401
Answers
601–700
Answers and Explanations
To determine which curve has the larger x values for y on the interval (0, 5), take a
point in the interval and substitute it into each equation. So if y = 1, then x = 1
2
– 1 = 0
and x = 4(1) = 4. Therefore, the integral to find the area of the region is
4
5
5
2
3
5 5
2
5
2
0
5
2
0
5
2
3
0
5
2
3
y y
y dy
y y dy
y
y
−
−
(
)
=
−
=
−
=
−
∫
∫
( )
( )
3 3
125
6
=
656.
18
Notice that you can easily solve the two equations for x, so integrating with respect to
y makes sense. If you were to solve the second equation for y in order to integrate with
respect to x, you’d have to complete the square to solve for y, which would needlessly
complicate the problem.
Begin by solving both equations for x to get x
y
=
−
2
6
2
and x = y + 1. Then set the
expressions equal to each other and solve for y to find the points of intersection:
y
y
y
y
y
y
y
y
y
+ =
−
+ =
−
− − =
−
+ =
= −
1
6
2
2
2
6
2
8 0
4
2 0
4 2
2
2
2
(
)(
)
,
To find which curve has the larger x values for y on the interval (–2, 4), take a point
in the interval and substitute it into each equation. So if y = 0, then x = − = −
0 6
2
3 and
x = 0 + 1 = 1. Therefore, the integral to find the area of the region is
(
)
(
)
y
y
dy
y
y
d
+ −
−
=
+ −
−
−
−
∫
∫
1
6
2
1
2
3
2
2
4
2
2
4
y y
y
y dy
y
y
y
=
+ −
(
)
=
+
−
=
+ −
(
) −
−
−
∫
4 1
2
2
4
1
6
16
2
16 1
6
64
2
2
2
4
2
3
2
4
( )
− − +
(
)
=
8 1
6
8
18
( )
Answers
601–700
Answers and Explanations
To determine which curve has the larger x values for y on the interval (0, 5), take a
point in the interval and substitute it into each equation. So if y = 1, then x = 1
2
– 1 = 0
and x = 4(1) = 4. Therefore, the integral to find the area of the region is
4
5
5
2
3
5 5
2
5
2
0
5
2
0
5
2
3
0
5
2
3
y y
y dy
y y dy
y
y
−
−
(
)
=
−
=
−
=
−
∫
∫
( )
( )
3 3
125
6
=
656.
18
Notice that you can easily solve the two equations for x, so integrating with respect to
y makes sense. If you were to solve the second equation for y in order to integrate with
respect to x, you’d have to complete the square to solve for y, which would needlessly
complicate the problem.
Begin by solving both equations for x to get x
y
=
−
2
6
2
and x = y + 1. Then set the
expressions equal to each other and solve for y to find the points of intersection:
y
y
y
y
y
y
y
y
y
+ =
−
+ =
−
− − =
−
+ =
= −
1
6
2
2
2
6
2
8 0
4
2 0
4 2
2
2
2
(
)(
)
,
To find which curve has the larger x values for y on the interval (–2, 4), take a point
in the interval and substitute it into each equation. So if y = 0, then x = − = −
0 6
2
3 and
x = 0 + 1 = 1. Therefore, the integral to find the area of the region is
(
)
(
)
y
y
dy
y
y
d
+ −
−
=
+ −
−
−
−
∫
∫
1
6
2
1
2
3
2
2
4
2
2
4
y y
y
y dy
y
y
y
=
+ −
(
)
=
+
−
=
+ −
(
) −
−
−
∫
4 1
2
2
4
1
6
16
2
16 1
6
64
2
2
2
4
2
3
2
4
( )
− − +
(
)
=
8 1
6
8
18
( )