Thomas Calculus 13th [Solutions]

404 Chapter 5 Integration
(c)
2
c
f ( x) g( x)
c x AL
[ f ( x) g( x)]
dx
cc 2
( )
c
c x dx 3 c
x
3/2
3/2
cx 2 c c
3
c
3
4 3/2
c . Again, the area of the whole shaded region can be found by setting c 4 A 32 . From the
3
3
3/2 2/3
condition A 2 A L , we get 4 c 32 c 4 as in part (b).
3 3
2
104. (a) Limits of integration: y 3 x and y 1
2 2
3 x 1 x 4 a 2 and b 2;
2 2
f ( x) g( x) (3 x ) ( 1) 4 x
2 3 2
2
A
2 (4 x ) dx 4 x x
3
2
8 8 8 8 16 16 32
3 3 3 3
2
(b) Limits of integration: let x 0 in y 3 x
y 3; f ( y) g( y) 3 y 3 y
1/2
2(3 y)
4
3
3 1/2
A 2 (3 y) dy 2
1
3/2
0 (3 1) 4 (8) 32
3 3
3 1/2
2(3 y)
(3 y) ( 1) dy ( 2)
1
3
3/2 3
1
105. Limits of integration: y 1 x and y 2
x
1 2
2
x , x 0 x x 2 x (2 x)
x
x 4 4x 2
x
2
x 5x
4 0
( x 4)( x 1) 0 x 1, 4 (but x 4 does not
satisfy the equation); y 2 and y x 2 x
x 4 x 4
8 x x 64
3
x x 4. Therefore,
1/2
AREA A1 A2 : f1( x) g1( x) 1 x x
4
A1 1 1/2
0 1 x
3/2
2
x dx x 2 x x
1 2 1 37
1/2
1 0 ; f
4
3 8 0
3 8 24 2 ( x) g2 ( x) 2x
x
4
A2 4 1/2
1 2 2 4
x x
1/2
dx 4x x 4 2 16 4 1 4 15 17 ; Therefore,
4
8 1
8 8 8 8
AREA A 37 17 37 51 88 11
1 A2 24 8 24 24 3
2
106. Limits of integration: ( y 1) 3 y
2 2
y 2y 1 3 y y y 2 0
( y 2)( y 1) 0 y 2 since y 0; also,
2 2
2 y 3 y 4y 9 6y y y 10y
9 0
( y 9)( y 1) 0 y 1 since y 9 does not satisfy
the equation;
AREA A1 A2
1/2
1 1/2 2
f1( y) g1( y) 2 y 0 2y
4
1 2 0
2 y
A y dy
3 3
;
0
2
2 2
f2 ( y) g2
( y) (3 y) ( y 1) A 2 [3 ( 1) ]
1 y y dy 2 3
2
3 y 1 y 1 ( y 1)
2 3 1
6 2 1 1
3 3 2
0 1 1 1 7 . Therefore, A 4 7 15 5
3 2 6
1 A2
3 6 6 2
3/2 1
Copyright
2014 Pearson Education, Inc.