chapter 6 differential equations and mathematical modeling · chapter 6 differential equations and...
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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
Δx (x + Δx, y + Δy)
(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
Δx (x + Δx, y + Δy)
(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
Δx (x + Δx, y + Δy)
(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
Δx (x + Δx, y + Δy)
(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
Δx (x + Δx, y + Δy)
(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
Δx (x + Δx, y + Δy)
(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
Δx (x + Δx, y + Δy)
(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
Δx (x + Δx, y + Δy)
(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
Section 6.2 Exercises
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
+ −
=
Section 6.3 Exercises
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
Section 6.4 Exercises
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
The animal died about 14,101 years before A.D. 2000, in12,101 B.C.
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
The animal died about 15,070 years before A.D. 2000, in13,070 B.C.
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.
25. (a) 1200 individuals.
(b) 600 individuals.
(c) dP
dt
( ). ( )( )
6000 0002 600 1200 600
72
= −
= individualls per year.
26. (a) 5000 individuals.
(b) 2500 individuals.
(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
−= −
= − += − − +
∫ ∫
Chapter 6 Review 305
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
306 Chapter 6 Review
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:
Chapter 6 Review 307
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
308 Chapter 6 Review
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
Chapter 6 Review 309
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
310 Chapter 6 Review
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.
Chapter 6 Review 311
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.
312 Chapter 6 Review
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
It will take about 11.3 years.
(b) 20 000 10 000 0 063, , .= e t
22 0 063
2
0 06311 002
0 063==
= ≈
et
t
t.
ln .ln
..
It will take about 11.0 years.
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