Part II: The Answers
400
Answers
601–700
The following figure shows the region bounded by the functions.
654.
10 2
3
5 2
−
Begin by finding the points of intersection of the two curves by setting the expressions
equal to each other and solving for y:
y
y
y
y
y
y
y y
y
2
4
4
3
0
1 0
0 1
=
=
− =
−
(
) =
= ,
To determine which expression has larger x values for y on the interval (0, 1), pick a point
in the interval and substitute it into each equation. So if y = 1
4
, then x = ( ) =
1
4
1
16
2
and
x =
=
1
4
1
2
. For y on the interval (0, 1), you have y y
>
2
. Similarly, you can show that
y
y
2
>
for y on the interval (1, 2). Therefore, the integrals to find the area of the region are
y y dy
y
y dy
y
y
y
y
−
(
) +
−
(
)
=
−

 

  +
−

 

 
∫
∫
2
0
1
2
1
2
3 2
3
0
1
3
3 2
1
2
3
3
3
2
3
2 2
3 2
5 2
5 2
2
3
1
3
8
3
2 2
3
1
3
2
3
10
3
2
3
10 2
3
= − +
−

 

  − −
( )






=
−
= −
( )
655.
125
6
Begin by finding the points of intersection of the two curves by setting the expressions
equal to each other:
y
y
y
y
y
y y
y
2
2
4
5
0
5 0
0 5
− =
−
=
− =
=
(
)
,