Thomas Calculus 13th [Solutions]

80. Let
2
u t du dt du dt
Section 5.6 Substitution and Area Between Curves 387
2
v cos t dt (cos u)( du) sin u C1 sin( t) C1;
at t 0 and v 8 we have 8 (0) C1 C1
8 v ds sin( t) 8 s ( sin( t) 8) dt
dt
sin u du 8t C2 cos( t) 8 t C 2;
at t 0 and s 0 we have 0 1 C2 C2
1
s 8t cos ( t) 1 s(1) 8 cos 1 10 m
81. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on the
left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover,
2 2
2
sin x C1 1 cos x C1 C2 1 C 1;
also cos x C cos 2x
1 1 1
2 C
2 2 2 C3 C2 C
2 1 .
2
1/60
82. (a) 1
1/60
V 1
max
1
max sin 120 t dt 60 V
0 0
max 120
cos(120 t) V
2
[cos 2 cos0]
60
0
Vmax
[1 1] 0
2
(b) Vmax 2Vrms
2(240) 339 volts
2
1/60 2 2 2 1/60
max
1/60
(c)
1 cos 240 V
Vmax sin 120 t dt V t
0
max dt (1 cos 240 t)
dt
0 2 2 0
V 1/60
max 1 max 1 1 1
max
t sin 240 t
V sin(4 ) 0 sin(0)
V
2 240 0 2 60 240 240 120
2 2 2
5.6 SUBSTITUTION AND AREA BETWEEN CURVES
1. (a) Let u y 1 du dy;
y 0 u 1, y 3 u 4
3 4 1/2 3/2
4
y 1 dy u du 2 u 2 (4) 3/2 2 (1) 3/2 2 (8) 2 (1) 14
0 1 3 1 3 3 3 3 3
(b) Use the same substitution for u as in part (a); y 1 u 0, y 0 u 1
0 1 1/2 3/2
1
y 1 dy u du 2 u 2 (1) 3/2
0 2
1 0 3 0 3 3
2
2. (a) Let u 1 r du 2 r dr 1 du r dr;
r 0 u 1, r 1 u 0
2
1 2 0
3/2
0
1 1 1 3/2
r 1 r dr u du u 0 (1) 1
0 1 2 3 1 3 3
(b) Use the same substitution for u as in part (a); r 1 u 0, r 1 u 0
1 2 0
r 1
1 1 r dr u du
0 2
0
2
3. (a) Let u tan x du sec x dx; x 0 u 0, x
4
u 1
/4 2 1
2 1
2
tan xsec x dx u du u 1 0 1
0 0 2
0
2 2
(b) Use the same substitution as in part (a); x
4
u 1, x 0 u 0
0 0
2 0
2
tan xsec x dx u du u 0 1 1
/4 1 2
1
2 2
4. (a) Let u cos x du sin x dx du sin x dx; x 0 u 1, x u 1
2 1 2 3 1 3 3
0 3cos x sin x dx
1
3 u du [ u ] 1 ( 1) ( (1) ) 2
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