Thomas Calculus 13th [Solutions]

Chapter 5 Practice Exercises 421
119. We want to evaluate
1
365
1
365
f ( x)
dx
2 37
365
37sin ( x 101) 25 dx
2 25
365
sin ( x 101) dx dx
365 0 0
365 0
365 365 0 365 365 0
Notice that the period of y sin 2 ( 101)
365
x is 2 365 and that we are integrating this function over an
2
365
365 365
interval of length 365. Thus the value of 37 sin 2 ( x 101) dx 25 dx is 37 0 25 365 25.
365 0 365 365 0 365 365
675 5 2
120. 1 (8.27 10 (26 1.87 ))
675 20 20 T T dT 2 3 675
1 8.27T
26T 1.87T
655 5 5
2 10 310 20
2 3 2 3
1
26(675) 1.87(675) 26(20) 1.87(20)
8.27(675) 8.27(20)
655 5 5 5 5
2 10 310 2 10 310
1
655 (3724.44 165.40) 5.43 the average value of C v on [20, 675]. To find the temperature T at
5 2
2
which C v 5.43, solve 5.43 8.27 10 (26T 1.87 T ) for T. We obtain 1.87T
26T
284000 0
2
26 (26) 4(1.87)( 284000) 26 2124996
T . So T 382.82 or T 396.72. Only T 396.72 lies in the
2(1.87) 3.74
interval [20, 675], so T 396.72 C.
121.
dy
dx
dy
122.
dx
3
2 cos
x
3 2 2 3 2
2 cos (7 x ). d (7 x ) 14x 2 cos (7 x )
dx
dy
123. d
x
6 dt 6
dx dx 1
4 4
3 t 3 x
dy d
2
1 d
sec x
1 1 d sec x tan x
dx dx sec x t 1 dx 2 t 1 sec x 1 dx
1 sec x
124. dt dt (sec x)
2 2 2 2
125.
2
y 0 ln
2 2
cost x cos t dy cos(ln x ) d 2 2 cos(ln x )
2
(ln )
ln x
e dt 0
e dt dx e dx x x
e
126.
x
e 2 dy 2 x x
x
2 x
y ln( t 1) dt ln e 1 d e e ln e 1
1 dx dx 2 x
127.
y
1
sin x
dt dy 1 d 1
(sin x)
1 1
0 1 2t dx
1 2(sin x) dx
1 2(sin x) 1 x
2 1 2 1 2 2
128.
1
1 tan
1
x
/4 t tan x t dy tan x
y e dt e dt e d (tan 1 x)
e
tan x
/4 dx
dx
1 x
1 2
129. Yes. The function f, being differentiable on [ a, b ], is then continuous on [ a, b ]. The Fundamental Theorem of
Calculus says that every continuous function on [ a, b ] is the derivative of a function on [ a, b].
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