Thomas Calculus 13th [Solutions]

428 Chapter 5 Integration
36.
y
3
x
ln t dt
x
dy 2/3 1/2
3
ln ln
ln d ln d ln 1 ln 1
x x
3 2
3 3 3 3
x x x x x x x x x
dx dx dx 3 2
3 x 2 x
37.
38.
39.
40.
41.
ln x ln
sin e t dt y (sin e x ) d (ln x)
sin x
0
dx
x
2x
3 2x 2x 4 x 4 x 2x 4 x
y 4 x ln t dt y (ln e ) d ( e ) ln e d e (2 x)(2 e ) 4 x e d 4 x
e
dx dx dx
2x 4 x 2 2x 4 x
4xe 4 xe 4xe 8e
x
f ( )
x 3
(5 )
x
d ( 3) (5 ) dx
dx
dx
( x 3)(2 x ) x (5 x )
6 x
2
x 5x 2
x 6 6 x . Thus f ( x) 0 6 6x 0 x 1. Also, f ( x) 6 0 x 1 gives
a maximum.
x
( x ) x
x x x 2
ln x x ln x and ln( x ) x ln x x ln x ; then
x 2
ln x = 0. ln x = 0 x = 1; x x x ln x 2ln x x 2. Therefore,
A 1
2 e
e 2log 2 x 2
e
ln x (ln x)
1 ;
1 x
ln 2 1 x
ln 2 ln 2
1
A2 2 e
e 2log 4 x 2
e
ln (ln x)
dx x dx 1
1 4 ln 4 1 x 2ln 2 2ln 2
1
A1 : A2
2 :1
x 2 x 2 x 2
x ln x x ln x ( x x ) ln x 0 x x or
x
( x ) x x
x ( x ) when x = 2 or x = 1.
df
x
42. (a) 2ln e x
e 2x
dx x
e
1
(b) f (0) 2ln t dt
1 t
0
df
2
(c) 2 x f ( x) x C;
f(0) = 0 C = 0
dx
2
f ( x)
x the graph of f(x) is a parabola.
43.
g( x) g( x)
0
f ( x) e f ( x) e g ( x ), where g ( x) x f (2) e 2 2
4
1 x
1 16 17
1 1 1 1
44. The area of the blue shaded region is sin x dx sin y dy , which is the same as the area of the region to
0 0
the left of the curve y = sin x (and part of the rectangle formed by the coordinate axes and dashed lines y = 1,
x 2 . The area of the rectangle is 1 1 /2
sin y dy sin x dx , so we have
2 0 0
1 1 /2 /2 1 1
sin x dx sin x dx sin x dx sin x dx.
2 0 0 0 2 0
45. (a) slope of L 1 ln ln
3 slope of L2 slope of L
b a
1
1
b b a a
(b) area of small (shaded) rectangle < area under curve < area of large rectangle
1 ( b a) b 1 dx 1 ( b a)
1 ln b ln a 1
b a x a b b a a
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