and Euler’s Method(pp. 321–330)

Exploration 1 Seeing the Slopes

1. Sincedy

dx= 0 represents a line with a slope of 0, we should

expect to see intervals with no change in y. We see this atodd multiples of π / .2

2. Since y is the dependent variable, I

t will have no effect on the value of dy

dxx= cos .

3. The graph of dy

dxwill look the same at all values of y.

4. When xdy

dxx= = =0 1, cos . This can be seen on the graph

near the origin. At that point, the change in y and change inx are the same.

5. When xdy

dxx= = = −π , cos .1 This can be seen in the

graph at x = π. At this point, the change in y is negative ofthe change in x.

6. This is true because each point on the graph has a negativeof itself.

Quick Review 6.1

1. Yes.d

dxe ex x=

2. Yes.d

dxe ex x4 44=

3. No.d

dxx e xe x ex x x( )2 22= +

4. Yes.d

dxe xex x2 2

2=

5. No.d

dxe xex x( )

2 2

5 2+ =

6. Yes.d

dxx

x x2

1

2 22

1

2= =( )

7. Yes.d

dxx x xsec sec tan=

8. No.d

dxx x− −= −1 2

9. y x x C= + += + += −

3 42

2 3(1) 4(1)5

2 CC

10. y x x C= − +2 3sin cos

4 2 sin(0) 3 cos(0)7

= − += −

CC

11. y e x Cx= + +2 sec

5 03

2 0= + +=

e CC

( ) sec( )

12. y x x= + − +−tan ln(2 1)1 C

ππ

= + − +

=

−tan (1) ln(2(1) 1)3

4

1 C

C

Section 6.1 Exercises

1. dy x x dx∫ ∫= −( sec )5 4 2

y x x C= − +5 tan

2. dy x x e dxx= −∫∫ (sec tan )

y x e Cx= − +sec

3. dy x e x dxx= − +−∫∫ (sin )8 3

y x e x Cx= − + + +−cos 2 4

4. dyx x

dx xx

C= −⎛⎝⎜

⎞⎠⎟

= + +∫∫1 1 1

2ln

5. dyx

dx x Cx x= ++

⎛⎝⎜

⎞⎠⎟

= + +−∫∫ 5 51

15

21ln tan

6. dyx x

dx x x C=−

−⎛

⎝⎜

⎞

⎠⎟ = − +−∫∫

1

1

12

2

1sin

7. dy t t dt t C= = +∫∫ ( cos( )) sin( )3 3 3

8. dy t e dtt= ∫∫ cos sin

= +e Ctsin

9. dy x x dx= ∫∫ (sec ( )( ))2 5 45

= +tan x C5

10. dy u udu= ∫∫ 4 3(sin ) cos

= +(sin )u C4

11. dy x dx x C= = − +∫∫ 3 3sin cos

2 3 0 5= − + =cos( ) ,C C

y x= − +3 5cos

12. dy e x dx e x Cx x= − = − +∫∫ 2 2cos sin

3 2 0 1

2 1

0= − + == − +

e C C

y e xx

sin( ) ,

sin

13. du x x dx x x x C= − + = − + +∫∫ ( )7 3 5 56 2 7 3

1 1 1 5 4

5 4

7 3

7 3= − + + = −= − + −

C C

u x x x

,

Section 6.1 265

14. dA x x x dx x x x x C= + − + = + − + +∫∫ ( )10 5 2 4 49 4 10 4 2

6 1 1 1 4 1 1

4 1

10 4 2

10 4 2= + − + + == + − + +

( ) ,C C

A x x x x

15. dyx x

dx x x x c= − − +⎛⎝⎜

⎞⎠⎟

= + + +− −∫∫1 3

12 122 4

1 3

3 1 1 12 1 11

12 1

1 3

1 3= + + + = −= + + −

− −

− −( ) ,C C

y x x x 11 0( )x >

16. dy x x dx x x c= −⎛⎝⎜

⎞⎠⎟

= − +∫∫ 53

252 3 2sec tan /

7 5 0 0 7

5 7

3 2

3 2= − + == − +

tan( ) ( ) ,

tan

/

/C C

y x x

17. dyt

dt t Ct t=+

+⎛⎝⎜

⎞⎠⎟

= + +∫∫ −1

12 2 2

21ln tan

3 0 2 2

2

1 0

1= + + == + +

−

−tan ( ) ,

tan

C C

y t Ct

18. dxt t

dt t t t C= − +⎛⎝⎜

⎞⎠⎟

= + + +∫∫ −1 16 6

21ln

0 1 1 6 1 7

6

1

1= + + + = −= + + −

−

−ln( ) ( ) ,

ln

C C

x t t t 77 0( )t >

19. dv t t e t dt t e t Ct t= + +( ) = + + +∫∫ 4 6 4 3 2sec tan sec

5 4 0 3 0 00 2= + + + =sec( ) ( ) ,e C C

V t e t tt= + + −4 32

2secπ < <<⎛

⎝⎜⎞⎠⎟

π2

20. ds t t dt t t C= − = − +∫∫ ( )3 2 3 2

0 1 1 03 2

3 2= − + == −

( ) ( ) ,C C

s t t

21.dy

dx

d

dxf t dt t dt

x

a

x= = ∫∫ ( ) sin( )2

1

y tx

= +∫ sin( ) 52

1dt

22.du

dx

d

dxf t dt t dt

x

a

x= = +∫∫ ( ) cos2

0

u t dtx

= + −∫ 2 30

cos

23. F xd

dxf t dt e dttx

a

x1

2( ) ( ) cos= = ∫∫

F x e dttx( ) cos= +∫ 9

2

24. ′ = = ∫∫G sd

dxf t dt t dt

x

a

x( ) ( ) tan3

0

G s t dtx

( ) tan= +∫ 30

4

25. Graph (b).

(sin0)

(sin1)

(sin(

2

2=

−

0

1 02>0

>))

26. Graph (c).

(sin 0)

(sin1)

(sin

3

3=

−

0

0

1 03>

<( ))

27. Graph (a).

(cos0)

(cos1)

(cos(

2

2>>

>

0

0

1 02− )

28. Graph (d).

(cos0)

(cos1)

(cos(

3

3>>

0

0

2 03− <))

29.

x

y

1

2

0−1 1

−1

30.

x

y

1

2

0−1

1

−1

31.

x

y

1

2

0

−1

1

−1

32.

x

y

1

2

0−1

1

−1

266 Section 6.1

33.

x

y

1

2

0

−1

1

−1

34.

x

y

1

2

0−1

1

−1

35.

36.

37.

38.

39.

40.

41.(x, y)

dydx

= x – 1 Δx Δy = dydx

Δx (x + Δx, y + Δy)

(1, 2) 0.0 0.1 0 (1.1, 2)

(1.1, 2) 0.1 0.1 0.01 (1.2, 2.01)

(1.2, 2.01) 0.2 0.1 0.02 (1.3, 2.03)

y = 2.03

42.(x, y)

dydx

= y – 1 Δx Δy = dydx


(1, 3) 2.0 0.1 0.2 (1.1, 3.2)

(1.1, 3.2) 2.2 0.1 0.22 (1.2, 3.42)

(1.2,3.42)

2.42 0.1 0.242 (1.3, 3.662)

y = 3.662

43.(x, y)

dydx

= 2x–y Δx Δy = dydx


(1, 2) 1.0 0.1 0.1 (1.1, 2.1)

(1.1, 2.1) 1.0 0.1 0.1 (1.2, 2.2)

(1.2, 2.2) 1.0 0.1 0.1 (1.3, 2.3)

y = 2.3

44.(x, y)

dydx

= 2x – y Δx Δy = dydx


(1, 0) 2.0 0.1 0.2 (1.1, 0.2)

(1.1, 0.2) 2.0 0.1 0.2 (1.2, 0.4)

(1.2, 0.4) 2.0 0.1 0.2 (1.3, 0.6)

y = 0.6

45.(x, y)

dydx

= 2 – x Δx Δ = Δydydx

x (x + Δx, y + Δy)

(2, 1) 0.0 –0.1 0.0 (1.9, 1)

(1.9, 1) 0.1 –0.1 –0.01 (1.8, 0.99)

(1.8, 0.99) 0.2 –0.1 –0.02 (1.7, 0.97)

46.(x, y)

dydx

= 1 + y Δx Δy = dydx


(2, 0) 1.0 –0.1 –0.1 (1.9, –0.1)

(1.9, –0.1) 0.9 –0.1 –0.09 (1.8, –0.19)

(1.8, –0.19) 0.81 –0.1 –0.081 (1.7, –0.271)

y = –0.271

47.(x, y)

dydx

= x – y Δx Δy = dydx


(2, 2) –0.0 –0.1 0 (1.9, 2.0)

(1.9, 2) –0.1 –0.1 0.01 (1.8, 2.01)

(1.8, 2.01) –0.21 –0.1 0.021 (1.7, 2.031)

y = 2.031

48.(x, y)

dydx

= x – 2y Δx Δy = dydx


(2, 1) 0.0 –0.1 0.0 (1.9, 1.0)

(1.9, 1) –0.1 –0.1 0.01 (1.8, 1.01)

(1.8, 1.01) –0.22 –0.1 0.022 (1.7, 1.032)

y = 1.032

Section 6.1 267

49. (a) Graph (b)

(b) The slope is always positive, so (a) and (c) can be ruledout.

x

�2

–1 10

y

(a)

x

p2

–1 10

(b)

y

(c)

x

p2

–1 10

y

50. (a) Graph (b)

(b) The solution should have positive slope when x isnegative, zero slope when x is zero and negative slope

when x is positive since slope = dy

dx = –x.

Graphs (a) and (c) don’t show this slope pattern.

x

y

0

(–1, 1)

(a)

x

y

0

(–1, 1)

(b)

x

y

0

(–1, 1)

(c)

51. There are positive slopes in the second quadrent of theslope field. The graph of y x= 2 has negative slopes in thesecond quadrent.

52. The slope of y = sin x would be +1 at the origin, while theslope field shows a slope of zero at every point on they-axis.

53.(x, y)

dydx

x= +2 1 Δx Δy = dydx


(1, 3) 3.0 0.1 0.3 (1.1, 3.3)

(1.1, 3.3) 3.2 0.1 0.32 (1.2, 3.62)

(1.2, 3.62) 3.4 0.1 0.34 (1.3, 3.96)

(1.3, 3.96) 3.6 0.1 0.36 (1.4, 4.32)

y = 4.32

Euler’s Method gives an estimate f (1.4) ≈ 4.32.The solution to the initial value problem is

f (x) = x2 + x + 1, from which we get f (1.4) = 4.36. The

percentage error is thus 4 36 4 32

4 360 9

. .

.. %.

− =

54.(x, y)

dydx

= 2x – 1 Δx Δy = dydx


(2, 3) 3.0 –0.1 –0.3 (1.9, 2.7)

(1.9, 2.7) 2.8 –0.1 –0.28 (1.8, 2.42)

(1.8, 2.42) 2.6 –0.1 –0.26 (1.7, 2.16)

1.7, 2.16) 2.4 –0.1 –0.24 (1.6, 1.92)

y = 1.92Euler’s Method gives an estimate f (1.6) ≈ 1.92. Thesolution to the initial value problem is f (x) = x2 – x + 1,from which we get f (1.6) = 1.96. The percentage error is

thus 1 96 1 92

1 962

. .

.%.

− =

55. At every (x, y), ( − − = − = −− −e e ex y y x( )/ ( )/)( ) ,2 2 0 1 so the

slopes are negative reciprocals. The slope lines aretherefore perpendicular.

56. Since the slopes must be negative reciprocals, g(x) = –cos x.

57. The perpendicular slope field would be produced bydy

dx = –sin x, so y = cos x + C for any constant C.

58. The perpendicular slope field would be produced bydy

dx = –x, so y = 0.5 x2 + C for any constant C.

59. True. They are all lines of the form y = 5x + C.

60. False. For example, f (x) = x2 is a solution of dy

dx = 2x,

but f x x− =1( ) is not a solution of dy

dx = 2y.

61. C. m = 42 – 42 = 0

62. E. y < 0, x2 > 0, therefore dy

dx < 0.

63. B. y e( )0 102

= =

dy

dxxe xyx= =2 2

2

.

64. A.

268 Section 6.1

65. (a) dy

dxx

x= − 1

2

dy

dxdx x x dx

yx

x Cx

xC

= −

= + + = + +

−

−

∫∫ ( )2

21

2

2 2

1

Initial condition: y(1) = 2

21

2

1

1

23

21

2

2

= + +

= +

=

C

C

C

Solution: yx

xx= + + >

2

2

1 1

20,

(b) Again, yx

xC= + +

2

12

.

Initial condition: y(–1) = 1

11

2

1

1

11

23

2

2

= − +−

+

= +

=

( )

( )C

C

C

−

Solution: yx

xx= + + <

2

2

1 3

20,

(c) For xdy

dx

d

dx x

xC< = + +

⎛

⎝⎜

⎞

⎠⎟0

1

2

2

1,

= − +

= −

1

1

2

2

xx

xx

.

For xdy

dx

d

dx x

xC> = + +

⎛

⎝⎜

⎞

⎠⎟0

1

2

2

2,

= − +

= −

1

1

2

2

xx

xx

.

And for x = 0,dy

dxis undefined.

(d) Let C1 be the value from part (b), and let C2 be the value

from part (a). Thus, C1 = 3

2and C2 = 1

2.

(e) y y( ) ( )2 1 2 2= − − =

− = + + =−

+ − +

− = + = +

11

2

2

22

1

2

2

2

15

22

3

2

2

2

2

1

2

C C

C C

( )

( )

11

2 17

2

1

2− = =C C

Thus, C1 = 1

2 and C2 = − 7

2.

66. (a) d

dxx C

xx(ln )+ = >1

0for

(b) d

dxx C

x

d

dxx

xln ( ) ( ) ( )− +⎡⎣ ⎤⎦ =

−− =

−⎛⎝⎜

⎞⎠⎟

− =1 11

1

xx

for x < 0

(c) For x x C x C> + = +0, ln ln , which is a solution to the

differential equation, as we showed in part (a). For

x x C x C< + = − +0, ln ln ) ,( which is a solution to the

differential equation, as we showed in part (b). Thus,d

dxx

xln = 1

for all x except 0.

(d) For x < 0, we have y = ln (–x) + C1, which is a solution

to the diferential equation, as we showed in part (a). For

x > 0, we have y = ln x + C2 , which is a solution to the

differential equation, as we showed part (b). Thus,dy

dx x= 1

for all x except 0.

67. (a) y x dx′ = +∫12 4

y x x C

y x x C dx

y x x C x C

′ = + += + +

= + + +∫6 4

6 4

2 2

21

21

3 21 2

(b) y e x dxx′ = +∫ sin

y e x C

y e x C dx

y e x C x C

x

x

x

′ = − += − +

= − + +∫

cos

cos

sin

1

1

1 22

(c) y x x dx′ = + −∫ 3 3

yx

xC

yx

xC dx

yx

xC

′ = − +

= − +

= + +

∫

4

2 1

4

2 1

5

1

4

1

2

4

1

2

20

1

2xx C+ 2

Section 6.2 269

68. (a) y x dx

y x x C

CC

′ = −

′ = − += − +=

∫ 24 10

8 10

3 8 1 10 15

2

3

3( ) ( )

yy x x dx

y x x x C

= − +

= − + += − +

∫8 10 5

2 5 5

5 2 1 5 1

3

4 2

4 2( ) ( ) 55 13

2 5 5 34 2

( ) +== − + +

CC

y x x x

(b) y x x dx

y x x CC

C

′ = −

′ = + += + +

∫ cos sin

sin cossin cos2 0 0

=== + += − + + += − +

∫1

1

0 0

y x x dx

y x x x C

sin cos

cos sincos ssin

cos sin

0 01

1

+ +== − + + +

CCy x x x

(c) y e x dx

y ex

C

e C

C

y ex

x

x

x

′ = −

′ = − +

= − +

= −

= −

∫2

02

2

2

00

21

2−−

= − − +

= − − +

=

= − −

∫ 1

6

10

60

0

6

3

03

3

dx

y ex

x C

e C

C

y ex

x

x

x

69. (a) y x

y x dxx

C

′ =

= = +∫2

2

(b) y x

y x dxx

C

′ = −

= − = − +∫2

2

(c) y y′ =d

dxCe Ce

y Ce

x x

x

( ) =

=

(d) y y′ = −d

dxCe Ce

y Ce

x x

x

( )− −

−

= −

=

(e) y xy′ =d

dxCe Cxe

y Ce

x x

x

2 2

2

2 2

2

/ /

/

( ) =

=

70. (a) ′′ =y x

′ = = +

= + = + +

∫

∫

y x dxx

C

yx

C dxx

C x C

2

1

2

1

3

1 2

2

2 6

(b) ′′ = −y x

y x dxx

C

yx

C dxx

C x C

′ = − = − +

= − + = − + +

∫

∫

2

1

2

1

3

1 2

2

2 6

(c) ′′ = −y xsin

y x dx x C

y x C dx x C x C

′ = − = +

= + = + +∫∫

sin cos

cos sin1

1 1 2

(d) ′′ =y y

d

dxC e C e C e C e y

d

dxC e C e

x x x x

x x

1 2 1 2

1 2

+( ) = − = ′

−(− −

− )) = + = ′′

= +

−

−

C e C e y

y C e C e

x x

x x

1 2

1 2

(e) ′′ = −y y

d

dxC x C x C x C x y

d

dxC

( sin cos ) cos sin

cos

1 2 1 2

1

+ = − = ′

xx C x C x C x

y C x C x

−( ) = − −

= +2 1 2

1 2

sin sin cos

sin cos

Section 6.2 Antidifferentiation bySubstitution (pp. 331–340)

Exploration 1 Are f u du( )∫ and f u dx( )∫ the

Same Thing?

1. f u du u du

uC

( )∫ ∫=

= +

3

4

4

2. u x x4 2 4 6

4 4 4= =( )

3. f u u x x

x dxx

( ) ( )= = =

=∫

3 2 3 6

67

7

4. No

270 Section 6.2

Quick Review 6.2

1. x dx x4

0

2 5

0

2 5 51

5

1

52

1

50

32

5∫ = ⎤⎦ = − =( ) ( )

2. x dx x dx x− = − = − ⎤⎦∫ ∫1 1

2

31

1

5 1 2

1

5 3 2

1

5( ) ( )/ /

= −2

34

2

303 2 3 2( ) ( )/ /

= =2

38

16

3( )

3. dy

dxx= 3

4. dy

dxx= 3

5. dy

dxx x x x= − + −4 2 3 3 43 2 3 2( ) ( )

6. dy

dxx x= − −2 5 4sin (4 5) cos (4 ) i

= − −8 5 5sin (4 cos (4x x) )

7. dy

dx xx x= − = −1

cossin tani

8. dy

dx xx x= =1

sincos coti

9. dy

dx x xx x x=

++1 2

sec tan(sec tan sec )i

= ++

sec tan sec

sec tan

x x x

x x

2

= ++

sec (tan sec )

sec tan

x x x

x x = sec x

10.dy

dx x xx x x=

+− −1 2

csc cot( csc cot csc )

= − ++

csc cot csc

csc cot

x x x

x x

2

= − ++

csc (cot csc )

csc cot

x x x

x x = − csc x


1. (cos )x x dx x x C− −3 2 3= +∫ sin

2. x dx x C− −= − +∫ 2 1

3. tt

dtt

t C22

311

3−⎛

⎝⎜⎞⎠⎟

= + +−∫

4. dt

tt

21

1tan +

+= −∫ C

5. ( sec ) tan3 23

54 3 2 5 2x x x dx x x x− + = + + +− −∫ C

6. ( sec tan sec /2 22

33 2e x x x dx e x xx x+ − = + − +∫ ) C

7. ( cot ) ( csc ) csc− + = − − =u u uC 1 2 2

8. ( csc ) ( csc cot ) csc cot− + = − − =u C u u u u1

9. 1

2

1

222

12 2e C e ex x x+⎛

⎝⎜⎞⎠⎟

= =( )

10.1

55

1

55 5 5

1

ln ln(ln )x x xC+

⎛⎝⎜

⎞⎠⎟

= =

11. (tan )− + =+

1 12

1

1u C

u

12. ( )sin1

1

1 1

2

−

−u C

u+ =

13. f u du u du u C x C( ) / /= = + = +∫∫2

3

2

33 2 3 2

f u dx u dx x dx x dx x C( ) = = = = +∫∫ ∫ ∫2 21

2

14. f u du u du u C x C( ) = = + = +∫∫ 2 3 151

3

1

3

f u dx u dx x dx x C( )∫ ∫ ∫= = = +2 10 111

11

15. f u du e du e C e Cu u x( )∫ ∫= = + = +7

f u du e dx e dx e Cu x x( ) = = = +∫ ∫∫ 7 71

7

16. f u du u du u C x C( ) sin cos cos= = − + = − +∫∫ 4

f u dx u dx x dx x C( ) sin sin cos∫ ∫∫= = = − +41

44

17. u x= 3du dx= 3

1

3du dx=

sin sin31

3x dx u du= ∫∫

= − +1

3cos u C

= − +1

33cos x C

Check: d

dxx C x x− +⎛

⎝⎜⎞⎠⎟

= − =1

33

1

33 3 3cos ( sin )( ) sin

Section 6.2 271

18. u x= 2 2

du x dx= 4

x dx du= 1

4

x x dx u ducos( ) cos21

42 = ∫∫

= +1

4sinu C

= +1

42 2sin( )x C

Check: d

dxx C x x x

1

42

1

42 4 22 2sin( ) cos( )( ) cos(+⎛

⎝⎜⎞⎠⎟

= = xx2)

19. u x= 2du dx= 2

1

2du dx=

sec tan sec tan2 21

2x x dx u u du= ∫∫

= +1

2secu C

= +1

22sec x C

Check:d

dxx C x x x

1

22

1

22 2 2 2 2sec sec tan sec tan+⎛

⎝⎜⎞⎠⎟

= =i xx

20. u x= −7 2du dx= 7

1

7du dx=

28 7 21

728 7 23 3 4 4( ) ( )x dx u du u C x C− = = + = − +∫∫

Check: d

dxx C x x( ) ( ) ( ) ( )7 2 4 7 2 7 28 7 24 3 3− +⎡

⎣⎤⎦ = − = −

21. ux=3

du dx= 1

33 du dx=

dx

x

du

u2 29

3

9 9+=

+∫∫

=+∫

3

9 12

du

u

=+∫

1

3 12

du

u

= +−1

31tan u C

= ⎛⎝⎜

⎞⎠⎟

+−1

3 31tan

xC

Check: d

dx

xC

x x

1

3 3

1

3

1

13

1

3

1

91

2tan− +⎛

⎝⎜⎞⎠⎟

=

+ ⎛⎝⎜

⎞⎠⎟

=+

i22

22. u r= −1 3

du r dr= −3 2

− =1

32du r dr

9

19

1

3

2

3

r dr

r

du

u−= −⎛

⎝⎜⎞⎠⎟ ∫∫

= − −∫3 1 2u du/

= − +3 2 1 2( ) /u C

= − − +6 1 3r C

Check:

d

dxr C

rr

r

r

− − +( ) = −−

⎛

⎝⎜

⎞

⎠⎟ −

=−

6 1 61

2 13

9

1

3

3

2

2

3

( )

23. ut= −12

cos

dut

dt= 1

2 2sin

22

dut

dt= sin

12 2

22

2−⎛⎝⎜

⎞⎠⎟

= ∫∫ cos sint t

dt u du

= +

= −⎛⎝⎜

⎞⎠⎟

+

2

32

31

2

3

3

u C

tCcos

Check: d

dx

tC

2

31

2

3

−⎛⎝⎜

⎞⎠⎟

+⎡

⎣⎢⎢

⎤

⎦⎥⎥

cos

= −⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

2 12 2

1

2

2

cos sint t

= −⎛⎝⎜

⎞⎠⎟

12 2

2

cos sint t

24. u y y= + +4 24 1

du y y dy= +( )4 83

du y y dy= +4 23( )

1

423du y y dy= +( )

272 Section 6.2

24. Continued

8 4 1 2 81

44 2 2 3 2( ) ( )y y y y dy u du+ + + = ⎛

⎝⎜⎞⎠⎟ ∫∫

= +

= + + +

2

32

34 1

3

4 2 3

u C

y y C( )

Check: d

dxy y C

2

34 14 2 3( )+ + +⎡

⎣⎢⎤⎦⎥

= + + +2 4 1 4 84 2 2 3( ) ( )y y y y

= + + +8 4 1 24 2 2 3( ) ( )y y y y

25. Let u x= −1du dx= −

dx

x

du

uu C

xC

( )1

1

1

2 2

1−

= −

= +

=−

+

∫ ∫−

26. Let u x= + 2du dx=

sec ( ) sec

tantan( )

2 22

2

x dx u du

u Cx C

+ == += + +

∫ ∫

27. Let u x= tan

du x dx= sec2

tan sec

(tan )

/

/

/

x x dx u du

u C

x

2 1 2

3 2

3 2

2

32

3

∫ ∫=

= +

= + CC

28. Let u = +θ π2

du d= θ

sec tan sec tanθ π θ π θ+⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

=∫ ∫2 2d u u du

== +

= +⎛⎝⎜

⎞⎠⎟

+

sec

sec

u C

Cθ π2

29. tan( )

tan

4 2

4 24

1

41

4

x dx

u xdu dx

du dx

u du

+= +

=

=

∫

∫== − + +

+ +

1

44 2

1

44 2

ln cos( )

ln sec( )

x C

x C

or

30. 32

(sin )x dx∫−

u x= sindu x dx= cos

du

xdx

cos=

3 2∫ −u dx

= − +3cot x C

31. Let u z= +3 4du dz= 3

1

3du dz=

cos( ) cos3 41

3z dz u du+ = ∫∫

= +1

3sinu C

= + +1

33 4sin( )z C

32. Let u x= cot

du x dx= − csc2

cot csc /x x dx u du∫ ∫= −2 1 2

= − +2

33 2u C/

= − +2

33 2(cot ) /x C

33. Let u x= ln

dux

dx= 1

ln66x

xdx u du∫ ∫=

= +1

77u C

= +1

77(ln )x C

34. Let ux= tan2

dux

dx= 1

2 22sec

tan sec

tan

7 2 7

8

8

2 22

21

81

4 2

∫ ∫=

= +

= +

x xdx u du

u C

xC

i

35. Let u s= −4 3 8/

du s ds= 4

31 3/

3

41 3du s ds= /

Section 6.2 273

35. Continued

s s ds u du

u C

1 3 4 3 83

43

43

4

/ /cos( ) cos

sin

s

∫ ∫− =

= +

= iin( )/s C4 3 8− +

36.dx

xx dx

sincsc

22

33= ∫∫

Let u x= 3du dx= 3

1

3du dx=

csc csc2 231

3∫ ∫=x dx u du

= − +1

3cotu C

= − +1

33cot( )x C

37. Let u t= +cos( )2 1

du t dt= − +sin( )( )2 1 2

− = +1

22 1du t dtsin( )

sin( )

cos ( )

2 1

2 1

1

21

21

2

22

1

t

tdt u du

u C

++

= −

= +

=

∫ ∫ −

−

ccos( )

sec( )

2 11

22 1

tC

t C

++

= + +

38. Let u t= +2 sindu t dt= cos

6

26

66

2

22

1

cos

( sin )

sin

t

tdt u du

u C

t

+=

= − +

= −+

+

∫ ∫ −

−

CC

39.dx

x xln∫u x= ln

dudx

x=

x du dx=du

uu x C= = +∫ ln ln(ln )

40. tan sec2 2∫ x x dx

u x

du x dxdu

xdx

==

=

tan

sec

sec

2

2

u du u C

x C

2 3

3

1

31

3

∫ = +

+tan

41.x dx

x2 1+∫u xdu x dxdu

xdx

= +=

=

2 12

21

2 1

1

21

21

2

2

du

xu C

x C

+= +

= + +

∫ ln

ln( )

42.40

252

dx

x +∫u xadu dx

==

=5

401

40

5 58

5

2 21

1 1

du

u a a

u

aC

xC

x+

= +

+ =

−

− −

∫ tan

tan tan ++ C

43. dx

x

x

xdx

cot

sin

cos3

3

3∫ ∫=

Letu x= cos3

du x dx= −3 sin3

− =

= −

= − +

= −

∫∫

1

33

3

1

3

1

1

31

3

du x dx

dx

x udu

u C

sin

cot

ln

lln cos3x C+

( ln sec .)Anequivalent expression is1

33x C+

44. Letu x= +5 8

du dx

du dx

dx

xu du

=

=

+= −∫∫

51

5

5 8

1

51 2/

= +

= + +

1

52

2

55 8

1 2i u C

x C

/

274 Section 6.2

45. sec secsec tan

sec tanx dx x

x x

x xdx= +

+⎛⎝⎜

⎞⎠⎟∫∫ i

=

sec sec tan

sec tansec tan

2 x x x

x xdx

u x x

du

++

= +=

∫Let

ssec tan sec

sec ln ln sec

x x x dx

x dxu

du u C x

+

= = + =∫∫

2

1 ++ +tan x C

46. csc csccsc cot

csc cotx dx x

x x

x xdx= +

+⎛⎝⎜

⎞⎠⎟∫∫

= ++∫

csc csc cot

csc cot

2 x x x

x xdx

Letu x x

du x x x dx

x dx

= += − −

= −

csc cot

csc cot csc

csc

2

1

uudu

u C

x x C

∫∫= − += − + +

ln

ln csc cot

47. sin (sin ) sin3 22 2 2x dx x x dx= ∫∫ = −∫ ( cos ) sin1 2 22 x x dx

u xdu x dx

== −cos

sin2

2

= −

= − +

= − +

∫ ( )

coscos

1

3

22

3

2

3

3

u du

uu

C

xx

C

48. sec (sec )sec4 2 2x dx x x dx= ∫∫

= +=

== +

∫ ( tan )sec

tan

sec

( )

1

1

2 2

2

2

x x dx

u x

du x dx

u du∫∫= + +

= + +

uu

C

xx

C

3

33

3tan

tan

49. 2 1 2 12 2sin ( sin )∫ ∫= + −x dx x dx

= +=

=

= +

= +

∫

∫

( cos )

( cos )

(

1 2

22

1

21

1

2

x dx

u xdu dx

u du

u ssin )

sin

u C

xx

C

+

= + +2

2

50. 4 2 1 2 22 2cos ( ( cos ) )x dx x dx= − − +∫∫

= − +=

=

= − +

=

∫

∫

( cos )

( cos )

2 2 2

22

1

22 2

1

x dx

u xdu dx

u du

222 2

2 2

( sin )

sin

− + +

= − +

u u C

x x C

51. tan tan (sec )4 3 2 1x x x dx= −∫∫= −

==

= −

∫ (tan sec tan )

tan

sec

(

3 2 2

2

2

x x x dx

u x

du x dx

u uu du

uu C

xx x C

)

tantan

∫= − +

= − + +

3

33

3

52. (cos sin )4 4x x dx−∫= + −

=∫ (cos sin )(cos sin )

( (cos ))

2 2 2 2

1 2

x x x x dx

x dx∫∫= +1

22sin x C

53. Let u ydu dy

= +=

1

y dy u du

u

+ =

= ⎤⎦

= −

∫ ∫1

2

32

34

2

31

0

3 1 2

1

4

3 2

1

4

3 2

/

/

/( ) ( ))

( )

/3 2

2

88

2

3

14

3= − =

54. Letu r= −1 2

du r dr

du r dr

r r dr u du

=

− =

− = −∫ ∫

21

2

11

22

0

1 1 2

1

0 /

= − ⎤⎦

= − + =

1

2

2

31

30

1

31

1

3

3 2

1

0i u /

( ) ( )

Section 6.2 275

55. Letu x= tan

du x dx

x x dx u du

=

=−− ∫∫

sec

tan sec

2

2

1

00

π /4

= ⎤⎦

= − −

= −

−

1

21

20

1

21

1

2

2

1

0

2

u

( ) ( )

56. Letu r= +4 2

du r dr

du r dr

r

rdr u du

=

=

+= =

−−∫ ∫

21

25

4

5

20

2 21

1 2

5

5

( )

57. Letu = +1 3 2� /

du d

du d

d

=

=

+=∫

3

22

310

1

2

3

1 2

1 2

3 2 20

1

� �

� �

�

��

/

/

/( )(110 2

1

2) u du−∫

= − ⎤⎦

= − −⎛⎝⎜

⎞⎠⎟

= − −⎛⎝⎜

⎞⎠⎟

=

−20

320

3

1

21

20

3

1

2

1

1

1

2u

00

3

58. Letu x= +4 3sin

du x dx

du x dx

x

xdx u

=

=

+=

−−∫

31

3

4 3

1

3

cos

cos

cos

sinπ

π 11 2

4

40/ du =∫

59. Letu t t= +5 2

du t dt

t t t dt u du

= +

+ + = ∫∫( )

( ) /

5 2

2 5 2

4

5 4 1 2

0

3

0

1

= ⎤⎦⎥

=

= =

2

32

33

2

327 2 3

3 2

0

3

3 2

u /

/( )

60. Let u = cos2�

du d

du d

d

= −

− =

= −−

2 21

22

2 21

23

sin

sin

cos sin

θ θ

θ θ

θ θ θ00

6 3

1

1 2π / /

∫ ∫ −u du

= − −⎛⎝⎜

⎞⎠⎟

⎤

⎦⎥

= ⎛⎝⎜

⎞⎠⎟

−⎛

⎝⎜⎜

−

−

1

2

1

2

1

4

1

21

2

1

1 2

2

i u/

⎞⎞

⎠⎟⎟

= =1

43

3

4( )

61. dx

x +∫ 20

7

u xdu dx

du

uu x

= +=

= = + = ⎛⎝⎜

⎞⎠⎟∫

2

29

20

7

0

7

0

7ln ln( ) ln == 1 504.

62. dx

x2 32

5

−∫u xdu dx

du

uu x

= −=

= = − =∫

2 32

1

2

1

2

1

22 3

12

5

2

5

2

5ln ln ( )

227 0 973ln( ) .=

63. dt

t −∫ 31

2

u tdu dt

du

uu t

= −=

= = − = ⎛⎝⎜

⎞⎠⎟∫

3

31

21

2

1

2

1

2ln ln( ) ln == −0 693.

64. cotcos

sinx dx

x dx

x= ∫∫ π

ππ

π

4

/4

4

/4 33

u x

du x dx==

sincos

du

uu xπ

π

π

π

π

π

4

3 4

4

3 4

4

3 40

/

/

/

/

/ln ln (sin )∫ = = =

65. x dx

x21

3

1+−∫u xdu x dx

= +=

2 12

1

2

1

2

1

21

1

25 0

1

3

1

3 2

1

3du

uu x

− − −∫ = = + = =ln ln( ) ln( ) .8805

66. e dx

e

x

x30

2

+∫u e

du e dx

x

x

= +=3

du

uu ex

0

2

0

2

0

23 0 954∫ = = + =ln ln( ) .

276 Section 6.2

67. Let u x du x dx= + =4 39 4, .

(a) x dx

xu du u

3

40

1

9

10 1 2 1 2

9

10

9

1

4

1

2+= = ⎤

⎦⎥∫ ∫ − / /

= −

= − ≈

1

210

1

29

1

210

3

20 081.

(b) x

xdx u du

3

41 2

9

1

4+=∫ ∫ − /

= +

= + +

1

21

29

1 2

4

u C

x C

/

x

xdx x

3

40

1 4

0

1

9

1

29

+= + ⎤

⎦⎥∫

= −

= − ≈

1

210

1

29

1

210

3

20 081.

68. Let 3u x du x dx= − =1 3 3cos , sin .

(a) ( cos )sin1 3 31

3

1

61

2 2

1

2

− = = ⎤⎦⎥

=

∫∫ x x dx udu uπ

π

/6

/3

11

62

1

61

1

22 2( ) ( )− =

(b) ( cos )sin11

3− = ∫∫ 3 3x x dx u du

= +

= − +

1

61

61 3

2

2

u C

x C( cos )

( cos )sin ( cos )1 3 31

61 3 2− = − ⎤

⎦⎥∫ x x dx xπ

π

π

π

/6

/3

/6

//3

= 1

62 2( ) −− =1

61

1

22( )

69. We show that and where′ = =f x x f( ) tan ( ) ,3 5

f x( ) lncos

cos.= +

3

35

′ = +

= −

⎛⎝⎜

⎞⎠⎟

f xd

dx xd

dx

( ) lncos

cos

(ln cos ln co

35

3 ss )

ln cos

cos( sin ) tan

x

d

dxx

xx x

+

= −

= − − =

5

1

f ( )cos

cos(ln )3 5 1 5 5= + = + =

3

3

70. yx= +ln

sin

sin26

u x vdu x dx

dyd

dx

u

v

dyd

= ==

= +⎛⎝⎜

⎞⎠⎟

=

sin sincos

ln

2

6

ddxu v

dydu

u

x

xx

f

(ln ln )

cos

sincot

( ) cot(

− +

= = =

=

6

2 22 6) =

71. False. The interval of integration should change from[ , / ]0 4π to [0,1], resulting in a different numerical answer.

72. True. Using the substitution u f x du f x dx= = ′( ), ( ) ,

we have ′ = =∫ ∫

f x dx

f x

du

uu

a

b

f a

f b

f a

f b( )

( ) ( )

( )

( )

( )

ln

= − =⎛⎝⎜

⎞⎠⎟

ln ln ln( ( )) ( ( ))( )

( ).f b f a

f b

f a

73. D.

74. E. e dxe ex

x2

0

2 2

0

2 4

2

1∫ = = −

2

75. B. F x a dx F a F a( ) ( ) ( )− = − − − =∫ 5 3 73

5

F x dx F a F aa

a( ) ( ) ( )= − − − =

−

−∫ 5 3 7

3

5

76. A. d

dxx xsin cos=

cos

cos( )

cos

−⎛⎝⎜

⎞⎠⎟

=

=⎛⎝⎜

⎞⎠⎟

=

π

π

20

0 1

20

77. (a) Letu x= +1

du dx=

x dx u du+ =∫ ∫1 1 2/

= +

= + +

2

32

31

3 2

3 2

u C

x C

/

/( )

Alternatively, ( ) ./d

dxx C x

2

31 13 2+ +⎛

⎝⎜⎞⎠⎟

= +

(b) By Part 1 of the Fundamental Theorem of Calculus,dy

dxx

dy

dxx1 21 1= + = +and , so both are

antiderivatives of x +1.

Section 6.2 277

77. Continued(c) Using NINT to find the values of y1 and y2 , we have:

x 0 1 2 3 4

y1 0 1.219 2.797 4.667 6.787

y2 –4.667 –3.448 –1.869 0 2.120

y y1 2− 4.667 4.667 4.667 4.667 4.667

C = 42

3(d) C y y= −1 2

= + − +

= + + +

= +

∫ ∫∫ ∫∫

x dx x dx

x dx x dx

x

x x

x

x

1 1

1 1

1

0 3

0

3

0

3ddx

78. (a) d

dxF x C f x[ ( ) ] ( ).+ should equal

(b) The slope field should help you visualize the solutioncurve y F x= ( ).

(c) The graphs of y F x y f t dtx

1 2 0= = ∫( ) ( )and should differ

only by a vertical shift C.

(d) A table of values for should show tha1 2y y− tty y C x1 2 for any value of− = in the appropriate

domain.

(e) The graph of f should be the same as the graph of NDERof F(x).

(f) First, we need to find ( ). LetF x u x du= + =2 1, 22x dx.

x

xdx u du

2

1 2

1

1

2+=∫ ∫ − /

=

= + +

u

x C

1 2

2 1

/

Therefore, we may let F x x( ) .= +2 1

a)d

dxx C

xx( ) ( )2

21

1

2 12+ + =

+

=+

=x

xf x

2 1( )

b)

c)

d)

x 0 1 2 3 4

y1 1.000 1.414 2.236 3.162 4.123

y2 0.000 0.414 1.236 2.162 3.123

y1 – y2 1 1 1 1 1

e)

79. (a) 2 2 2 2sin cos sinx x dx udu u C x C= = + = +∫∫(b) 2 2 2 2sin cos cosx x dx udu u C x C= − = − + = − +∫∫(c) Since sin ( cos ) ,2 2 1x x− − = the two answers differ by a

constant (accounted for in the constant of integration).

80. (a) 2 22 2 2sec tan tanx x dx udu u C x C= = + = +∫∫(b) 2 22 2 2sec tan secx x dx udu u C x C= = + = +∫∫(c) Since sec tan ,2 2 1x x− = the two answers differ by a

constant (accounted for in the constant of integration).

81. (a) dx

x

u du

u

u du

udu

1 11

2 2 2−=

−= = ∫∫∫

cos

sin

cos

cos.

( cos , cos cos cos .)Note sou u u u> = =0 2

(b) dx

xdu u C x C

11

2

1

−= = + = +∫ ∫ −sin

82. (a) dx

x

udu

u

udu

udu

1 11

2

2

2

2

2+=

+= =∫ ∫ ∫∫

sec

tan

sec

sec

(b) dx

xdu u C x C

112

1

+= = + = +−∫∫ tan

83. (a) xdx

x

y y y dy

y1

2

1 20

1

1

1

−=

−−

− sin sin cos

sinsin

sin / i22

0

1 2

∫∫/

= =∫ ∫2

22

0

4 2

0

4sin cos

cossin

/ /y y dy

yy dy

π π

(b) xdx

xy dy

12 2

0

4

0

1 2

−= ∫∫ sin

// π

= − = −∫ ( cos ) [ ( / )sin ] //1 2 1 2 2 0

4

0

4y dy y y ππ

= −( ) /π 2 4

278 Section 6.3

84. (a) dx

x

u du

u1 12

2

20

3

0

3

1

1

+=

+−

−

∫∫sec

tantan

tan

= =∫ ∫sec

secsec

/ /2

0

3

0

3u du

uu du

π π

(b) dx

xu du u u

1 2 0

3

0

3

0

3

+= = +⎡⎣ ⎤⎦∫∫ sec sec tan

/ /ln

π π

= +( ) − + = +( )ln ln ln2 3 1 0 2 3( )

Section 6.3 Antidifferentiation by Parts(pp. 341–349)

Exploration 1 Choosing the Rightu and dv1. u du

dv x x v x x dx

= == = ∫

1 0

cos cos

Using 1 for u is never a good idea because it places usback where we started.

2. u x x du x x x

dv dx

= = −=

cos cos sin

v dx= =∫ 1

The selection of u x x= cos will place a more difficult

integral into vdu∫ .

3. u x u x

dv x dx v x dx x

= = −= = =∫

cos sind2

The selection of dv x dx= will place a more difficult

integral into vdu∫ .

4. u x= and dv x dx= cos are good choices because the

integral is simplified.

Quick Review 6.3

1. dy

dxx x x x= +( )(cos )( ) (sin )( )3 22 2 2 3

= +2 2 3 23 2x x x xcos sin

2. dy

dxe

xx ex x=

+⎛⎝⎜

⎞⎠⎟

+ +( ) ln ( )( )2 23

3 13 1 2

=+

+ +3

3 12 3 1

22e

xe x

xx ln ( )

3. dy

dx x=

+1

1 22

2( )i

=+

2

1 4 2x

4. dy

dx x=

− +

1

1 3 2( )

5. y x= −tan 1 3

tan

tan

y x

x y

=

=

31

3

6. y xy xx y

= += += −

−cos ( )cos

cos

1 11

1

7. sin cosππ

πx dx x= − ⎤⎦⎥∫

1

0

1

0

1

= − +

= − − + =

1 10

11

1 2π

ππ

π π π

cos cos

( )

8. dy

dxe x= 2

dy e dxx= 2

Integrate both sides.

dy e dx

y e C

x

x

=

= +

∫∫ 2

21

2

9. dy

dxx x= + sin

dy x x dx= +( sin )

Integrate both sides.

dy x x dx

y x x C

y CC

y

= +

= − +

= − + ==

∫∫ ( sin )

cos

( )

1

20 1 2

3

2

== − +1

232x xcos

10. d

dxe x xx1

2(sin cos )−⎛

⎝⎜⎞⎠⎟

= + + −

= +

1

2

1

21

2

1

e x x x x e

e x

x x

x

(cos sin ) (sin cos )

cos22

1

2

1

2e x e x e x

e x

x x x

x

sin sin cos

sin

+ −

=


1. x x dxsin∫

dv x dx v x dx x

u x du dx

x x

= = = −= =

− − −

∫sin sin cos

cos cos xx dx x x x C= − + +∫ cos sin

2. xe dxx∫dv e dx v e dx e

u x du dx

xe e dx xe e C

x x x

x x x x

= = == =− = − +

∫

∫

Section 6.3 279

3. 3 2t e dtt∫dv e dt v e dt

e

u t du dt

te e

t tt

t t

= = =

= =

−

∫2 22

2 2

23 3

32

32

ddt te e Ct t= − +∫3

2

3

42 2

4. 2 t t dtcos ( )3∫dv t dt v t dt

t

u t du dt

t

= = =

= =∫cos cos ( )

sin

si

3 33

32 2

2nn cos ( )

sin cos ( )3

32

3

3

2

33

2

93

t tdt t t t C− = − +∫

5. x x dx2 cos∫dv x dx v x dx x

u x du x dx

x x x

= = =

= =−

∫cos cos sin

sin

2

2

2

2 ssin

sin sin cos

x dx

dv x dx v x dx x

u x du dx

x

∫∫= = = −

= =2 222

2

2 2

2 2

sin cos cos

sin cos sin

x x x xdx

x x x x x

+ −

= + − +∫

CC

6. x e dxx2∫ −

dv e dx v e dx e

u x du x dx

x e x e

x x x

x

= = = −

= =− − −

− − −

−

∫2

2

2

2 −−

− −

−

∫∫= = = −

= =− −

x

x x x

x

dx

dv e v e dx e

u x du dx

x e x

2 2

22 ee e dx x e xe e Cx x x x x− − − − −− − = − − − +∫ 2 2 22

7. 3 2 2x e dxx∫dv e dx v e dx

e

u x du x

xe

xe

x xx

x

= = =

= =

−

∫2 22

2

22 2

23 6

32

6xx

x x

x xx

dx x e x e dx

dv e v e dxe

u

2

3

23

2

2 2 2

2 22

= −

= = =

∫∫

∫== =

− − = −

3 3

3

2

3

23

2

3

2

3

22 2 2

22 2

x du dx

x e xee

dx x ex xx

x xxe e Cx x2 23

4+ +∫

8. xx

dx2

2∫⎛⎝⎜

⎞⎠⎟

cos

dvx

dx vx

dxx= ⎛

⎝⎜⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠

cos cos sin2 2

22⎟⎟

= =⎛⎝⎜

⎞⎠⎟

− ⎛⎝⎜

⎞⎠⎟

∫

∫u x du x dx

xx

xx

2

2

2

22

42

sin sin ddx

dvx

vx

dxx= ⎛

⎝⎜⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

= − ⎛⎝∫sin sin cos

2 22

2⎜⎜⎞⎠⎟

= =⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞⎠⎟

u x du dx

xx

xx

4 4

22

82

2 sin cos

−− ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

+ ⎛⎝⎜

⎞8

22

28

22cos sin cos

xdx x

xx

x⎠⎠⎟

− ⎛⎝⎜

⎞⎠⎟

+∫ 162

sinx

C

9. y y dy

dv y dy v y dyy

u y duy

dy

y y

ln

ln

ln

∫∫= = =

= =

−

2

2

21

1

2

yy

ydy y y

yC

22

2

2

1 1

2 4= − +∫ ln

10. t t dt

dv t dt v t dtt

u t dut

dt

t

2

2 23

3

31

1

3

∫∫= = =

= =

ln

ln

lnn lntt

tdt t t

tC− = − +∫

33

3

3

1 1

3 9

11. dy x x dx= +( )∫∫ ( )sin2

dv x dx v x dx x

u x du dx

x

= = = −= + =

− +

∫sin sin cos

( )cos

2

2 xx x dx x x x C− − = − + + += − +

∫ cos ( )cos sin

( )cos( )

2

2 0 2 0 ++ += − +== − + + +

sin( )

( )cos sin

02 2

42 4

CC

Cy x x x

[–4, 4] by [0, 10]

280 Section 6.3

12. dy xe dxx= −∫∫ 2

dv e v e dx e

u x du dx

xe e d

x x x

x x

= = = −= =

− − −

− − −

− −

∫

∫2 2

2 2 xx xe e C

e e CC

y

x x= − − +

= − − +== −

− −

− −

2 2

3 2 0 25

0 0( ) ( ) ( )

22 2 5xe ex x− −− +

[–2, 4] by [0, 10]

13. du x x dx=∫ ∫ sec2

dv x dx v x dx x

w x dw dx

x x

= = == =

−

∫ ∫sec sec tan

tan tan

2 2

xx dx x x x C= + += + +

∫ tan ln | cos |

tan ( ) ln | cos ( ) |1 0 0 0 CCCu x x x

== + +

11tan( ) ln | cos( ) |

[–1.2, 1.2] by [0, 3]

14. dz x x dx= ∫∫ 3 ln

dv x v x dxx

u x dux

dx

xx

x

xdx

= = =

= =

− =

∫3 34

4 4

41

4 4

1

ln

lnxx

xx

C

C

C

z

4 4

4 44 16

51

41

1

1681

16

ln

( )ln ( )

( )

− +

= − +

=

∫

== − +xx

x4 4

4 16

81

16ln

[0, 5] by [0, 100]

15. dy x x dx= −∫∫ 1

dv x v x dx x

u x du dx

= − = − = −

= =∫( ) ( ) ( )/ / /1 1

2

311 2 1 2 3 2

22

31

2

31

31

4

15

3 2 3 2

3 2

x x x dx

x x

( ) ( )

( ) (

/ /

/

− − −

= 2 − −

∫xx C

C

C

y

− +

= − − − +

=

1

22

31 1 1

4

151 1

2

5 2

3 2 5 2

)

( ) ( ) ( )

/

/ /

== − − − +2

31

4

151 23 2 5 2x x x( ) ( )/ /

[1, 5] by [0, 20]

16. dy x x dx= +∫∫ 2 2

dv x v x dx x

u x du

= + = + = +

= =∫( ) ( ) ( )/ / /2 2

2

32

2 2

1 2 1 2 3 2

ddx

x x x dx

x x

4

32

4

32

4

32

8

1

3 2 3 2

3 2

( ) ( )

( )

/ /

/

+ − +

= + −

∫

552

04

31 1 2

8

151 2

5 2

3 2 5 2

( )

( ) ( ) ( )

/

/ /

x C

C

+ +

= − − + − + +

CC

y x x x

=

= + − + +

28

154

32

8

152

28

153 2 5 2( ) ( )/ /

[–2, 4] by [–3, 25]

17. e x dxx∫ sin

dv e dx v e dx e

u x du x dx

e x e

x x x

x x

= = == =

−

∫sin cos

sin coos

cos sin

si

xd

dv e dx v e dx e

u x du x dx

e

x x x

x

∫∫= = =

= = −nn sin ( cos sin )

sin

x dx e x e x e x dx

e x dxe

x x x

x

= − − −

=

∫∫xx

x x2

(sin cos )− +∫ C

Section 6.3 281

18. e x dxx−∫ cos

dv x dx v x dx x

u e du e dx

e

x x

x

= = =

= = −∫

− −

−

cos cos sin

sin xx e x dx

dv x dx v x dx x

u e

x

x

− −

= = = −

=

−

−

∫∫

sin

sin sin cos

ddu e dx

e x dx e x e x e

x

x x x x

= −= − − −

−

− − − −∫ cos sin ( cos coos )

cos (sin cos )

x dx

e x dxe

x xxx

∫∫ −

−= − +

2C

19. e x dxx cos 2∫dv x dx v x dx x

u e du e dx

e

x x

x

= = =

= =∫cos cos sin2 2 22

2 ssin sin

sin sin c

2 2 2

2 2 2 2 4

x x e dx

dv x v x dx

x−

= = = −∫

oos

cos

2

2

x

u e du e dx

e x dx

x x

x

∫

∫= =

= − − − −∫2 2 4 2 4 2e x e x e dx x dx

e

x x x

x

sin ( cos cos )

coos ( sin cos )25

2 2 2x dxe

x xx

= + +∫ C

20. e x dxx−∫ sin2

dv x dx v x dx x

u e du e dx x

= = = −

= = −∫

− −

sin sin cos2 2 22

xx

e x e x dx

dv x dx v

x x− −

= =

− −∫2 2 2 2

2 2

cos cos

cos cos xx dx x

u e du e dx

e x dx e

x x

x

=

= = −= −

∫

∫− −

− −

2

2 2

sin

sin xx

x

x

e x

cos

( sin

2

2− −− −−

= − +

−

−−

∫∫

2

25

2 2

e x dx

e x dxe

x

x

xx

sin )

sin ( cos sinn )2x + C

21. Use tabular integration with f x x( ) = 4 and g x e x( ) .= −

x e dx

x e x e x e xe e

x

x x x x

4

4 3 24 12 24 24

−

− − − − −∫= − − − − − xx

x

C

x x x x e C

+= − + + + + +−( )4 3 24 12 24 24

22. Let u x x dv e dxx= − =2 5

du x dx v e

x x e dx

x

x

= − =− =∫

( )

( )

2 5

52 (( ) ( )x x e e x dxx x2 5 2 5− − −∫Let u x dv e dxx= − =2 5

du dx v ex= =2

( ) ( )

( ) ( )

x x e e x dx

x x e x e

x x

x x

2

2

5 2 5

5 2 5 2

− − −

= − − − +∫

ee dx

x x e x e e C

x x e

x

x x x∫

= − − − + += − +

( ) ( )

( )

2

25 2 5 2

7 7 xx C+

23. Use tabular integration with f x x( ) = 3 and g x e x( ) .= −2

x e dx

x e x e xe e

x

x x x

3 2

3 2 2 2 2 21

2

3

4

3

4

3

8

−

− − − −

∫= − − − − xx

x

C

x x xe C

+

= − + + +⎛

⎝⎜

⎞

⎠⎟ +−

3 22

2

3

4

3

4

3

8

24. Use tabular integration with f x x( ) = 3 and g x x( ) .= cos 2

xx

xx

xx x C

3 2

22

3

22

3

42

3

82sin cos sin cos+ − − +

282 Section 6.3

25. Use tabular integration with f x x( ) = 2 and g x x( ) .= sin 2

x x dx x x x x x C2 221

22

1

22

1

42

1

sin cos sin cos∫ = − + + +

= − 22

42

22

2

2

2

0

2

xx

xx C

x x dx

⎛

⎝⎜

⎞

⎠⎟ + +

∫

cos

/

sin

sinπ

== −⎛

⎝⎜

⎞

⎠⎟ +

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

1 2

42

22

1

2

0

2x

xx

xcos

/

sin

π

−− ⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟

− + − ⎛⎝⎜

⎞⎠⎟

22

41 0

1

41

2π

( ) ( ) −−

= − ≈

0

8

1

20 734

2π.

Check: NINT x x x2 2 02

0 734sin , , , .π⎛

⎝⎜⎞⎠⎟

≈

26. Use tabular integration with f x x( ) = 3 and g x x( ) .= cos 2

x x dx x x x x x

x

3 3 221

22

3

42

3

4

23

8

cos sin cos

sin

= + −

−

∫ccos

sin cos

2

2

3

42

3

4

3

82

3 2

x

x xx

x= −⎛

⎝⎜

⎞

⎠⎟ + −

⎛

⎝⎜

⎞

⎠⎟ xx C+

x x dxx x

xx3

0

2 3 2

22

3

42

3

4

3

8cos sin

/π∫ = −

⎛

⎝⎜

⎞

⎠⎟ + −

⎛

⎝⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= + −⎛

⎝⎜

⎞

⎠⎟ −

cos

(

/

2

03

16

3

81

0

2

2

x

π

π)) ( )

.

− − −⎛⎝⎜

⎞⎠⎟

= − ≈ −

03

81

3

4

3

161 101

2π

Check: NINT x x x3 2 02

1 101cos , , , .π⎛

⎝⎜⎞⎠⎟

≈ −

27. Let u e dv x dxx= =2 3cos

3

cos 3

du e dx v x

e x dx e

x

x

= =

=∫

21

32

2 2

sin

( xx x x dx) )1

3

1

3sin 3 sin 3 (2 2⎛

⎝⎜⎞⎠⎟

− ⎛⎝⎜

⎞⎠⎟∫ e x

3 3= −1

3

2

32 2e x e xx xsin sin ddx∫

Let u e dv x dxx= =2 3sin

du e dx v x

e x dx e

x

x

= = −

=∫

21

33

31

3

2

2 2cos

cos

xx

x

x

e x

sin

( )

3

2

3

1

332 cos− −⎛

⎝⎜⎞⎠⎟

⎡

⎣⎢ − −−⎛

⎝⎜⎞⎠⎟

⎤

⎦⎥∫

1

33cos (2 2x dxe x )

== + − ∫1

93 3 2 3

4

93

13

9

2 2

2

e x x e x dx

e

x x( sin cos ) cos

xx x

x

x dx e x x

e

cos

c

31

93 3 2 32

2

∫ = +( sin cos )

oos

cos

31

133 3 2 3

3

2

2

x dx e x x

e x

x

x

∫ = +( sin cos )

ddx e x xx

−−

∫ = +⎡⎣⎢

⎤⎦⎥

=

2

3 2

2

31

133 3 2 3

1

1

( sin cos )

333 9 2 96[ ( sin cos )e +

− − + −

= +

−e

e

4

6

3 6 2 61

132 9

( sin( ) cos ( )]

[ ( cos 33 9

24

sin )

( co− −e ss sin )].

6 3 618 186

−≈ −

Check: NINT e x xx2 2 3 18 186cos , , , .3 −( ) ≈ −

28. Let u e dv x dxx= =−2 2sin

2

sin 2

du e dx v x

e x dx

x

x

= − = −−

−

21

22

2

cos

∫∫ = −⎛⎝⎜

⎞⎠⎟

−( ) cose xx2 1

22

2 ( 2 2− −⎛⎝⎜

⎞⎠⎟∫

1

2cos )x dx− −e x

2 2= − −− −∫1

22 2e x e x dxx xcos cos

Section 6.3 283

28. Continued

Let u e dv x dxx= =−2 2cos

du sin 2

sin 2

= − =

= −

−

−∫

21

21

2

2

e v x

e x dx

x

x

22

1

2

1

2

2 2e x e xx x− −− ⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

−

cos ( )2 sin 2

sin 22 2

2

2x dx

e xx

⎛⎝⎜

⎞⎠⎟

⎤

⎦⎥

= − +

∫−

( )

(cos sin

− −e x

1

22 22 sin 2

sin 2

x e x dx

e x dx e

x

x x

) −

= −

−

− −

∫∫

2

2 221

2((cos sin )

(cos

2 2

sin 2 2

x x C

e x dxe

xxx

+ +

= −−−

∫ 22

4++ +

= − +−−

−

∫

sin )

(cos sin

2

sin 2 2 2

x C

e x dxe

xxx

2

3

2 2

4xx

e

e

)

(cos sin )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= − +

+

−−

3

2

4

64

4

4 4

[cos(( ) sin ( )]

(cos sin )

(cos

− + −

= − +

+

−

6 6

4

4

4

6

4 4e

e6 6

125 028

−

≈

sin )

.

Check: NINT e x xx− −( ) ≈2 3 2 125 028sin , , , .2

29. y x e dxx= ∫ 2 4

Let

2

2 4u x dv e dx

du x dx v e

x= =

= = 1

444x

y x e e x dx

x e

x x= ⎛⎝⎜

⎞⎠⎟

− ⎛⎝⎜

⎞⎠⎟

=

∫( )1

4

1

4(2 )2 4 4

21

444 41

2x xxe dx− ∫

Let 4u x dv e dx

du dx

x= =

= v e x= 1

44

y x e x e e dxx x x= − ⎛⎝⎜

⎞⎠⎟

− ⎛⎝⎜

⎞⎠⎟

⎡

⎣∫

1

4

1

2

1

42 4 4 4( )

1

4⎢⎢⎤

⎦⎥

= − + +

= − +⎛

y x xe e C

yx x

x x x1

4e

1

32

8

1

32

2 4 4 4

2

1

8

4⎝⎝⎜

⎞

⎠⎟ +e4x C

30. y x x dx= ∫ 2 ln

Let ln1

x

1

3

2

3

u x dv x dx

du dx v x

= =

= =

y x x xx

dx

y

= ⎛⎝⎜

⎞⎠⎟

− ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=

∫(ln )1

3

1

33 3 1

1

331

3

1

9

x x x dx

y x x x C

3 2

3 3

ln1

3

ln

−

= − +

∫

31. y d= −∫θ θ θsec 1

Let sec

12

u dv d

du du v

= =

=−

=

−1

21 1

2

� � �

� ��

Note that we are told � >1, so no absolute value is needed

in the expression for du.

y d= ⎛⎝⎜

⎞⎠⎟

− ⎛⎝⎜

⎞⎠⎟ −

⎛

⎝⎜

⎞−(sec )1

21 2 2

2

1

2

1

1θ θ θ

θ θθ

⎠⎠⎟

= −−

= −

∫

∫−yd

w dw

θ θθ θ

θθ

2

sec2

Let

2

1

4 11

1

2

2

, ==

= −

= −

− −

−

∫2

sec

sec

θ θθ θ

θ θ

d

y w dw

y w

21 1 2

21

2

1

4

2

1

211 2

sec

+

= − − +−

C

yθ θ θ

21 2

2

1

21 C

32. y d= ∫� � � �sec tan

Let sec tansec

u dv ddu d v

= == =

� � � ��

y d

y C

= −= − + +

∫θ θ θ θθ θ θ θ

sec sec

sec ln sec tan

Note : In the last step, we used the result of Exercise 29 inSection 6.2.

33. Let sinu x dv x dx= =du dx v x= = − cos

x x dx x x x dx

x x

sin cos cos

cos sin

= − += − +

∫∫x C+

(a) x x dx x x dxsin sin0 0

π π∫ ∫=

cos sin= − +⎡⎣ ⎤⎦= − −

x x x0

1

π

π ( )) + + −=

0 0(1) 0π

284 Section 6.3

33. Continued

(b) x x dx x x dxsin sin22

= −∫∫ π

π

π

π

= −⎡⎣ ⎤⎦= − − − +=

x x xcos sin

2 (1) 0 ( 03

2

ππ

π ππ

1)

(c) x x dx x x dx x x dxsin sin sin22

= + ∫∫∫ π

ππ

π

π

0

= + =π π π3 4

34. We begin by evaluating ( 1)2x x e dx.x+ + −∫Let 1

(2 1)

(

2

2

u x x dv e dx

du x dx v e

x x

= + + == + = −

+

−

−

x

x

++

= − + + + +

=

−

− −∫

∫1)

( 1) (2 1)

Let

2

e dx

x x e x e dx

u

x

x x

22 1

2

( 1)2

x dv e dx

du dx v e

x x e dx

+ == = −

+ +

= −

−

−

−∫

x

x

x

(( 1) (2 1) 2

( 1)

2

2

x x e x e e dx

x x e

+ + − + +

= − + +

− − −

−∫x x x

xx x x

x

− + − += − + + +

− −

−(2 1) 2

( 3 4)2x e e C

x x e C

The graph shows that the two curves intersect at x = k,where k ≈ 1.050. The area we seek is

( 1)2 2

00x x e x dx

kk+ + −− ∫∫ x

= − + +⎡⎣

⎤⎦ − ⎡

⎣⎢⎤⎦⎥

≈ − +

−( 3 4)

( 4

2 3x x e xx kk

00

1

32 888. )) (0.386 )

0.726− −

≈0

35. First, we evaluate e t dt−∫ t cos .

Let cos

sin

u e dv t dt

du e dt v t

e t

= == − =

−

−

−

t

t

cos sin sin

Let

t dt e t t e dt

u e d

t t

t

= +

=

− −

−∫∫vv t dt

du e dt v t

e t dt e

t

t

== − = −

=

−

− −

sin

cos

cos tt t t

t

t e t e t dt

e t dt

sin cos cos

2 cos

− −

=

− −

−∫∫

ee t t C

e t dt e t

t

t t

−

− −

− +

=

∫ (sin cos )

cos (sin1

2−− +∫ cos )t C

Now we find the average value of y e tt= −2 cos for

0 2≤ ≤t π.

Average value =

=

=

−

−

∫∫

1

22 cos

cos

0

2

0

2π

π

π

π

π

e t dt

e t dt

e

t

t1

1

2−−

−

− ⎤⎦⎥

= − − −⎡⎣

t t t

e e

(sin cos )

( (0

0

2

21

21 1

π

π

π) )⎤⎤

⎦

= − ≈−1

2

2e π

π0.159

36. True. Use parts, letting u = x, dv = g(x)dx, and v = f (x).

37. True. Use parts, letting u = x2 , dv = g(x)dx, and v = f (x).

38. B. x x dx x x x x x C2 2 2 2cos sin sincos= + − +∫See problem 5.

See

2 2 2x x dx x x x C∫ = − + +sin cos sin

problem 1.

h x x x C( ) sin= +2

39. B. x x dx

dv x dx v x dx

∫∫= = = −

sin ( )

sin( ) sin( )

5

5 51

5ccos

cos( ) cos( )

5

1

55

1

55

1

x

u x du dx

x x x dx

= =

− − − = −∫ 551

255

x x

x

cos

sin( )+

40. C. x x dx∫ csc2

dv x dx v x dx x

u x du dx

x

= = = −= =

−

∫csc csc cot

cot

2 2

xx x dx x x x C− − = − + ⏐ ⏐+∫ cot cot ln sin

41. C. dy x x dx∫ ∫= 4 ln

dv x dx v x dx x

u x dux

dx

x x x

= = =

= =

−

∫4 4 2

1

2 2

2

2

ln

ln 22 2 212

xdx x x x C= − +∫ ln

42. (a) Let u x dv e dxx= =du dx v e

xe dx xe e dx

xe e C

x e

x

x x x

x x

= == −

= − += −

∫ ∫

( )1 xx C+

(b) Using the result from part (a):

Let u x dv e dx

du x dx v e

x e dx x e xe

x

x

x x

= == =

= −∫

2

2 2

2

2 xx

x x

x

dx

x e x e C

x x e C

∫= − − += − + +

2

22 1

2 2

( )

( )

Section 6.3 285

42. Continued(c) Using the result from part (b):

Let

3

3

3

2

3 3 2

u x dv e dx

du x dx v e

x e dx x e x e

x

x

x x

= == =

= − xx

x x

x

dx

x e x x e C

x x x e

∫∫= − − + += − + − +

3 2

3 23( 2 2)

( 3 6 6) CC

(d) xd

dxx

d

dxx

d

dxx e Cn n n n

n

nn x− + − + −

⎡

⎣⎢⎢

⎤

⎦⎥⎥

+2

1� ( )

or x nx n n xn n n− + − −⎡⎣

− −1 ( 1) 2

�+ − + − ⎤⎦ +−( 1) ( )! ( 1) ( !)1n n xn x n e C

(e) Use mathematical induction or argue based on tabulatintegration.

Alternately, show that the derivative of the answer topart (d) is x en x:

d

dxx nx n n x

n x

n n n

n

− + − −(⎡⎣

+ − + −

− −

−

1 2

1

( 1)

( 1) ( !) (� 11) !

( 1)

( )

1 2

n x

n n n

n

n e C

x nx n n x

) + ⎤⎦

= − + − −+ −

− −

−[

� 1 11

1

( !) ( 1) !]

( 1)

n x n e

ed

dxx nx n n n

n x

x n n n

+ − +

− + −− −2 −−⎡⎣

+ − + −= − + −

−

−� ( 1) ( !) ( 1) !]

( 1

1

1

n n

n n

n x n

x nx n n[ ))

( 1) !) ( 1) !

(

2

1

1

x

n x n e

nx n n

n

n n x

n

−

−

−

−+ − + + −+ −

� ( ]

−−⎡⎣

+ − − −+ −

=

−

−

−

1)

( 1)( 2)

( ) !

2x

n n n x

n e

n

n

n x

3

11� ]

xx en x

43. Let . Then2

so 2 2 .w x dwdx

xdx x dw w dw= = = =,

sin (sin )(2 ) 2 sin

Let

x dx w w dw w w dw

u

= ==

∫∫∫ww dv w dw

du dw v w

w w dw w

== = −

= −

sincos

sin cos cos

cos sin

sin 2

w w dw

w w w C

x dx

+= − + +

=

∫∫

ww w dw

w w w C

x x

sin

2 cos 2 sin

2 cos

∫∫= − + += − ++ +2 sin x C

44. Let 3 9. Then1

2 3 9(3) , so

2

33

w x dwx

dx

dx x

= + =+

= ++ =

= ⎛⎝⎜

⎞⎠⎟

=+ ∫

92

3.

( )2

33 9

dw w dw

e dw e w dwwx 2

3ww e dww∫∫

Let u w dv e dw

du dw v e

w e dw w e e

w

w

w w w

= == =

= − dw

w e e

w e

e dx w e dw

w w

w

x w

∫∫

∫∫

= −= −

=

=

+

( 1)2

32

3 9

33( 1)

2

3( 3 9 1) 3

w e

x e C

w

x

−

= + − ++9

45. Let Then 2 .2w x dw x dx= =.

x e dx x e x dx w e dwx x w7 2 3 32 2

( ) .= = ∫∫∫1

2

Use tabular integration with f x w( ) 3= and g w ew( ) = .

w e dw w e w e w e e C

w w w

w w w w w3 3 2

3 2

3 6 6

( 3 6

= − + − +

= − + −∫

66)

1

2( 3 6 6

7 3

3 2

2

e C

x e dx w e dw

w w w

w

x w

+

=

= − + −

∫∫1

2

))

( 3 6 6)

2

6 4 2 2

e C

x x x eC

w

x

+

= − + − +

46. Let ln . Then1

, and soy r dyr

dr= = = =dr r dy ey dy.

Using the result of Exercise 13, we have:

sin (ln ) (sin

(sin cos

r dr y dy

y

=

= −

∫∫ )e

e

y

y1

2yy

r r

r

)

sin (ln ) cos (ln )

2s

ln r

+

= − +

=

C

e C1

2[ ]

[ iin (ln ) cos (ln )r r− +] C

47. Let cosu dx= =x dv xn

du = =−nx dx v xn 1 sin

x x nxn ncos sin

sin

x dx x x dx

x x

n

n

(sin )( )= −

= −

−∫∫ 1

nn x xn−∫ 1 sin dx

286 Section 6.3

48. Let u = =x dv x dxn sin

du nx dx v xn= = −−1 cos

n x nx dxx n nsin )( cos cosx dx x x( ) ( )( )= − − −

= −

−∫∫ 1

xx x n x xn ncos cos+ −∫ 1 dx

49. Let u = =x dv e dxn ax

du nx dx va

en ax= =−1 1

x ea

ea

en ax ax ax( (dx x nxn n= ⎛⎝⎜

⎞⎠⎟

− ⎛⎝⎜

⎞⎠⎟

−)1 1 1 ddx

dx a

)

, 0

∫∫

∫= − ≠−x e

a

n

ax e

n axn ax1

50. Let (ln )u x n= =dv dx

(ln )du

x n

= =−n

xdx v x

1

(ln ) (ln ) ( )x dx x xn n= −⎡

⎣⎢⎢

⎤

⎦⎥⎥

−

∫ xn x

x

n(ln ) 1

ddx

x x n xn n

∫∫= − −(ln ) (ln ) 1 dx

51. (a) Let Then soy x dx= = = ′−f x f y f y dy1( ). ( ), ( ) .

Hence, f dx dy− = ′⎡⎣ ⎤⎦ = ′1( ) ( ) ( ) ( )x y f y y f y dy∫∫∫∫(b) Let u = = ′y dv f y dy( )

du dy v f y= = ( )

y y f y f y dy

f x x f y

( )f y dy′ = −

= −∫∫

−

( ) ( )

( )( ) ( )1 ddy

x dx y f y dy

x f x

∫∫∫ −

−

= ′

= −

Hence, f 1

1

( ) ( )

( ) ff y dy( ) .∫52. Let u f x dv dx= =−1( )

dud

dxf x dx v x= ⎛

⎝⎜⎞⎠⎟

=−1( )

f x dx xf x xd

dxf x dx− − −= − ⎛

⎝⎜⎞⎠⎟∫∫ 1 1 1( ) ( ) ( )

53. (a) Using and ( )y f y y= = =− −f x x1 1( ) sin sin ,

− ≤ ≤π π2 2

y , we have:

sin− −

−

= −

= + +=

∫∫ 1 1

1

x dx x x y dy

x x y C

sin sin

sin cos

xx x x Csin cos )− −+ +1 1(sin

(b) sin sin sin− − −= − ⎛⎝⎜

⎞⎠⎟∫∫ 1 1 1x dx x x x

d

dxx dx

= −−

− ∫x x xx

dxsin 1

2

1

1

u x du x dx= − = −1 22,

= +

= + +

= + −

− −

−

−

∫x x u du

x x u C

x x

sin

sin

sin

1 1 2

1 1 2

1

1

2

1 xx C2 +

(c) cos (sin )− = −1 21x x

54. (a) Using and ( )y f y= = =− −f x x y1 1( ) tan tan ,

− < <

= −− −

π π2 2

1 1

y

x dx x x y dy

,

tan tan

we have:

tan ∫∫∫= − +−x x y Ctan ln sec

(

1

Section 6.2, Example 5))

= + += + +

−

− −

x x y C

x x x

tan ln cos

tan ln cos (tan )

1

1 1 CC

(b) tan tan tan− − −= − ⎛⎝⎜

⎞⎠⎟∫∫ 1 1 1x dx x x x

d

dxx dx

= −+

⎛⎝⎜

⎞⎠⎟

= + =

− ∫x x xx

u x du x dx

tan

,

12

2

1

11 2

dx

= −

= − +

− −

−

∫x x u du

x x u C

tan

tan ln

1 1

1

1

21

2

(1= − + +−x x x Ctan ln )1 21

2

(c) ln cos ) ln ln )(tan (1− =+

= − +1

2

21

1

1

2x

xx

55. (a) Using andy = =− −f x x1 1( ) cosf y x x

x dx x

( ) cos , ,

cos c

= ≤ ≤= −− −0

1 1

π we have:

cos oos

cos sin

cos sin

y dy

x x y C

x x (cos

∫∫= − += −

−

−

1

1 −− +1x C)

(b) cos cos cos− − −= − ⎛⎝⎜

⎞⎠⎟∫∫ 1 1 1x dx x x x

d

dxx dx

= − −−

⎛

⎝⎜

⎞

⎠⎟

= − = −

=

− ∫x x xx

x du x dx

cos

,

1

2

2

1

11 2

dx

u

xx x u du

x x u C

x x

cos

cos

cos

− −

−

−

−

= − +

= −

∫1 1 2

1 1 2

1

1

2

1−− +x C2

(c) sin )(cos− = −1 21x x

Section 6.4 287

56. (a) Using and ( ) we havey f y= = =−f x x y12 2( ) log ,

log log

logln

log

2 2

2

2

2

2

2

x dx x x dy

x x C

x

y= −

= − +

=

∫∫

xx x− 12 2

lnlog

2

(b) log log log2 2 2x dx x x xd

dxx= − ⎛

⎝⎜⎞⎠⎟∫∫ dx

= − ⎛⎝⎜

⎞⎠⎟

= −

=

∫

∫

x x xx

x xd

x

log

logln

l

2

2

1

ln 2

2

dx

oogln2

1x C− ⎛

⎝⎜⎞⎠⎟

+2

(c) 2 2log x x=

Quick Quiz Section 6.1–6.3 1. E.

2. C.

3. A. xe dxx2∫dv e dx v e dx

u x du dx

xe

2

edx

x xx

x x

= = =

= =

−

=

∫

∫

2 22

2 2

2

2

e

xxe

2

eC

x x2 2

4− +

4. (a) y

x

(b) Let 2 and 2 in the differential eqdy

dxy x b= = + uuation:

2 2(2 ) 42 2

1

= + −==

x b xb

b

(c) First, note thatdy

dx= − =2(0) 4(0) 0 at the point (0, 0).

Also, d y

dx

d

dxy x

dy

dx

2

2(2 4 ) 2 4,= − = − which is –4 at the

point (0, 0).

By the Second Dervative test, g has a local maximum at(0, 0).

Section 6.4 Exponential Growth and Decay(pp.350–361)

Exploration 1 Choosing a Convenient Base

1. ht

= =1 1

5. h is the reciprocal to the doubling period.

2. 3 25 3

25 3

27 925

=

=

= =

ht

ht

t

log

loglog

log. years.

3. ht

= =1 1

10. h is the reciprocal to the tripling period.

4. 2 32

310 2

36 3093

=

=

= =

ht

ht

t

log

loglog

log. years.

5. ht

= =1 1

15. h is the reciprocal to the half life.

6. .

log(. )

log

log(.

101

210

12

15

= ⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=

ht

ht

110

12

49 83)

log.

⎛⎝⎜

⎞⎠⎟

= =t years.

Quick Review 6.4

1. a eb=

2. c d= ln

3. ln( )x

x e

x e

+ =+ =

= −

3 2

3

3

2

2

4. 100 600

62 6

1

26

2

2e

ex

x

x

x

===

=

ln

ln

5. 0 85 2 5

0 85 2 50 85 2 5

2 5

. .

ln . ln .ln . ln .

ln .

l

x

x

x

x

===

=nn .

.0 85

5 638≈ −

6. 2 3

2 31 2 3

2 3

1

1

k k

k k

k kk

+

+==

+ == −

ln ln( ) ln ln

ln (ln ln 222

3 21 710

)ln

ln ln.k =

−≈

288 Section 6.4

7. 1 1 10

1 1 101 1 10

10

1 1

1

.

ln . lnln . ln

ln

ln . l

t

t

t

t

===

= =oog .

.1 1

24 159≈

8. e

t

t

t− =

− = ⎛⎝⎜

⎞⎠⎟

= − ⎛⎝⎜

⎞⎠⎟

= =

2 1

4

21

41

2

1

4

1

24

ln

ln ln lln2

9. ln( )y x

y e

y e

x

x

+ = ++ =

= − +

+

+

1 2 3

1

1

2 3

2 3

10. ln y t

y e

y e

y e

t

t

t

+ = −+ =+ = ±

= − ±

−

−

−

2 3 1

2

2

2

3 1

3 1

3 1


1. y dy x dx∫ ∫=

y x2 2

22 2

1

= +

= +=

C

CC

(2)3

2 ( )

y x= 2 + 3, valid for all real numbers

2. y dy x dx∫ ∫= −

y x2 2

22 2

4

= − +

= − +=

C

CC

(3)25

2 ( )

y x= 25 2− , valid on the interval (–5, 5)

3. 1 1

y xdy dx∫ ∫=

ln ln

2 20

y x Cy x C

CC

= += += +=

y = x, valid on the interval ( , )0 ∞

4. 1

2y

xdy dx∫ ∫=

ln y x

y x x

= += =+

2

2 2

C

e e eC C

y x x= ± =Ae e2 2

3 , valid for all real numbers

5. dy

dxy

x+

+5

( 2)∫ ∫=

ln

/

/

(yx

x

y

y

x x

x x

+ = + +

= −= −

+ +

+

5)2

2 2

2 2

22

52

2

C

e

e e

C

c 55 52 2 2= −+Aex x/

y e= −+6 x x2 2 2 5/ , valid for all real numbers

6. dy

ydx

cos2∫ ∫=

tan

tan (0)0

y x CC

C

= += +=

0

y = tan ,−1 x valid for all real numbers.

7. dy

dxx e ey x= cos sin

e dy x e dxy x− = ∫∫ cos sin

− = +− = +

= −

−e e C

e e CC

y xsin

sin 00

2

y e x= − −ln ( ),sin2 valid for all real numbers.

8. dy

dxe ey x=

e dy e dx

e e C

C e e e

y x

y x

−

−∫ ∫=

− = += − = −2 0 2 1

y e ex= + −ln ( ),2 1 valid for all real numbers.

9. 1

2y

x

C

C

C

2dy dx

y x

∫ ∫=

− = − +

= +

=

−

−

1 2

1

3

1

.25

yx

=+1

32, valid for all real numbers.

10. dy

dx

y x

x=

4 ln

dydx

u x

dux

dx

y u du

y

y

x

x

4 ln∫ ∫

∫

=

=

=

=

=

ln1

2 4

2 22

10

2

4

u Cy x C

CC

+= += +=

(ln )(ln )e

y x= (ln ) ,4 valid on the interval ( , ).0 ∞

Section 6.4 289

11. y t y

y t

kt

t

( )

( ) .

==

01 5100

e

e

12. y t y

y t

kt

t

( )

( ) .

== −

00 5200

e

e

13. y t y

y t

y

kt

kt

k

k

( )

( )

( )

=== ==

0

5

5

50

5 100 50

22

e

e

e

eln ==

==

50 2 2

50 0 2 2

kk

y t t

.

( ) ( . )ln

Solution : olne rr y t t( ) .= 50 20 2i

14. y t y

y t

y

kt

kt

k

k

( )

( )

( )

=== =

=

0

10

10

60

10 30 601

2

e

e

e

e

lnn

ln ln

1

210

0 11

20 1 2

=

= = −

k

k

y t

. .

( )Solution: == =− −60 60 20 1 2 10e ( . ) /( )ln ort ty t i15. Doubling time:

A t A rt

t

t

( )

.

.

.

====

00 086

0 0862000 1000

22 0 0

e

e

eln 886

28 06

t

t = ≈ln

0.086. yr

Amount in 30 years:

A = ≈1000 13 197 140 086 30e( . )( ) , .$

16. Annual rate:

A t A

r

r

rt

r

r

( )( )( )

====

=

015

154000 2000

22 15

e

e

eln

lln

15

20 0462 4 62≈ =. . %

Amount in 30 years:

A t A

A

rt( )[( )/ ]( )

===

02 15 30

22000

2000

e

e

e

ln

ln 22

22000 28000

==

i$

17. Initial deposit:

A t A

A

A

rt( )

..

( . )( )

==

=

0

00 0525 30

0

2898 442898 44

e

e

ee1 575600 00

.$ .≈

Doubling time:

A t A rt

t

t

( )

.

.

.

====

00 0525

0 05251200 600

22 0

e

e

eln 00525

213 2

t

t = ≈ln

0.0525. years

18. Annual rate:

A t A rt

r

( )

, ..

( )( )

==

=

03010 405 37 1200

104 0537

12

e

e

ee30

104 0537

1230

1

30

104 0537

120 0

r

r

r

ln

ln

.

..

=

= ≈ 772 7 2= . %

Doubling time:

A t A rt

t

t

( )

.

.

.

====

00 072

0 0722400 1200

22 0 0

e

e

eln 772

29 63

t

t = ≈ln

0.072. years

19. (a) Annually:

2 1 04752

214 94

==

= ≈

.ln ln

ln

ln.

t

t

t

1.0475

1.0475yeears

(b) Monthly:

2 10 0475

12

2 12 10 0475

12

12

= +⎛⎝⎜

⎞⎠⎟

= +⎛⎝

.

ln ln.

t

t ⎜⎜⎞⎠⎟

=+⎛

⎝⎜⎞⎠⎟

≈tln

ln.

. year2

12 10 0475

12

14 62 ss

(c) Quarterly:

2 10 0475

42 4

2

4

= +⎛⎝⎜

⎞⎠⎟

=

=

.

ln lnln

ln

t

t

t

1.011875

4 11.011875years≈ 14 68.

(d) Continuously:

22 0 0475

2

0 047514 59

==

= ≈

et

t

t0 0475.

ln .ln

.. yearss

290 Section 6.4

20. (a) Annually:

2 1 08252 1 0825

2

1 08258

==

= ≈

.ln ln .

ln

ln ..

t

t

t 774 years

(b) Monthly:

2 10 0825

12

2 12 10 0825

12

12

= +⎛⎝⎜

⎞⎠⎟

= +⎛⎝

.

ln ln.

t

t ⎜⎜⎞⎠⎟

=+⎛

⎝⎜⎞⎠⎟

≈tln

ln.

. years2

12 10 0825

12

8 43

(c) Quarterly:

2 10 0825

42 4

2

4

= +⎛⎝⎜

⎞⎠⎟

=

=

.

ln lnln

ln

t

t

t

1.020625

4 11.020625years≈ 8 49.

(d) Continuously:

22

28 40

==

= ≈

et

t

t0 0825

0 0825

0 0825

.

ln .ln

.. years

21. dy

dty= −0 0077.

10 0077

ydy dt= −∫∫ .

ln .ln ( / )

.

y t

t

= −

=−

=

0 00771 2

0 007790 years

22. dy

dtky=

1

1 2

650 01067

ydy dt

y kt

k

k

=

=

=

=

∫∫ k

lnln ( / )

.23. (a) Since there are 48 half-hour doubling times in 24 hours,

there will be 2 2 8 1048 14≈ ×. bacteria.

(b) The bacteria reproduce fast enough that even if manyare destroyed there are still enough left to make theperson sick.

24. Using y y ekt= 0 , we have10 000 03, = y e k and

40 000 05, .= y e k Hence

40 000

10 0000

5

03

,

,,=

y e

y e

k

kwhich gives

e k2 4= , or k = ln .2 Solving10 000 03 2, ,ln= y e we have

y0 1250= . There were 1250 bacteria initially. We could

solve this more quickly by noticing that the population

increased by a factor of 4, i.e. doubled twice, in 2 hrs, so thedoubling time is 1 hr. Thus in 3 hrs the population wouldhave doubled 3 times, so the initial population was

10 000

21250

3

,.=

25. 0 9 0 18. .= −e t

ln . .ln .

..

0 9 0 180 9

0 180 585

= −

= − ≈

t

t days

26. (a) Half-life = = ≈ln ln

.. days

2 2

0 005138 6

k

(b) 0 05 00. .= e t−0 5

ln .ln

..

0.050.05

days

= −

= − ≈

0 005

0 005

t

t 599 15

The sample will be useful for about 599 days.

27. Since y y0 0 2= =( ) , we have:

y e

e

kt

k

===

2

5 2 2( )( )

ln ln5 2 2+ k

k = =ln ln. ln

5 22.5

−2

0 5

Function: y e t= ≈2 or(0.5 ln 2.5) y e t2 0 4581.

28. Since y y0 0 1 1= =( ) . , we have:

y e

e

kt

k

===

−1 1

3 1 1 3.

.ln ln

( )( )

3 1.1− 3k

k = 1

3(ln ln1.1 3)−

Function: 1.1 3) /3y e t= ≈ −1 1 1 1. .(ln ln− or y e 00 3344. t

29. At time tk

= 3, the amount remaining is

y e y e y e ykt k k0 0

30

300 0499− − −= = ≈( / ) . . This is less than 5%

of the original amount, which means that over 95% hasdecayed already.

30. T T T T es skt− = − −( )0

35 65 65

50 65 650

10

020

− = −− = −

−

−( )

( )

( )( )

( )(

T e

T e

k

k ))

Dividing the first equation by the second, we have:

21

102

10=

=

e

k

k

ln

Substituting back into the first equation, we have:

− = −

− = − ⎛⎝

−30 65

30 651

2

02 10 10

0

( )

( )

[(ln )/ ]( )T e

T ⎜⎜⎞⎠⎟

− = −=

60 655

0

0

TT

The beam’s initial temperature is 5°F.

Section 6.4 291

31. (a) First, we find the value of k.

T T T T e

e

e

s skt

k

− = −− = −

=

−

−

−

( )

( ) ( )( )0

1060 20 90 204

7110

1

10

k

k = − ln4

7When the soup cools to 35°, we have:

35 20 90 20

15 70

1 10

1 1− = −

=( ) [( / ) ln )]

[( /e

e

t(4/7

00

3

14

1

10

4

7

103

14

) ln )]

ln ln

ln

(4/7 t

t

t

= ⎛⎝⎜

⎞⎠⎟

=

⎛⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

≈ln

. min47

27 53

It takes a total of about 27.53 minutes, which is anadditional 17.53 minutes after the first 10 minutes.

(b) Using the same value of k as in part (a), we have:

T T T T e

es s

kt− = −− − = − −

−( )

( ) [ ( )] [( / )0

1 1035 15 90 15 (4/7

(4/7

ln )]

[( / ) ln )]

ln

t

te50 10510

21

1 10=

== ⎛⎝⎜

⎞⎠⎟

=

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

1

10

4

7

101021

47

ln

ln

ln

t

t ≈≈ 13 26.

It takes about 13.26 minutes

32. First, we find the value of k.Taking “right now” as t 0,= 60° above room temperaturemeans T Ts0 60− = . Thus, we have

T T T T e

e

e

k

s skt

k

k

− = −=

=

=

−

− −( )

( )( )0

20

20

70 607

61

20lln

7

6

(a) T T T T e es skt− = − = ≈− −( ) ( ( / ) ln( / ))( )

01 20 7 6 1560 533 45.

It will be about 53.45°C above room temperature.

(b) T T T T e es skt− = − = ≈− −( ) ( ( / ) ln( / ))( )

01 20 7 6 12060 223 79.

It will be about 23.79° above room temperature.

(c) T T T T es skt− = − −( )0

10 601

6

1

20

7

6

1 20=

= −⎛⎝⎜

⎞⎠⎟

−e t( ( / ) ln ))

ln ln

(7/6

tt

t = − ≈20

7 6232 47

ln

ln ( / ). min

(1/6)

It will take about 232.47 min or 3.9 hr.

33. (a) T Tst− = 79 47 0 932. ( . )

(b) T t= +10 79 47 0 932. ( . )

(c) Solving T =12 and using the exact values from theregression equation, we obtain t ≈ 52 5. sec.

(d) Substituting t = 0 into the equation we found in part(b), the temperature was approximately 89.47°C.

34. (a) Newton’s Law of Cooling predicts that the differencebetween the probe temperature (T ) and the surroundingtemperature (Ts ) is an exponential function of time,but in this case Ts 0,= so T is an exponential function oftime.

(b) T = ×79 96 0 9273. . t

[–0, 40] by [0, 86]

(c) At about 37 seconds.

(d) 76.96°C

35. Use k = ln 2

5700(see Example 3).

e

kt

tk

kt− =− =

= − = −

0 4450 445

5700 0 4

.ln .

ln ln .0.445 445

26658

ln≈ years

Crater Lake is about 6658 years old.

36. Use k = ln 2

5700(see Example 3).

(a) e kt− = 0 17.− =

= − = − ≈

kt

tk

ln .ln ln .

ln,

0.170 17

5700 0 17

214 5711 years

The animal died about 14,571 years before A.D. 2000, in12,571 B.C.

(b) e kt− = 0 18.− =

= − = − ≈

kt

tk

ln .ln ln .

ln,

0.180 18

5700 0 18

214 1011 years


292 Section 6.4

36. Continued

(c) e kt− = 0 16.− =

= − = − ≈

kt

tk

ln .ln . ln .

ln,

0 160 16 5700 0 16

215 070 yyears


37. 1

3= −e kt

k

e

t

kt

= =

=

=−

=

−

ln( / ).

ln( / )

..

1 3

50 22

1

21 2

0 223 15 yeears

38. 3 = ert

r

e

t

rt

= =

=

= =

ln ( ).

ln( )

..

3

100 11

44

0 1112 62 years

39. y y e kt= −0

800 1000

0 80 8

1010

10

10==

= −

=

−

−e

e

k

t

k

k

( )( )

.ln .

At ++ ===

− −14 24

1000

1000

0 8 10 24

2 4 0

h:( ln . / )

. lny e

e .. .8 585 4≈ kg

About 585.4 kg will remain.

40. 0 2 0 1. .= −e t

ln . .ln . .

0 2 0 110 0 2 16 09

= −= − ≈

tt yr

It will take about 16.09 years.

41. (a) dp

dhkp=

dp

pk dh

dp

pk dh

p kh C

e e

p e e

p kh C

C kh

=

=

= +

==

∫∫

+

lnln

pp Aekh=Initial condition: p p h= =0 0when

p AeA p

00

0

==

Solution: p p ekh= 0

Using the given altitude-pressure data, we

have p0 1013= millibars, so:

p e

e

e

kh

k

k

==

=

1013

90 101390

1013

20

20

( )( )

k = ≈ − −1

20

90

10130 121 1ln . km

Thus, we have p e h≈ −1013 0 121.

(b) At 50 km, the pressure is

1013 2 3831 20 90 1013 50

e/ ln /

.( ) ( )( )( ) ≈ millibarss.

(c) 900 1013= ekh

900

1013= ekh

hk

= = ≈1 900

1013

20 900 1013

90 10130 9ln

ln ( / )

ln ( / ). 777 km

The pressure is 900 millibars at an altitude of about0.977 km.

42. By the Law of Exponential Change, y e t= −100 0 6. . At t =1hour, the amount remaining will be

100 54 880 6 1e− ≈. ( ) . .grams

43. (a) By the Law of Exponential Change, the solution is

V V e t= −0

1 40( / ) .

(b) 0 1 1 40. ( / )= −e t

ln .

ln . . sec

0 140

40 0 1 92 1

= −

= − ≈

t

t

It will take about 92.1 seconds.

44. (a) A t A et( ) = 0

It grows by a factor of e each year.

(b) 3 = et

ln3 = t

It will take ln . .3 1 1≈ yr

(c) In one year your account grows from A0 to A0e, so youcan earn A e A e0 0 1− −, ( )or times your initial amount.This represents an increase of about 172%.

45. (a) 90 100= e r( )( )

ln

ln. . %

90 10090

1000 045 4 5

=

= ≈

r

r or

(b) 131 100= e r( )( )

lnln

. . %

131 100131

1000 049 4 9

=

= ≈

r

r or

Section 6.4 293

46. (a) 2y y ert0 0=

2 = ert

ln2 = rt

tr

= ln2

(b)

(c) ln . ,2 0 69≈ so the doubling time is0 69.

rwhich is almost

the same as the rules.

(d)70

514

72

514 4= =years or years.

(e) 3

33

3

0 0y y e

ert

tr

rt

rt

===

=

lnln

Since ln . ,3 1 099≈ a suitable rule is 108

100

108

r ior .

(We choose 108 instead of 110 because 108 has morefactors.)

47. False. The correct solution is y ekx C= + , which can bewritten (with a new C ) as y Cekx= .

48. True. The differential equation is solved by an exponentialequation that can be written in any base. Note that

Ce C k2t kt= ( ) =3 2when 3/ (ln ).

49. D. A A( )t ert= 0

2 1 75= e

r

t

= =

= =

ln( ).

ln( )

..

2

70 099

3

0 09911 1

50. C. A A= ⎛⎝⎜

⎞⎠⎟0

1

2

t r/

1 1001

2

199

= ⎛⎝⎜

⎞⎠⎟

/r

ln(. ) ln(. )

ln(. )

ln(.

01199

55

199 5

0

=

=r11

30)

=

51. D.

52. E. T e kt− = − −68 425 68( )

195 68 357 30= + −e k

e

k

k− = =

=−

=

30 127

3570 356

0 356

300344

.

..

100 68 357100 68

357

0

0 0344= +

=

−⎛⎝⎜

⎞⎠⎟

−

−e

t

t( . )

ln

.0034470

70 30 40

=

− =

min

53. (a) Since acceleration isdv

dt, we have Force = = −m

dv

dtkv.

(b) From mdv

dtkv= − we get

dv

dt

k

mv= − , which is the

differential equation for exponential growth modeled byv Ce k m t= −( / ) . Since v v t= =0 0at , it followsthat C v= 0.

(c) In each case, we would solve 2 = −e k m t( / ) . If k isconstant, an increase in m would require an increase in t.The object of larger mass takes longer to slow down.Alternatively, one can consider the equation

dv

dt

k

mv= − to see that v changes more slowly for larger

values of m.

54. (a) s t v e dtv m

ke Ck m t k m t( ) ( / ) ( / )= = − +− −∫ 0

0

Initial condition: s( )0 0=

0 0

0

0 0

0

= − +

=

= − +

=

−

v m

kC

v m

kC

s tv m

ke

v m

kv m

k

k m t( ) ( / )

11−( )−e k m t( / )

(b) lim ( ) lim ( / )

t t

k m ts tv m

ke

v m

k→∞ →∞

−= −( ) =0 01

55. v m

k0 = coasting distance

( . )( . ).

0 80 49 901 32

998

33

k

k

=

=

We know that v m

k

k

m0 1 32

998

33 49 9

20

33= = =.

( . ).and

Using Equation 3, we have:

s tv m

ke

e

k m t

t

( )

.

.

( / )

/

= −( )= −( )≈

−

−

0

20 33

1

1 32 1

1 32 11 0 606−( )−e t.

294 Section 6.4

55. ContinuedA graph of the model is shown superimposed on a graphof the data.

56. v m

k0 = coasting distance

( . )( . ).

.

( ) ( (

0 86 30 840 97

27 343

10

kk

s tv m

ke k

=

≈

= − − // )

( . / . )

)

( ) . ( )

( ) .

m t

ts t e

s t

= −=

−0 97 1

0

27 343 30 84

997 1 0 8866( ).− −e t

A graph of the model is shown superimposed on a graphof the data.

[0, 3] by [0, 1]

57. (a) x1

1+⎛⎝⎜

⎞⎠⎟x

x

10 2.5937100 2.7048

1000 2.716910,000 2.7181

100,000 2.7183e ≈ 2 7183.

Graphical support:

yx

y ex

1 211= +⎛

⎝⎜⎞⎠⎟

=,

(b) r = 2

x 12+⎛

⎝⎜⎞⎠⎟x

x

10 6.1917100 7.2446

1000 7.374310,000 7.3876

100,000 7.3889

e2 ≈ 7 389.

Graphical support:

yx

y ex

1 212= +⎛

⎝⎜⎞⎠⎟

=, 2

r = 0 5.

x 10 5+⎛

⎝⎜⎞⎠⎟

.

x

x

10 1.6289100 1.6467

1000 1.648510,000 1.6487

100,000 1.6487

e0.5 ≈ 1 6487.

Graphical support:

yx

y ex

1 210 5= +⎛

⎝⎜⎞⎠⎟

=., 0.5

(c) As we compound more times, the increment of timebetween compounding approaches 0. Continuouscompounding is based on an instantaneous rate ofchange which is a limit of average rates as the incrementin time approaches 0.

58. (a) To simplify calculations somewhat, we may write:

v tmg

k

e e e

e e emg

k

e

e

at at at

at at at

at

a

( ) = −+

= −

−

−

2

2

1tt

at

at

at

mg

k

e

emg

k e

+

= + −+

= −+

⎛⎝⎜

⎞⎠⎟

11 2

1

12

1

2

2

2

( )

The left side of the differential equation is:

mdv

dtm

mg

ke ae

mamg

ke

at at

at

= +

=

−( )( ) ( )

(

2 1 2

4

2 2 2

2 ++

= +

=

−

−

1

4 1

4

2 2

2 2 2

2

) ( )

( ) ( )

e

mgk

m

mg

ke e

mge

at

at at

aat

ate( )2 21+

Section 6.5 295

58. Continued

The right side of the differential equation is:

mg kv mg kmg

k e

mg

at− = − ⎛

⎝⎜⎞⎠⎟

−+

⎛⎝⎜

⎞⎠⎟

= − −

22

2

12

1

1 122

1

1 14

1

4

2

2

2 2

e

mge e

at

at a

+⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= − ++

−( tt

at

at

a

mge

emg e

+⎛⎝⎜

⎞⎠⎟

= + −+

=

1

4 1 4

14

2

2

2 2

2

)

( )

( )tt

ate( )2 21+Since the left and right sides are equal, the differentialequation is satisfied.

And vmg

k

e e

e e( ) ,0 0

0 0

0 0= −

+= so the initial condition is

also satisfied.

(b) lim ( ) limt t

at at

at at

at

v tmg

k

e e

e e

e

e→∞ →∞

−

−

−

−= −+

iaat

⎛

⎝⎜

⎞

⎠⎟

= −+

⎛

⎝⎜

⎞

⎠⎟

= −+

⎛⎝⎜

→∞

−

−limt

at

at

mg

k

e

e

mg

k

1

1

1 0

1 0

2

2

⎞⎞⎠⎟

= mg

k

The limiting velocity is mg

k.

(c) mg

k= ≈160

0 005179

./ secft

The limiting velocity is about 179 ft/sec,or about 122 mi/hr.

Section 6.5 Logistic Growth (pp. 362–376)

Exploration 1 Exponential Growth Revisited

1. 100 2 40960012( ) =

2. 100 2 4 97 1012 24 90( ) .( . ) = ×

3. No. This number is much larger than the estimated numberof atoms.

4. 500 000 100 25000

212 29

, ( )log

log.

=

= =

x

x hours

Exploration 2 Learning From theDifferential Equation

1. dp

dtwill be close to zero when P is close to M.

2. P is half the value of M at its vertex.

3. The graph begins a downward trend at half the carryingcapacity, causing a decline in growth rates.

4. When the initial population is less than M, the initial growthrate is positive.

5. When the initial population is more than M, the initialgrowth rate is negative.

6. When the initial population is equal to M, the growth rate isat a maximum.

7. lim ( ) . lim ( )t t

P t M P t→∞ →∞

= depends only on M.

Quick Review 6.5

1. )x x

x

x x

xx

xx

−+

−

−

+ +−

1

1

1

1

11

1

2

2

2. )x x

x

x

2 2

2

2

41

4

4

14

4

−

−

+−

3. )x x x x

x x

x x

2

2

2

2 1

2

3

13

2

+ − + +

+ −

++ −

12

4. )x x

x

x x

x

xx

x

2 3

3

3

1 5

55

5

− −

−−

+ −−

5. ( , )−∞ ∞

6. lim ( ).x

f xe→∞ − ∞( )=

+=60

1 560

0 1

7. lim ( )( . ( ))x

f xe→∞ − −∞=

+=60

1 50

0 1

8. ye

( )( . ( ))

060

1 510

0 1 0=

+=−

9. From problems 6 and 7, the two horizontal asymptotes arey = 0 and y = 60.

296 Section 6.5

10. y

x

60

10

0

Section 6.5 Exercises 1. A x B x x( ) ( )− + = −4 12

x BB

x AA

= = −= −

= − + − = −=

4 4 82

1 1 4 2 1 1 123

,

, ( ) ( )( )

2. A x B x x( ) ( )− + + = +2 3 2 16

x BBB

x A

= + = +==

= − − − =

2 2 3 2 2 165 20

43 3 2 2

, ( ) ( )

, ( ) (( )− +− =

= −

3 165 10

2AA

3. A x B x x( ) ( )+ + − = −5 2 16

x BBB

x A

= − − − = − −− =

= −= + =

5 5 2 16 57 21

32 2 5

, ( ) ( )

, ( ) 116 27 14

2

−==

AA

4. A x B x( ) ( )+ + − =3 3 3

x BBB

x AA

= − − − =− =

= −= + =

=

3 3 3 36 3

1 23 3 3 3

6 3

, ( )

/, ( )

AA = 1 2/

5. See problem 1.

x

xdx

x xdx

x

−−

= + −−

⎛⎝⎜

⎞⎠⎟

= −

∫∫12

4

3 2

43 2

2

ln ln (xx C

x

xC

− +

=−

⎛

⎝⎜⎜

⎞

⎠⎟⎟

+

4

4

3

2

)

ln( )

6. See problem 2.

2 16

6

2

3

4

22

2

x

x xdx

x xdx

x

++ −

= −+

+−

⎛⎝⎜

⎞⎠⎟

= −

∫∫ln ( ++ + − +

= −+

⎛

⎝⎜

⎞

⎠⎟ +

3 4 2

2

3

4

2

) ( )

ln( )

( )

ln x C

x

xC

7. )x x

x2 34 2

2

−

2 8

8

28

4

42

3

2

2

2

x x

x

xx

xdx

u xdu x dx

x

−

+−

⎛⎝⎜

⎞⎠⎟

= −=

+

∫

44 4

4

2

2 2 4

du

ux u C

x x C

= + +

= + − +∫ ln

ln( )

8. )x x

x

2 2

29 6

9

3

1− −

−

13

9

3 33 3 3

3

2∫

∫

+−

= ++

+−

− + + ==

xdx

xA

x

B

xdx

A x B xx

( ) ( ),, ( )

/, ( )

//

BB

x AA

x

3 3 31 2

3 3 3 31 2

1 2

+ ==

= − − − == −

+ −xx x

dx

xx

xC

++

−

= + −+

+

∫ 3

1 2

33

3

/

ln

9. 21

22

1dx

xx C

+= +−∫ tan

10. 29

232

1dx

x

xC

+= ⎛

⎝⎜⎞⎠⎟

+−∫ tan

11.7

2 5 32x xdx

− −∫A

x

B

xA x B xx B

2 1 37

3 2 1 73 2 3 1

++

−=

− + + == + =( ) ( )

, ( ( ) ) 771

1 2 1 2 3 72

2

2 1

1

3

Bx A

A

x x

== − − − =

= −−

++

−⎛⎝⎜

⎞

/ , ( / )

⎠⎠⎟

= −+

⎛⎝⎜

⎞⎠⎟

+

∫ dx

x

xCln

3

2 1

Section 6.5 297

12.1 3

3 5 32

−− −∫

x

x xdx

A

x

B

xx

A x B x xx B

3 2 11 3

1 31 3 1

−+

−= −

− + − = −=( 1) (3 2)

, ( ( )) ) ( )

,

− = −= −

= −⎛⎝⎜

⎞⎠⎟

= − ⎛⎝⎜

⎞⎠

2 1 3 12

2

3

2

31 1 3

2

3

B

x A ⎟⎟

− = −

=

−+ −

−⎛⎝⎜

⎞⎠⎟

= − −

∫

1

31

33

3 2

2

1

3 2 2

A

A

x xdx

xln( ) lln( )

ln( )

x C

x

xC

− +

= −−

⎛⎝⎜

⎞⎠⎟

+

1

3 2

1 2

13.8 7

2 32

x

x x

−− −∫

A

x

B

xx

A x B x x

x B

++

−= −

− + + = −

= +

1 2 38 7

2 3 1 8 73

2

3

2

( ) ( )

, 11 83

27

5

25

21 2 3 8

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

−

=

== − − =

B

Bx A x x, ( ) −−

− − = −− = −

=

++

−⎛⎝⎜

⎞⎠⎟∫

72 3 8 7

5 153

3

1

2

2 3

AAA

x xdx

( )

== + + − +

= + −( ) +

3 1 2 3

1 2 33

ln ln

ln

x x C

x x C

14.5 14

72

x

x xdx

++∫

A

x

B

xx

A x Bx xx B

++

= +

+ + = += − − = − +

75 14

7 5 147 7 5 7

( ), ( ) 114

7 213

0 0 7 5 0 147 14

2

2 3

− = −=

= + = +==

+

BB

x AAA

x

, ( ) ( )

xxdx

x x C

x x C

+⎛⎝⎜

⎞⎠⎟

= + + +

= +( ) +

∫ 72 3 7

72 3

ln ln ( )

ln

15. dyx

x xdx∫ ∫= −

−2 6

22

A

x

B

xx

A x Bx xx B

B

+−

= −

− + = −= = −

= −

22 6

2 2 62 2 2 2 6

2

( ), ( )

221

0 0 2 2 0 62 6

33 1

2

Bx A

AA

x x

= −= − = −

− = −=

+ −−

⎛⎝⎜

, ( ) ( )

⎞⎞⎠⎟

= − − +

=−

+

∫ dx

y x x C

yx

xC

3 2

2

3

ln ln

ln

16. dux

dx∫ ∫=−

2

12

A

x

B

xA x B xx B

BB

++

−=

− + + == + =

=

1 12

1 1 21 1 1 2

2 2

( ) ( ), ( )

=== − − − =

− == −

= −+

+−

⎛⎝⎜

⎞⎠

11 1 1 2

2 21

1

1

1

1

x AAA

ux x

, ( )

⎟⎟

= − + + − +

= −+

+

∫ dx

u x x C

ux

xC

ln ln

ln

1 1

1

1

17. F x dxx x

dx′∫ ∫=−

( )2

3

A

x

B

x

C

xA x x Bx x Cx x

++

+−

=

+ − + − + + =1 1

2

1 1 1 1 2( )( ) ( ) ( )xx C

Cx B

Bx A

A

x x

= ==

= − ==

= − == −

− ++

1 2 21

1 2 21

0 22

2 1

,

,

,

11

1

12 1 1

+−

⎛⎝⎜

⎞⎠⎟

= − + + + − +

∫ xdx

F x x x x C

F x

( ) ln ln ln

( )) ln=−⎛

⎝⎜⎜

⎞

⎠⎟⎟

+x

xC

2

2

1

298 Section 6.5

18. G t dtt

t tdt′( )∫ ∫=

−2 3

3

)t t t

t tt

t

t tdt

tt

dt

A t

3 3

3

3

2

2

2 22

2

22

22

1

−−

= +−

= +−

∫

∫( ++ + − == − − − =

− == −

=

1 1 21 1 1 2

2 21

1 1

) ( ), ( )

, (

B tt B

BB

t A ++ ===

1 22 2

1

)AA

G t tt t

dt

t t

( )( ) ( )

ln ln

= + −−

++

⎛⎝⎜

⎞⎠⎟

= − − +

∫21

1

1

1

2 1 tt C

tt

tC

+ +

= + +−

+

1

21

1ln

19.2

42

x

xdx

−∫u xdu x dx

du

uu C x C

= −=

= + = − +∫

2

2

42

4ln ln

20.4 3

2 3 12

x

x xdx

−− +∫

u x xdu x dx

du

uu C

x x

= − += −

= +

= − + +

∫

2 3 14 3

2 3 1

2

2

( )

ln

ln CC

21.x x

x xdx

2

2

1+ −−∫

)x x x x

x x

xx

x xdx

u x

2 2

2

2

2

1

2 1

1

12 1

− + −−

−

+ −−

⎛⎝⎜

⎞⎠⎟

= −

∫xx

du x dx

xdu

ux u C

x x x C

= −

+ = + +

= − +

∫( )

ln

ln

2 1

2

22.2

1

3

2

x

xdx

−∫

)x x

x x

x

x

xx

xdx

u xdu

2 3

3

2

2

1 2

2 2

2

2

22

1

1

−−

+−

⎛⎝⎜

⎞⎠⎟

= −

∫

==

+

= + += + − +

∫2

1

2

2

2 2

x dx

xdu

ux u C

x x C

ln

ln

23. (a) 200 individuals.

(b) 100 individuals.

(c)dP

dt

( ). ( )( )

1000 006 100 200 100

60

= −

= individuals per year.

24. (a) 700 individuals.(b) 350 individuals.

(c) dP

dt

( ). ( )( )

3500 0008 350 700 350

98

= −

= individualss per year.



(c) dP

dt

( ). ( )( )

6000 0002 600 1200 600

72

= −

= individualls per year.



(c) dP

dt

( )( )( )

.

250010 2500 5000 2500

62 5

5= −

=

−

individduals per year.

27.dP

dtP P= −. ( )006 200

dP

P Pdt

A

P

B

PA P BP

( ).

( )

200006

2001

200 1

−=

+−

=

− + =

∫∫

PP BB

P AAA

= ==

= − ===

200 200 10 005

0 200 0 1200 1

0

,.

, ( )

... .

.

0050 005 0 005

2000 006

1 1

2

P PdP t

P

+−

⎛⎝⎜

⎞⎠⎟

=

+

∫

0001 2

200 1 2

−⎛⎝⎜

⎞⎠⎟

=

− − = +

∫ PdP t

P P t C

.

ln ln ( ) .

Section 6.5 299

27. Continued

ln .

.

2001 2

2001

200

1 2

−⎛⎝⎜

⎞⎠⎟

= − −

− =

=

− −

P

Pt C

Pe e

P

t c

11

200

81

24200

1

1 2

1 2 0

+

= +

=

=+

− −

− −

−

e e

e e

e

P

t c

c

c

.

. ( )

224 1 2e t− .

[–1, 7] by [0, 200]

28.dP

dtP P

dP

P Pdt

A

P

= −

−=

+

∫∫

0 0008 700

7000 0008

. ( )

( ).

BB

PA P BPP B

BP

7001

700 1700 700 1

0 0014

−=

− + == =

=

( ),

.== − =

=

+−

⎛⎝

0 700 0 10 0014

0 0014 0 0014

700

, ( ).

. .

AA

P P⎜⎜⎞⎠⎟

=

+−

⎛⎝⎜

⎞⎠⎟

=

∫

∫

dP t

P PdP t

P

0 0008

1 1

7000 56

.

.

ln −− − = +ln ( ) .700 0 56P t C

ln .

.

7000 56

7001

700

0 56

−⎛⎝⎜

⎞⎠⎟

= − −

− = − −

P

Pt C

Pe et c

PPe e

e e

e

P

t c

c

c

= +

= +

=

=

− −

− −

−

1

700

101

69

0 56

0 56 0

.

. ( )

7700

1 69 0 56+ −e t.

[–1, 15] by [0, 700]

29.dPdt

P P= −0 0002 1200. ( )

dP

P Pdt

A

P

B

PA P

12000 0002

12001

1200

−( ) =

+−

=

−( )

∫∫ .

++ == =

== −( ) =

BPP B

BP A

11200 1200 1

0 000830 1200 0

,.

, 111200 1

0 00083

0 00083 0 00083

700

AA

P P

==

+−

⎛⎝⎜

⎞.

. .⎠⎠⎟

=

+−

⎛⎝⎜

⎞⎠⎟

=

−

∫

∫

dP t

P PdP t

P

0 0002

1 1

12000 24

.

.

ln lln .

ln .

1200 0 24

12000 24

−( ) = +

−⎛⎝⎜

⎞⎠⎟

= − −

P t C

P

Pt C

112001

12001

1200

20

0 24

0 24

Pe e

Pe e

t c

t c

− =

= +

− −

− −

.

.

== +

=

=+

− −

−

−

1

591200

1 59

0 24 0

0 24

e e

e

Pe

c

c

t

. ( )

.

[–1, 30] by [0, 1200]

30.dP

dtP P= −( )−10 50005

dP

P Pdt

A

P

B

P

A P B

500010

50001

5000

5

−( ) =

+−

=

−( ) +

−∫∫

PPP B

BP A

== =

== −( ) =

15000 5000 1

0 00020 5000 0 1

50

,.

,000 1

0 0002

0 0002 0 0002

5000

AA

P PdP

==

+−

⎛⎝⎜

⎞⎠⎟∫

.

. . ==

+−

⎛⎝⎜

⎞⎠⎟

=

−

∫

10

1 1

50000 5

5 t

P PdP t.

300 Section 6.5

30. Continued

ln ln .

ln .

P P t C

P

Pt

− −( ) = +−⎛

⎝⎜⎞⎠⎟

= − −

5000 0 5

50000 5 CC

Pe e

Pe e

t c

t c

50001

50001

5000

50

0 5

0 5

− =

= +

=

− −

− −

.

.

11

995000

1 99

0 5 0

0 5

+

=

=+

− −

−

−

e e

e

Pe

c

c

t

. ( )

.

[–1, 200] by [0, 5000]

31. (a) P te t

( ). .

=+ −

1000

1 4 8 0 7

=

+

=+

−

−

1000

1

1

4 8 0 7eM

Ae

t

kt

. .

This is a logistic growth model with k = 0.7 andM = 1000.

(b) Pe

( ).

01000

18

4 8=

+≈

Initially there are 8 rabbits.

32. (a) P te t

( ).

=+ −200

1 5 3

=

+

=+

−

−

200

1

1

5 3e eM

Ae

t

kt

.

This is a logistic growth model with k = 1 and M = 200.

(b) Pe

( ).

0200

11

5 3=

+≈

Initially 1 student has the measles.

33. (a) dP

dtP P= −0 0015 150. ( )

= −

= −

=

0 225

150150

0 225

.( )

( )

.

P P

k

MP M P

k MThus, and ==

=+

=+

−

−

150.

PM

Ae

Ae

kt

t

1150

1 0 225.

Initial condition: P

AeAA

( )0 6

6150

11 25

0

=

=+

+ == 224

150

1 24 0 225Formula: P

e t=

+ − .

(b) 100150

1 24

1 243

2

24

0 225

0 225

0 225

=+

+ =

−

−

−

e

e

e

t

t

.

.

. tt

te

t

t

=

=

− = −

= ≈

−

1

21

480 225 48

48

0 2251

0 225.

. lnln

.77 21

125150

1 24

1 24

0 225

0 225

.

.

.

weeks

=+

+ =

−

−

e

e

t

t 66

5

241

51

1200 225 120

0 225

0 225

e

e

t

t

t

−

−

=

=

− = −

.

.

. ln

tt = ≈ln

..

120

0 22521 28

It will take about 17.21 weeks to reach 100 guppies, andabout 21.28 weeks to reach 125 guppies.

34. (a) dP

dtP P= −0 0004 250. ( )

= −

= −

0 1

250250

.( )

( )

P P

k

MP M P

Thus, andk M

PM

Ae

Ae

kt

t

= =

=+

=+

−

−

0 1 250

1250

1 0 1

. .

.

Initial condition: where represenP t( ) ,0 28 0= = ttsthe year 1970.

28250

128 1 250

250

281

111

147 928

0=

++ =

= − = ≈

AeA

A

( )

. 66

Formula:/

or approximatP te t

( ) ,.

=+ −

250

1 111 140 1eely

P te t

( ). .

=+ −

250

1 7 9286 0 1

Section 6.5 301

34. Continued

(b) The population P(t) will round to 250 whenP t

e t

( ) . .

.

.

.

≥

=+

+

−

249 5

249 5250

1 111 14

249 5 111

0 1 /

11

14250

249 5 111

140

0 1

0 1

e

e

t

t

−

−

⎛

⎝⎜

⎞

⎠⎟ =

=

.

.( . )( ).55

14

55 389

0 114

55 38910 55

0 1e

t

t

t− =

− =

=

.

,

. ln,

(ln ,3389 14 82 8− ≈ln ) .

It will take about 83 years.

35. dP

dtkP M P= −( )

dP

P M Pk dt

Q

P

R

M PRP Q M P

( )

( )

−=

+−

=

+ − =

∫∫1

1

P MQ

QM

P M MR

RM

= =

=

= =

=

0 11

11

,

,

1 1

1 1

MP

MM P

dP kt C

P M P

+−

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

= +

+−

⎛⎝⎜

⎞⎠

∫ ( )

( )⎟⎟ = +

−⎛⎝⎜

⎞⎠⎟

= − −

− =

∫

− −

dP Mkt C

M P

PMkt C

M

Pe eMkt C

ln

1

MM

Pe A

PM

Ae

Mkt

Mkt

= +

=+

−

−

1

1

36. (a) dP

dtk M P= −( )

dP

M Pk dt

−= ∫∫

− − = +− =

== −

− −

−

−

ln( )M P kt C

M P e e

e A

P M Ae

kt c

c

kt

(b) lim ( )t

kP t M Ae M→∞

− ∞= − =

(c) When t = 0.

(d) This curve has no inflection point. If the initialpopulation is greater than M, the curve is alwaysconcave up and approaches y = M asymptotically fromabove. If the initial population is smaller than M, thecurve is always concave down and approaches y = Masymptotically from below.

37. (a) The regression equation is

Pe t

=+ −

232739 9

1 14 582 0 101

.

. .

[–5, 70] by [–24000, 260000]

(b) Pe

( ).

.,

. ( )∞ =

+≈− ∞

232739 9

1 14 582232 740

0 101peoplle.

(c) Pe t

=+ −

232739 9

1 14 582 0 101

.

. .

14 58232739 9

225 0001

0

0 101

0 101

..

,.

.

.

e

e

t

t

−

−

= 2 −

= 00246 05

0 10160t = −

−≈.

.or 2010.

(d) dP

dtkP M P P P= − = × −−( ) ( . ) ( . ).4 352 10 232739 97

38. (a) The regression equation is

Pe t

=+ −

458791 8

1 18 771 0 113

.

. .

(b) Pe

( ).

.,

. ( )∞ =

+≈− ∞

458791 8

1 18 771458 792

0 113

(c) Pe t

=+ −

458791 8

1 18 771 0 113

.

. .

18 771458791 8

450 0001

0

0 113

0 113

..

,.

.

.

e

e

t

t

−

−

= −

= 00016 87

0 11360t = −

−≈.

.or 2010.

(d) dP

dtkP M P P P= − = × −−( ) ( . ) ( . )2 4626 10 458791 87

39. False. It does look exponential, but it resembles the solution

to dP

dtkP k P= − =( ) ( ) .100 10 90

302 Section 6.5

40. True. The graph will be a logistic curve withlim ( ) lim ( ) .t t

P t P t→∞ →−∞

= =100 0and

41. D. 600

2300= .

42. B. dy

dt e

( ) ..

. ( )

∞ =+

=− ∞0 9

1 450 9

0 15

( . . )1 0 0 9 100 10− = %

43. D. 3

1 22

3

( )( )x xdx

− +∫A

x

B

xA x B xx B

B

−+

+=

+ + − == − − − =

−

1 23

2 1 32 2 1 3

3

( ) ( ), ( )

=== −

= + ===

1−

+ −+

⎛⎝⎜

⎞⎠⎟

31

1 1 2 33 3

1

1

1

2

Bx A

AA

x xd

, ( )

xx

x

x

∫

= −+

⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

ln ln1

2

8

52

3

44. B.

45. (a) Note that k > 0 and M > 0, so the sign of dPdt

is the

sameas the sign of ( )( ).M P P m− − For m < P < M, bothM P P m− −and are positive, so the product is positive.For P m P M< >or , the expressions M – P and P – mhave opposite signs, so the product is negative.

(b) dP

dt

k

MM P P m= − −( )( )

dP

dt

kP P= − −

12001200 100( )( )

1200

1200 1001100

1200 100

( )( )

( )(

− −=

− −

P P

dP

dtk

P P ))( ) ( )

( )( )

dP

dtk

P P

P P

=

− + −− −

11

12100 1200

1200 100

ddP

dtk

P P

dP

dtk

=

−+

−⎛⎝⎜

⎞⎠⎟

=

11

121

1200

1

100

11

12

1

12200

1

100

11

12

1200

−+

−⎛⎝⎜

⎞⎠⎟

=

− − +

∫∫ P PdP k dt

Pln ln PP kt C

P

Pkt C

P

− = +

−−

= +

−

10011

12100

1200

11

12100

1

ln

2200100

1200

11 12

11 12

−= ±

−−

=

Pe e

P

PAe

C kt

kt

/

/

P Ae APe

P Ae

kt kt

kt

− = −+

100 1200

1

11 11

11

/12 /12

/12( )) = +

= ++

1200 100

1200 100

1

11

11Ae

PAe

Ae

kt

kt

/12

/12

111kt /12

(c) 3001200 100

1

0

0= +

+Ae

Ae300 1 1200 100300 100 1200 300

200 900

( )+ = +− = −

=

A AA A

AA

A

P te

e

kt

k

=

= ++

2

91200 2 9 100

1 2 9

11

11( )

( )

( )

/

/

/12

tt

kt

ktP t

e

e

/12

/12

/12( )

( ) ( )= ++

1200 2 100 9

9 2

11

11

PP te

e

kt

kt( )

( )= ++

300 8 3

9 2

11

11

/12

/12

(d)

[0, 75] by [0, 1500]Note that the slope field is given bydP

dtP P= − −0 1

12001200 100

.( )( ).

(e) dP

dt

k

MM P P m= − −( )( )

M

M P P m

dP

dtk

M

M m

M m

M P P m

dP

dtk

P

( )( )

( )( )(

− −=

−−

− −=

− mm M P

M P P m

dP

dt

M m

Mk

M P P m

) ( )

( )( )

+ −− −

= −

−+

−⎛⎝⎜

⎞⎠

1 1⎟⎟ = −

−+

−⎛⎝⎜

⎞⎠⎟

= −

−

∫ ∫

dP

dt

M m

Mk

M P P mdP

M m

Mk dt

M

1 1

ln −− + − = − +

−−

= − +

−−

= ±

P P mM m

Mkt C

P m

M P

M m

Mkt C

P m

M P

ln

ln

ee e

P m

M PAe

P m M P Ae

C M m kt M

M m kt M

( )

( )

(( )

−

−−−

=

− = −

/

/

MM m kt M

M m kt M M m kt MP Ae AMe m

PA

−

− −+ = +

=

)

( ) ( )( )

/

/ /1

MMe m

Ae

PAMe m

Ae

M m kt M

M m kt M

( )

( )

( )

−

−+

+

= ++

/

/1

01

0

0== +

+AM m

A1

Chapter 6 Review 303

45. Continued

(e) P A AM mA P M m P

Am P

P

( )( )( ( ) ) ( )

( )

( )

0 10 0

0

0

+ = +− = −

= −− MM

P m

M P= −

−( )

( )

0

0Therefore, the solution to the differential equation is

PAMe m

AeA

PM m kt M

M m kt M= +

+= −−

−

( )

( )

( )/

/where

1

0 mm

M P− ( ).

0

46. (a) 1 1

a

x

aCtan− ⎛

⎝⎜⎞⎠⎟

+

(b) 1

2a

x a

x aCln

+−

+

(c) −+

+1

x aC

47. (a) 5 315

3ln x

xC+ +

++

(b) −+

++

+5

3

15

2 3 2x xC

( )

48. (a) This is true since

A

x

B

x

C

x

A x B x C

x−+

−+

−= − + − +

−1 1 1

1 1

12 3

2

3( ) ( )

( ) ( )

( )

(b) A x B x C x x( ) ( )− + − + = + +1 1 3 52 2

x A B Cx Cx A B

ABB

= − + == == − + =

=− + =

=

0 51 92 2 3

11 9 5

5

,,,

(c) ln( )

xx x

C− −−

−−

+15

1

9

2 1 2

Quick Quiz 1. C.

2. C.

3. A. dx

x x( )( )− +∫ 1 3

A

x

B

xA x B xx B

Bx

−+

+=

+ + − == − − =

= −

1 31

3 1 13 4 1

1 4

( ) ( ),

/== =

=

−+ −

+⎛⎝⎜

⎞⎠⎟

= −

∫

1 4 11 4

1 4

1

1 4

3

1

4

,/

/ /

ln

AA

x xdx

x 11

3xC

++

4.dP

dt

P P= −⎛⎝⎜

⎞⎠⎟5

10

10

dp

P Pdt

A

P

B

PA P BPP

( )

( )

10

1

50

101

10 110

−=

+−

=

− + ==

∫∫

,,.

,.

10 10 1

0 10 10 1

BB

P AA

==

= ==

0 1 0 1

10

1

50

101 5

. .

ln /

P PdP t C

P

Pt C

+−

⎛⎝⎜

⎞⎠⎟

= +

− = − −

∫

PPe e

PAe

A

t C=

+

= =+

=

− −

−

10

1

0 310

12 33

1 5

1 5 0

/

/ ( )( )

.

liim ( ). / ( )t

P te→∞ − ∞=

+=10

1 2 3310

1 5

(b) PAe

( )/ ( )

0 2010

1 1 5 0= =

+ −

A

P tet

=

=+

=→∞ − ∞

5 6610

1 5 6610

1 5

.

lim ( ). / ( )

(c) Separate the variables.

dY

Y

tdt

Yt t

C

Y Cet t

= −⎛⎝⎜

⎞⎠⎟

= − +

= −

1

51

10

5 100

2

1

5

ln

/ 22 100 1

0

5 2 100

3 3

3

/

/ /

where C e

Ce C

Y e

C

t t

== ⇒ =

= −

(d) lim lim limt

t t

t

t

t te

e

e→∞

−

→∞ →∞= =3

35 21005

2 100

//

/

330

5

5 20

e

e

t

t t

/

/ /( )=

Chapter 6 Review Exercises

1. sec tan tan tan/2

0

30 3θ θ θ ππ π/3

0 3∫ = ⎤⎦ = − =d

2. xx

dx x x+⎛⎝⎜

⎞⎠⎟

= −⎡⎣⎢

⎤⎦⎥∫ −1 1

221

2 2 1

1

2

= −⎛⎝⎜

⎞⎠⎟

− −⎛⎝⎜

⎞⎠⎟

= +

= =

1

24

1

2

1

21

3

2

1

24

22

( )

304 Chapter 6 Review

3. Let u x= +2 1

du dx= 2

1

236

2 118

130

1

31

3

du dx

xdx

udu

=

+=∫ ∫( )

= −⎛⎝⎜

⎞⎠⎟

⎤

⎦⎥

= − −⎛⎝⎜

⎞⎠⎟

= − −⎛⎝⎜

⎞

−181

2

91

91

98

9

2

1

3

u

⎠⎠⎟

= 8

4. Let u x= −1 2

du x dx= −2

− =

− = − =−∫ ∫

du x dx

x x dx udu

2

2 1 02

1

1

0

0sin( ) sin

5. Let u x= sin

du x dx= cos

5 53 2

0

3 2

0

1sin cos/ /x x dx u du

π /2

∫ ∫=

= ⎤⎦⎥

= −=

52

52 1 02

5 2

0

1

i u /

( )

6. x x

xdx x dx x

2

1 2

4

1 2

433 0

+ = + ≠∫ ∫/ /( ) ( )

= +⎛

⎝⎜⎞⎠⎟

⎤

⎦⎥

= +⎛⎝⎜

⎞⎠⎟

−

1

23

1

216 3 4

1

2

1

2

1 2

4

x x/

( ) ( )44

3

2

⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

= − +⎛⎝⎜

⎞⎠⎟

= −

=

201

8

12

8

2013

8147

8 7. Let u x= tan

du x dx= sec2

e x dx e du

e

e ee

x u

u

tan/sec2

0

14

0

1

1 0

1

=

= ⎤⎦

= −= −

∫∫0

π

8. Let u r= ln

dur

dr= 1

ln

( )

/

/

r

rdr u du

u

e

1

1 2

0

1

3 2

0

12

32

31 0

2

3

∫ ∫=

= ⎤⎦⎥

= −

=

9. x

x xdx

20

1

5 6+ +∫x

x x( )( )+ +3 2

A

x

B

xx

A x B x xx B

B

++

+=

+ + + == − − + = −

=

3 22 3

2 2 3 2( ) ( )

, ( )−−

= − − + = −− = −

=

++ −

+

=

∫

23 3 2 3

33

3

3

2

2

x AAA

x xdx

,

ln(

( )

xx x+ − + = ⎛⎝⎜

⎞⎠⎟

3 2256

2433 2

0

1) ln( ) ln

10. 2 6

321

2 x

x xdx

+−∫

2 6

3

x

x x

+−( )

A

x

B

xx

A x Bx xx B

Bx

+−

= +

− + = += = +

=

3( )

4

2 6

3 2 63 3 2 3 6, ( )

== − = +− =

= −− +

−= −

∫

0 33 6

22 4

32

1

2

,

l

AAA

x xdx

(0 ) 2(0) 6

nn ln( ) lnx x+ − = −4 3 6 21

2

11. Let u xdu x dxdu x dx

= −= −

− =

2 sincos

cos

cos

sinln

ln sin

x

xdx

udu

u C

x C

2

1

2

−= −

= − += − − +

∫ ∫


12. Let u x= +3 4

du dx

du dx

=

=

31

3dx

xu du

u C

x

3 4

1

31

3

3

21

23 4

31 3

2 3

2 3

+=

= +

= +

∫ ∫ − /

/

/( )

i

++ C

13. Let u t= +2 5

du t dt

du t dt

=

=

21

2t dt

t udu u C

t C

t

2

2

2

5

1

2

1 1

21

25

1

2

+= = +

= + +

=

∫ ∫ ln

ln

ln( ++ +5) C

14. Let u = 1

θ

du d= − 12θ

θ

1 1 1

1

2θ θ θθ∫ ∫= −

= − +

= −

sec tan sec tan

sec

sec

d u u du

u C

θθ+ C

15. Let u y= ln

duy

dy= 1

tan(ln )tan

sin

cos

y

ydy u du

u

udu

∫ ∫

∫

=

=

Let w udw u du

wdw

w C

u C

== −

= −

= − += − +

∫

cossin

ln

ln cos

1

== − +ln cos(ln )y C

16. Let u ex=

du e dxx=e e dx u du

u u C

e

x x

x

sec( ) sec

ln sec tan

ln sec(

== + +=

∫∫

)) tan( )+ +e Cx

17. Let u x= ln

dux

dx

dx

x x udu

u C

x C

=

=

= += +

∫ ∫

1

1

lnln

ln ln

18. dt

t t

dt

t∫ ∫=3 2/

=

= − +

= − +

−

−∫ t dt

t C

tC

3 2

1 222

/

/

19. Use tabular integration with ( ) and ( )3f x x g x= == cos .x

x x dx3∫ cos

= + − − +x x x x x x x C3 23 6 6sin cos sin cos

20. Let u x dv x dx= =ln 4

dux

dx v x= =1 1

55

x x dx x x xx

dx

x x

4 5 5

5

1

5

1

5

1

1

5

1

5

∫ ∫= − ⎛⎝⎜

⎞⎠⎟

= −

ln ln

ln xx dx

x x x C

4

5 51

5

1

25

∫= − +ln


21. Letu e dv x dxx= =3 sin

du e dx v x

e x dx e x x e

x

x x

= = −= − +∫

3

3

3

3 3 3

cos

sin cos cos xx dx∫Integrate by parts again

Letu e dv x dxx= =3 3 cos

du e dx v xx= =9 3 sin

e x dx e x e x e x dxx x x x3 3 3 33 9∫ ∫= − + −sin cos sin sin

10 33 3 3e x dx e x e x Cx x xsin cos sin= − + +∫e x dx e x e x C

x

x x x3 3 31

103

3

10

∫ = − + +

=

sin [ cos sin ]

sin −−⎛⎝⎜

⎞⎠⎟

+cos xe Cx

103

22. Let u x dv e dxx= = −2 3

du x dx v e x= = − −21

33

x e dx x e e x dxx x x2 3 2 3 31

3

2

3− − −∫ ∫= − +

Let u x dv e dxx= = −3

du dx v e

x e xe e

x

x x x

= = −

= − + − +

−

− − −∫

1

31

3

2

3

1

3

1

3

3

2 3 3 3 ddx

x e xe e dx

x

x x x

⎡⎣⎢

⎤⎦⎥

= − − +

= −

− − −∫1

3

2

9

2

91

3

2 3 3 3

2ee xe e C

x xe

x x x− − −− − +

= − − −⎛

⎝⎜

⎞

⎠⎟

3 3 3

2

2

9

2

27

3

2

9

2

27−− +3x C

23. 25

252xdx

−∫25

5 5( )( )x x+ −A

x

B

xA x B xx B

++

−=

− + + == + =

5 525

5 55

10

( ) ( ) 25(5 5) 25,

BBB

x AAA

==

= − − − =− =

= −

255 2

5 5 5 2510 25

5 2

/, ( )

/

−+

+−

⎛⎝⎜

⎞⎠⎟

= −+

+

∫5 2

5

5 2

55

2

5

5

/ /

ln

x xdx

x

xC

24. 5 2

2 12

x

x xdx

++ −∫5 2

2 1 1

15 2

1 1

x

x xA

x

B

xx

A x B x

+− +

−+

+= +

+ + −

( )( )

( )2 1

(2 )) 5 21,

= += − − − = − +

− = −=

=

xx B

BB

x

( ( ) ) ( )2 1 1 5 1 23 3

11

2,, A S

A

A

x x

1

21 2

3

2

9

2

1

1

+⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

+

=

=

−+

+

1

2

33

2 111

22 1 13 3

⎛⎝⎜

⎞⎠⎟

= − + +

∫ dx

x x Cln ( ) ( )

25. dy

dxx

x= + +12

2

dy xx

dx

dy xx

dx

y x

= + +⎛

⎝⎜

⎞

⎠⎟

= + +⎛

⎝⎜

⎞

⎠⎟

= +

∫∫

12

12

1

2

2

22

1

60 1

6 21

2 3

3 2

x x C

y C

yx x

x

+ +

= =

= + + +

( )

Graphical support:


26. dy

dxx

x= +⎛

⎝⎜⎞⎠⎟

12

dy xx

dx= +⎛⎝⎜

⎞⎠⎟

12

dy xx

dx

y xx

dx

y x

∫ ∫

∫

= +⎛⎝⎜

⎞⎠⎟

= + +⎛⎝⎜

⎞⎠⎟

=

1

21

1

3

2

22

3 ++ − +

= + − + =

+ =

= −

= +

−2

11

32 1 1

4

31

1

3

32

1

3

x x C

y C

C

C

yx

( )

xxx

− −1 1

3Graphical support:

27. dy

dt t=

+1

4

dyt

dt

dyt

dt

y t Cy C

=+

=+

= + +− = + =

∫ ∫

1

41

44

3 1ln

( ) ln( ) 222

4 2Cy t

== + +ln( )

Graphical Support:

28. dy

dθθ θ= csc cot2 2

dy d

dy d

==∫ ∫csc cot

csc cot

2 2

2 2

θ θ θθ θ θ

y C

y C

C

y

= − +

⎛⎝⎜

⎞⎠⎟

= − + =

=

= −

1

22

4

1

21

3

21

22

csc

csc

θ

π

θ ++ 3

2

29. d y

dxx

x

( )′ = −212

d y xx

dx

d y xx

dx

y

( )

( )

′

′

= −⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟∫ ∫

21

21

2

2

′′′

′

= + += + == −= + −=

−

−

∫

x x Cy C

C

y x x

dy x

2 1

2 1

1 2 11

1

( )

( 22 1

3

1

1

3

11

30 1 0

+ −

= + − +

= + − + =

−∫ x dx

y x x x C

y C

)

ln

( )

− + =

=

= + − +

2

30

2

3

3

2

3

3

C

C

yx

x xln

Graphical support:

Let

Wefirst show thegraphof

f xx

x x

y

( ) ln .= + − +3

3

2

3== = + − >−f x x x x′( ) , ,2 1 1 0

along with theslope fieldd for y f x xx

′ ″= = −( ) .212


29. ContinuedWe now show the graph of along with thy f x= ( ) ee slope field

for y f x x x′ ′= = + −−( ) .2 1 1

30. d r

dtt

( )cos

″ = −

d r t dt

d r t dt

r t Cr

( ) cos

( ) cos

sin( )

″″″

″

= −= −= − +

∫ ∫0 == = −

= − −= − −= −

∫ ∫

Cr t

d r t dt

r t

11

1

″′′

sin

( ) ( sin )

cos tt Cr C

Cr t t

dr t t

+= + = −= −= − −= − −∫

′

′

( )

cos

(cos

0 1 12

2

22

22

0 1

22

2

2

)

sin

( )

sin

dt

r tt

t C

r C

r tt

t

∫= − − +

= = −

= − − −−1

Graphical support:

We first show the graph of y r t= ″ = − −sin 1

along with the slope field for ′ = ′″ = −y r tcos .

Next, we show the graph ofalo

y r t t= = − −′ cos 2nng with the slope field for ′ = = − −y r t″ sin .1

Finally we show the graphof y r tt

t= = − −sin2

22 −−

= =

1

along with the slope field for y r t′ ′ cos −− −t 2.

31. dy

dxy= + 2

dy

ydx

dy

ydx

y x C

y Ce

y Cey

x

x

+=

+=

+ = ++ == −

∫ ∫2

22

2

2

ln

(00 2 24

4 2

) = − === −

CC

y ex

Graphical support:

32.dy

dxx y= + +( )( )2 1 1

dy

yx dx

dy

yx dx

y x x C

y

+= +

+= +

+ = + +

∫∫1

2 1

12 1

1 2

( )

( )

ln

++ =

= −− = − =

=

= −

+

+

+

1

11 1 1

2

2

2

2

2

Ce

y Cey C

C

y e

x x

x x

x x

( )

11

Graphical support:

33.dy

dty y= −( )1

dy

y ydt

A

y

B

yA y Byy By A

( )

( ),,

1

11

1 11 10

−=

+−

=

− + == == == 1


33. Continued

1 1

11

1 1

11

1

y y

y ydy dt

y y t C

y

y

+−

=

+−

=

− − = +− =

∫ ∫ln ln

ln −− −

− =

− =

=+

=

− −

− −

−

t C

y

ye e

ye e

yAe

y

t c

t c

t

1

11

1

1

0 0( ) .111

19

1

1

0 1=

+=

=+

−AeA

ye

.

9 -0.1

34.dy

dxy y= −0 001 100. ( )

dy

y ydx

A

y

B

y

A y

0 001 100

0 001 1001

100

. ( )

.( )

−=

+−

=

− + BB yy B

By A

A

( . ), (. )

,.

0 001 1100 1 1

100 100 1

0

== =

== =

= 0010 01

0 001

10

100

0 001

0 00

.

.

.

.

y ydy x C+

−⎛⎝⎜

⎞⎠⎟

= +∫

11

1

1000 1

100 0 1

y ydy x C

y y x

+−

⎛⎝⎜

⎞⎠⎟

= +

− − = +

∫ .

ln ln . CC

y

yx C

ye e

yAe

x c

ln .

.

1000 1

1001

100

1

0 1

− = − −

− =

=+

− −

−−

−

−

= =+

=

=+

0 1

0 1 0

0

0 5100

119

100

1 19

.

. ( )( )

x

yAe

A

ye ..1x

35. y t dtx

= +∫ sin3

45

36. y t dtx

= + +∫ 1 24

1

37.

x

y

1

2

0

−1

1−1

38.

x

y

1

2

0−1

1

−1

39. Graph (b).

40. Graph (d).

41. Graph (c).

42. Graph (a).

43.

(x,y)dy

dxx y= + − 1 Δx Δ = Δy

dy

dxx ( , )x x y y+ Δ + Δ

(1,1) 1.0 0.1 0.1 (1.1, 1.1)

(1.1, 1.1) 1.2 0.1 0.12 (1.2, 1.22)

(1.2, 1.22) 1.42 0.1 0.142 (1.3, 1.362)

y = 1.362

44.

(x,y)dy

dxx y= − Δx Δ = Δy

dy

dxx ( , )x x y y+ Δ + Δ

(1,2) –1.0 –0.1 0.1 (0.9, 2.1)

(0.9, 2.1) –1.2 –0.1 0.12 (0.8, 2.22)

(0.8, 2.22) –1.42 –0.1 0.142 (0.7, 2.362)

y = 2.362

45. We seek the graph of a function whose derivative is sin

.x

x

Graph (b) is increasing on [ , ],−π π where sin x

x is positive,

and oscillates slightly outside of this interval. This is thecorrect choice, and this can be verified by graphing NINT

sin, , , .

x

xx x0

⎛⎝⎜

⎞⎠⎟

46. We seek the graph of a function whose derivative is e x− 2.

Since e x− 2 > 0 for all x, the desired graph is increasing for

all x. Thus, the only possibility is graph (d), and we may

verify that this is correct by graphing NINT ( , , , ).e x xx− 2 0


47. (iv) The given graph looks the graph of y x= 2, which

satisfies dy

dxx= 2 and y(1) = 1.

48. Yes, y = x is a solution.

49. (a) dv

dtt= +2 6

dv t dt

v t t Cv

= +

= + +=

∫∫ ( )2 6

2 34

2

Initial condition: wwhen tC

C

v t t

== +== + +

04 04

2 3 42

(b) v t dt t t dt( ) ( )0

1 2

0

12 3 4∫ ∫= + +

= + +⎡⎣

⎤⎦

= −=

t t t2 3

0

14

6 06

The particle moves 6 m.

50.

[–10, 10] by [–10, 10]

51. (a) Half-life = ln 2

k

2 645

2

2

2 6450 262059

.ln

ln

..

=

= ≈

k

k

(b) Mean life years= ≈13 81593

k.

52. T T T T es skt− = − −( )0

T eT t

kt− = −= =

−40 220 40180 15

( )Use the fact that and tto find k.

180 40 220 40180

140

9

71

15

15

15

− = −

= =

=

−( ) ( )( )e

e

k

k

k

lln

( )

(

(( / ) ln ( / ))

9

740 220 40

70 40

1 15 9 7T e t− = −− =

−

2220 40 1 15 9 7

1 15 9 7

− −) (( / ) ln ( / ))

(( / ) ln ( / ))

e

e

t

t == =

⎛⎝⎜

⎞⎠⎟

=

= ≈

180

306

1

15

9

76

15 6

9 71

ln ln

ln

ln ( / )

t

t 007 min

It took a total of about 107 minutes to cool from 220°F to70°F. Therefore, the time to cool from 180°F to 70°F wasabout 92 minutes.

53. T T T T es skt− = − −( )0

We have the system:

39 46

33 46

10

20

− = −− = −

⎧⎨⎪

⎩⎪

−

−T T e

T T es s

k

s sk

( )

( )

Thus, and

Si

39

4610

33

4610 20−

−=

−−

=− −T

T

T

Tes

s

k s

s

k

nnce this means:

39

( ) ,e e

T

T

k k

s

s

− −=

−−

⎛

⎝⎜⎞

10 2 20

46 ⎠⎠⎟=

−−

− = − −−

2

2

33

46

39 33 46

1521

T

T

T T T

s

s

s s s( ) ( )( )

778 1518 793

2 2T T T TT

s s s s

s

+ = − += −

The refrigerator ttemperature was − °3 C.

54. Use the method of Example 3 in Section 6.4.

e

kt

tk

kt− =− =

= − = −

0 9950 9951

0 9955700

2

.ln .

ln .ln

lnn . .0 995 41 2≈

The painting is about 41.2 years old.

55. Use the method of Example 3 in Section 6.4.Since 90% of the carbon-14 has decayed, 10% remains.

ekt

tk

kt− =− =

= − = − ≈

0 10 11

0 15700

20 1 1

.ln .

ln .ln

ln . 88 935,

The charcoal sample is about 18.935 years old.

56. Use yearst = − =1988 1924 64 .

250 7500

303030 30

640 05

e

ert

rt

rt

rt

===

= = ≈

lnln ln

. 33

The rate of appreciation is about 0.053, or 5.3%.

57. Using the Law of Exponential Change in Section 6.4 withappropriate changes of variables, the solution to thedifferential equation is L( ) ,x L e kx= −

0 where L L0 0= ( ) isthe surface intensity. We know 0 5 18. ,= −e k so

k =−

ln .0 5

18and our equation becomes

L x L e Lxx

( ) .(ln . )( / )/

= = ⎛⎝⎜

⎞⎠⎟0

0 5 180

181

2We now find the depth

where the intensity is one-tenth of the surface value.


57. Continued

0 11

2

0 118

1

218 0

18

.

ln . ln

ln .

/

= ⎛⎝⎜

⎞⎠⎟

= ⎛⎝⎜

⎞⎠⎟

=

x

x

x11

0 559 8

ln ..≈ ft

You can work without artificial light to a depth of about59.8 feet.

58. (a) dy

dt

kA

Vc y= −( )

dy

c y

kA

Vdt

c ykA

Vt C

c ykA

Vt C

c y

−=

− − = +

− = − −

− =

∫∫ln

ln

ee

c y e

y c e

y

kA V t C

kA V C

kA V t C

− −

− −

− −− = ±

= ±

( / )

( / )

( / )

== + −c De kA V t( / )

Initial condition y y t= =0 0when

y c D

y c D0

0

= +− =

Solution: y c y c e kA V t= + − −( ) ( / )0

(b) lim ( ) lim[ ( ) ]( / )

t t→∞ →∞

−= + − =y t c y c e ckA V t0

59. (a) p te e et t

( ). .

=+

=+− −

150

1

150

14 3 4 3

This is pM

AeM A e k

kt=

+= = =−1

150 14 3where and, , ..

Therefore, it is a solution of the logistic differentialequation.

dp

dt

k

MP M P

dp

dtP P= − = −( ), ( ).or

1

150150 The

carrying capacity is 150.

(b) Pe

( ).

0150

12

4 3=

+≈

Initially there were 2 infected students.

(c) 150

1125

4 3+=−e t.

6

51

1

55 4 3

4 3 5 5 9

4 3

4 3

= +

=

− = −= + ≈

−

−

e

e

tt

t

t

.

.

ln .. ln . days.

It took about 6 days.

60. Use the Fundamental Theorem of Calculus.

yd

dxt dt

d

dxx x

x

x′ = ⎛

⎝⎜⎞⎠⎟ + + +( )

= +

∫ sin

(sin )

2

0

3

2

2

(( )

(sin )

(cos )( )

3 1

3 1

2 6

2

2 2

2

x

yd

dxx x

x x x

+

= + +

= +=

″

22 62x x xcos( ) +Thus, the differential equation is satisfied. Verify the initialconditions:

y

y t dt

′( ) (sin ) ( )

( ) sin( )

0 0 3 0 1 1

0 0 0

2 2

2 3

= + + =

= + + ++ =∫ 2 20

0

61.dP

dtP

P= −⎛⎝⎜

⎞⎠⎟

0 002 1800

.

dP

dtP

P

P PdP

= −⎛⎝⎜

⎞⎠⎟

−=

0 002800

800800

8000 00

.

( ). 22

800

8000 002

dt

P P

P PdP dt

( )

( ).

− +−

=

1 1

8000 002

p PdP dt+

−⎛⎝⎜

⎞⎠⎟

= ∫∫ .

ln ln .

ln .

ln

P P t C

P

Pt C

− − = +

−= +

−

800 0 002

8000 002

800 PP

Pt C

P

Pe

P

Pe e

t C

C

= − −

− =

− = ±

− −

− −

0 002

800

800

0 002

.

.

00 002

0 002

0 002

8001

800

1

.

.

.

t

t

t

PAe

PAe

− =

=+

−

−

Initial condition: P( )0 50=

50800

11 16

15

0=

++ =

=

AeAA

Solution: Pe t

=+ −

800

1 15 0 002.

62. Method 1–Compare graph of y x12= ln x with

yx x x

2

3 3

3 9= −

⎛

⎝⎜

⎞

⎠⎟NDER

ln. The graphs should be the same.

Method 2–Compare graph of y x x12= NINT( ln )

with yx x x

2

3 3

3 9= −ln

. The graphs should be the same or

differ only by a vertical translation.


63. (a) 20 000 10 000 1 063, , ( . )= t

2 1 0632 1 063

2

1 06311 345

==

= ≈

.ln ln .

ln

ln ..

t

t

t


(b) 20 000 10 000 0 063, , .= e t

22 0 063

2

0 06311 002

0 063==

= ≈

et

t

t.

ln .ln

..


64. (a) f xd

dxu t dt u x

x′( ) ( ) ( )= =∫0

g xd

dxu t dt u x

x′( ) ( ) ( )= =∫3

(b) C f x g x= −( ) ( )

= −

=

∫∫∫

u t dt u t dt

u t dt + u t dt

xx

x

x

( ) ( )

( ) ( )

30

0

3

∫∫∫= u t dt( )

0

3

65. (a) The regression equation is ye t

=+ −

272286 4

1 302 69 0 2095

.

..

.

The graph is shown below.

(b) ye

( ).

.,

. ( )∞ =

+=− ∞

272286 4

1 302 69272 286

0 2095peoople.

(c) dp

dtP P= × −−7 694 10 272286 47. ( . )

(d) The carrying capacity drops to 267,312.6, which isbelow the actual 2003 population. The logisticregression is strongly affected by points at the extremesof the data, especially when there are so few data pointsbeing used. While the fit may be more dramatic for asmall data set, the equation is not as reliable.

66. (a) T t= 79 961 0 9273. ( . )

[–1, 33] by [–5, 90]

(b) Solving T t( ) = 40 graphically, we obtain t ≈ 9 2. sec.The temperature will reach 40° after about 9.2 seconds.

(c) When the probe was removed, the temperature wasabout T ( ) . .0 79 76≈ °C

67. (a) 1

2 of the town has heard the rumor when it is spreading

the fastest.

(b) dy

y ydt

1 2 1. ( )−=∫

A

y

B

yA y y B1 2 1

1

1 1 2 1.( ) .

+−

=

− + =y By A

y ydy t

= == =

+−

⎛⎝⎜

⎞⎠⎟

= +

1 0 830 11

1 2

0 83

1

, .,

.

.CC

yt C

∫− = − −ln.

.1

1 21 2

y

1

1 21

1

01

1

1 2

1 2

1 2

− =

=+

=+

− −

−

−

y

ye e

yAe

yAe

t C

t

.

( )

.

.

. (( )

.

0

1 2

91

1 9

A

ye t

=

=+ −

(c) 1

2

1

1 9 1 2=

+ −e t.

Solve for t to obtain

t = ≈5 3

31 83

ln. .days

68. (a) dp

dtk P= −( ).600 Separate the variables to obtain

dP

Pkdt

dP

Pkdt

P kt C

P

600

600600

6001

−=

−= −

− = − +− =

ln

CCe

Ce C

P e

P t

kt

kt

−

−− = ⇒ = −− = −

=

200 600 400

600 400

0

( ) 6600 400− −e kt

(b) 500 600 400 2= − −e k i

1 42 0 693

2/ln .

== ≈

−ek

k

(c) lim( ).

t

te→∞

−− =600 400 6000 693

Section 7.1 313

69. (a) Separate the variables to obtaindv

vdt

v t C

v Ce

Ce

t

+= −

+ = − ++ =

− + =

−

172

17 2

17

47 17

12

ln

00

2

2

30

17 30

30 17

⇒ = −+ = −

= − −

−

−

C

v e

v e

t

t

(b) lim( )t

te→∞

−− − = −30 17 172 feet per second

(c) − = − −−20 30 172e t

t = ≈ln.

10

21 151 seconds

Chapter 7Applications of Definite Integrals

Section 7.1 Integral as Net Change(pp. 378–389)

Exploration 1 Revisiting Example 2

1. s t tt

dtt

tC( )

( )= −

+

⎛

⎝⎜

⎞

⎠⎟ = +

++∫ 2

2

38

1 3

8

1

s C C

s tt

t

( )

( ) .

00

3

8

0 19 1

3

8

11

3

3

= ++

+ = ⇒ =

= ++

+Thus,

2. s( ) .11

3

8

1 11

16

3

3

= ++

+ = This is the same as the answer we

found in Example 2a.

3. s( ) .55

3

8

5 11 44

3

= ++

+ = This is the same answer we found

in Example 2b.

Quick Review 7.1

1. On the interval, sin 2x = 0 when x = − π π2

02

, , .or Test one

point on each subinterval: for x x= − =3

22 1

π, sin ; for

x x x x= − = − = =π π4

2 14

2 1, sin ; , sin ;for and for

x x= − = −3

42 1

π, sin . The function changes sign at

− π π2

02

, , .and The graph is

– –++

–3 20 �2

– �2

f (x)x

2. x2 – 3x + 2 = (x – 1)(x – 2) = 0 when x = 1 or 2. Test one

point on each subinterval: for x = 0, x2 – 3x + 2 = 2; for

x x x= − + = −3

23 2

1

42, ; and for x = 3, x2 – 3x + 2 = 2.

The function changes sign at 1 and 2. The graph is

+ +–

–2 1 2 4

f (x)x

3. x2 – 2x + 3 = 0 has no real solutions, since b2 – 4ac =

( )−2 2 – 4(1)(3) = –8 < 0. The function is always positive.

The graph is

+

–4 2

f (x)x

4. 2 3x – 3 2x + 1 = ( )x −1 2 (2x + 1) = 0 when x = − 1

2 or 1.

Test one point on each subinterval: for x = –1,

2 3x – 3 2x + 1 = –4; for x = 0, 2 3x – 3 x2 + 1 = 1; and

x = − 3

2, 2 3 1 13 2x x− + = . The function changes sign at

− 1

2. The graph is

+ +–

–2 – 2112

f (x)x

5. On the interval, x cos 2x = 0 when x = 04

3

4

5

4, , , .π π π

or

Test one point on each subinterval: for x = π8

,

x x x x x xcos , , cos ,22

16 22

22= = = − =π π π πfor for

x x x x xcos ; , cos . .2 4 2 0 58= = ≈−π and for The function

changes sign at π π π4

3

4

5

4, , .and The graph is

– –++

0 4�4

3�4

5�4

f (x)x

6. xe x− = 0 when x = 0. On the rest of the interval, xe x− isalways positive.

7. x

xx

2 10 0

+= =when . Test one point on each subinterval:

for xx

xx

x

x= −

+= − =

+=1

1

1

21

1

1

22 2, ; , .for The function

changes sign at 0. The graph is

+–

–5 0 30

f (x)x

Chapter 6 Differential Equations and Mathematical Modeling

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Transcript of Chapter 6 Differential Equations and Mathematical Modeling