Part II: The Answers
392
Answers
601–700
1
2
2
2
3
1
2
1 2
2
2 1 2
3
1
2
1
1 2
2
3
1
1 2
2
3
− −
(
)
=
−
−
= −
( ) − ( ) − −
−
−
∫
y
y dy
y
y
y
− −
−
−
−
=
( )
( )
1
2
2 1
3
9
8
2
3
The following figure shows the region bounded by the given curves:
644.
36
Begin by finding the points of intersection by setting the functions equal to each other
and solving for x:
2
8
2
8 0
4
2 0
4 2
2
2
x
x
x
x
x
x
x
= −
+
− =
+
− =
= −
(
)(
)
,
To determine which function is larger on the interval (–4, 2), take a point inside the interval and substitute it into each function. If you let x = 0, then y = 2(0) = 0 and y = 8 – 0
2
= 8.
Therefore, 8 – x
2
> 2x on (–4, 2), so the integral for the area of the bounded region is
392
Answers
601–700
1
2
2
2
3
1
2
1 2
2
2 1 2
3
1
2
1
1 2
2
3
1
1 2
2
3
− −
(
)
=
−
−
= −
( ) − ( ) − −
−
−
∫
y
y dy
y
y
y
− −
−
−
−
=
( )
( )
1
2
2 1
3
9
8
2
3
The following figure shows the region bounded by the given curves:
644.
36
Begin by finding the points of intersection by setting the functions equal to each other
and solving for x:
2
8
2
8 0
4
2 0
4 2
2
2
x
x
x
x
x
x
x
= −
+
− =
+
− =
= −
(
)(
)
,
To determine which function is larger on the interval (–4, 2), take a point inside the interval and substitute it into each function. If you let x = 0, then y = 2(0) = 0 and y = 8 – 0
2
= 8.
Therefore, 8 – x
2
> 2x on (–4, 2), so the integral for the area of the bounded region is