393
Answers
601–700
Answers and Explanations
8
2
2
8
3
8
2
4
2
2
4
2
3
2
4
2
−
(
) −
=
− − +
(
)
= − − +
−
−
−
∫
∫
x
x dx
x
x
dx
x
x
x
( )
= =
−
− +
− −
−
− −
+ −
= − − + −
+
( )
( )
( ) ( )
( )
2
3
2 8 2
4
3
4
8 4
8
3
4 16 64
3
16
3
2
3
2
+ +
= −
=
32
60 24
36
The following figure shows the region bounded by the given curves:
645.
16 2
3
In this example, integrating with respect to y makes sense. You could integrate with
respect to x, but you’d have to solve both x = 2 – y
2
and x = y
2
– 2 for y and then use two
integrals to compute the area, because the “top function” and the “bottom function”
aren’t the same for the entire region. (You could actually reduce the region to a single
integral by using symmetry, but that isn’t possible in general.)
Answers
601–700
Answers and Explanations
8
2
2
8
3
8
2
4
2
2
4
2
3
2
4
2
−
(
) −
=
− − +
(
)
= − − +
−
−
−
∫
∫
x
x dx
x
x
dx
x
x
x
( )
= =
−
− +
− −
−
− −
+ −
= − − + −
+
( )
( )
( ) ( )
( )
2
3
2 8 2
4
3
4
8 4
8
3
4 16 64
3
16
3
2
3
2
+ +
= −
=
32
60 24
36
The following figure shows the region bounded by the given curves:
645.
16 2
3
In this example, integrating with respect to y makes sense. You could integrate with
respect to x, but you’d have to solve both x = 2 – y
2
and x = y
2
– 2 for y and then use two
integrals to compute the area, because the “top function” and the “bottom function”
aren’t the same for the entire region. (You could actually reduce the region to a single
integral by using symmetry, but that isn’t possible in general.)