Thomas Calculus 13th [Solutions]

Section 5.6 Substitution and Area Between Curves 405
2
107. Area between parabola and
a y a : A 2
2 2
0 ( a x ) dx 2 2 3
a
3
a x 1 x 3 4
3
3 2 a a
0
3 0 a
3
;
2 3
Area of triangle AOC: 1 (2 )( ) ;
2 a a a limit of ratio 3
lim a 3 which is independent of a.
4
3 4
a 0 a
b
b
108. A 2 f ( x)
dx ( ) 2
a
a f x dx b
a f ( x)
dx b
a f ( x)
dx b
a f ( x) dx 4
109. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the regions
upper and lower bounding curves at x 0. The area of the shaded region is actually
0 1 0 1
A x ( x) dx x ( x) dx 2x dx 2x dx 2.
1 0 1 0
110. It is sometimes true. It is true if f ( x) g( x ) for all x between a and b. Otherwise it is false. If the graph of f lies
below the graph of g for a portion of the interval of integration, the integral over that portion will be negative
and the integral over [ a, b ] will be less than the area between the curves (see Exercise 71).
111. Let u 2x du 2 dx 1 du dx; x 1 u 2, x 3 u 6
2
3 sin 2 x 6
dx sin u 1
6
sin
6
du u du [ F( u)] 1 x 2 2 2
2 F(6) F(2)
u
u
2
112. Let u 1 x du dx du dx; x 0 u 1, x 1 u 0
1
0 0 1 1
f (1 x)
dx f ( u)( du) f ( u) du f ( u) du f ( x)
dx
0
1 1 0 0
3
113. (a) Let u x du dx ; x 1 u 1, x 0 u 0
f odd f ( x) f ( x ). Then
0
1 f ( x)
dx 0
f ( u)( du)
1
1
f ( u) du 3
0
(b) Let u x du dx ; x 1 u 1, x 0 u 0
0
f even f ( x) f ( x ). Then
1 f ( x)
dx 0
f ( u)( du)
1
0
1
f ( u) ( du)
0
f ( u)
du
1
0
f ( u)
du
1
1
f ( u) du 3
0
0
114. (a) Consider f ( x)
dx when f is odd. Let u x du dx du dx and x a u a and
a
0
x 0 u 0. Thus ( )
a f x dx 0
0
f ( u)
du ( )
a
a f u du a a f ( u ) du f ( x ) dx . Thus
0
0
( )
aa f x dx 0
( )
a f x dx a a a f ( x ) dx f ( x ) dx f ( x ) dx 0.
0
0
0
(b)
/2 /2
sin
/2 x dx [ cos x ] /2 cos cos 0 0 0.
2 2
115. Let u a x du dx ; x 0 u a, x a u 0
a f ( x)
dx 0 f ( a u)
I
0 f ( x) f ( a x)
( ) ( ) ( du
a f a u f u
) a f ( a u)
du a f ( a x)
dx
0 f ( u) f ( a u)
0 f ( x) f ( a x)
a f ( x) dx a f ( a x)
dx a f ( x) f ( a x)
a a
I I
dx dx [ x] 0 f ( x) f ( a x) 0 f ( x) f ( a x)
0 ( ) ( ) 0
0 a 0 a.
f x f a x
Therefore, 2 I a I a .
2
Copyright
2014 Pearson Education, Inc.