403
Answers
601–700
Answers and Explanations
To determine which function is larger on the interval (–4, 0), pick a point inside the
interval and substitute it into each equation. So if x = –3, then y = − + =
3 4 1 and
y = − + =
3 4
2
1
2
. Because x
x
+ > +
4
4
2
on (–4, 0), the integral to find the area is
(
)
(
)
(
)
x
x
dx
x
x
dx
x
+
− +

 

 
=
+
− −

 

 
=
+
−
−
∫
∫
4
4
2
4
2
2
2
4
1 2
4
0
1 2
4
0
3 2
3 3
4
2
2
3
4
0
4
2 0
0
4
4
2 4
4
2
4
0
3 2
2
2
−
−






=
−
−
− −
−
− −






=
−
x
x
( )
( )
( )
( )
3 3
659.
18
To find the points of intersection, begin by noting that y
x
x
=
=
2
2 on the interval [0,
∞). To find the point of intersection on [0, 8), set the functions equal to each other and
solve for x:
2
3
0
2
3
0
3
1
3 1
2
2
x x
x
x
x
x
x
=
−
=
− −
=
−
+
= −
(
)(
)
,
Because the interval under consideration is [0, 8), use the solution x = 3.
Likewise, on (–∞, 0), you have y
x
x
=
=−
2
2 . Again, find the point of intersection by
setting the functions equal to each other and solving for x:
− =
−
=
+
−
=
+
−
= −
2
3
0
2
3
0
3
1
3 1
2
2
x x
x
x
x
x
x
(
)(
)
,
Because the interval under consideration is (–∞, 0), keep only the solution x = –3. (You
could have also noted that because both y
x
= 2 and y = x
2
– 3 are even functions, then if
there’s a point of intersection at x = 3, there must also be a point of intersection at x = –3.)
On the interval (–3, 3), you have 2
3
2
x x
>
− . Therefore, the integral to find the area is
2
3
2
3
3
x
x
dx
−
−
(
)
(
)
−
∫
By symmetry, you can rewrite the integral as
2 2
3
2
0
3
x x
dx
−
−
(
)
(
)
∫