397
Answers
601–700
Answers and Explanations
649.
2 4
2
+ π
Notice that the functions y x
= − π
2
and y = cos x intersect when x = π
2
. On the interval
0 2
, π
( ) , you have cos x x
≥
−
( )
π
2
, and on the interval π π
2
,
( ) , you have x
x
− ≥
π
2
cos .
Therefore, the integrals to find the area of the region are
co
c
s
o s
sin
x x
dx
x
x dx
x x
− −
( )
+
−
( ) −
=
−
+
∫
∫
π
π
π
π
π
π
2
2
2
0
2
2
2
2 2
2 2
1 1
2 2
2 2
0
2
2
2
2
x
x
x
x
+
−
−
= − ( ) + ( ) +
π
π
π
π
π
π π
sin
π π
π
π
π
π
2
2
2
2
2
2
2
0
1
2 2
2
1
2 4
−
−
−
( ) − ( ) −
= +
650.
9
2
Begin by finding the points of intersection by setting the functions equal to each other
and solving for x:
x
x
x
x
x
x x
x
3
3
2
2
3
0
3 0
0
3
− =
−
=
−
(
) =
= ±
,
To determine which function is larger on the interval −
(
)
3 0
, , take a point in the
interval and substitute it into each function to determine which is larger. If x = –1,
then y = (–1)
3
– (–1) = 0 and y = 2(–1) = –2; therefore, x
3
– x > 2x on −
(
)
3 0
, . In a similar manner, check which function is larger on the interval 0 3
,
( ) . By letting x = 1,
you can show that 2x > x
3
– x on 0 3
,
( ) . Therefore, the integral to find the area of
the region is
x
x
x dx
x
x
x dx
x
x dx
x
3
3
0
3
3
0
3
2
2
3
3
−
(
) −
+
−
−
(
)
=
−
+
−
∫
∫ −
( )
( )
x x dx
x
x
x
x
3
0
3
3
0
4
2
3
0
2
4
0
3
4
3
2
3
2
4
0 0
9
4
(
)
=
−
+
−
= − −
∫
∫ −
−
(
)
− −
( ) + −
( )
=
−
=
=
9
2
9
2
9
4
18
2
18
4
18
4
9
2
Answers
601–700
Answers and Explanations
649.
2 4
2
+ π
Notice that the functions y x
= − π
2
and y = cos x intersect when x = π
2
. On the interval
0 2
, π
( ) , you have cos x x
≥
−
( )
π
2
, and on the interval π π
2
,
( ) , you have x
x
− ≥
π
2
cos .
Therefore, the integrals to find the area of the region are
co
c
s
o s
sin
x x
dx
x
x dx
x x
− −
( )
+
−
( ) −
=
−
+
∫
∫
π
π
π
π
π
π
2
2
2
0
2
2
2
2 2
2 2
1 1
2 2
2 2
0
2
2
2
2
x
x
x
x
+
−
−
= − ( ) + ( ) +
π
π
π
π
π
π π
sin
π π
π
π
π
π
2
2
2
2
2
2
2
0
1
2 2
2
1
2 4
−
−
−
( ) − ( ) −
= +
650.
9
2
Begin by finding the points of intersection by setting the functions equal to each other
and solving for x:
x
x
x
x
x
x x
x
3
3
2
2
3
0
3 0
0
3
− =
−
=
−
(
) =
= ±
,
To determine which function is larger on the interval −
(
)
3 0
, , take a point in the
interval and substitute it into each function to determine which is larger. If x = –1,
then y = (–1)
3
– (–1) = 0 and y = 2(–1) = –2; therefore, x
3
– x > 2x on −
(
)
3 0
, . In a similar manner, check which function is larger on the interval 0 3
,
( ) . By letting x = 1,
you can show that 2x > x
3
– x on 0 3
,
( ) . Therefore, the integral to find the area of
the region is
x
x
x dx
x
x
x dx
x
x dx
x
3
3
0
3
3
0
3
2
2
3
3
−
(
) −
+
−
−
(
)
=
−
+
−
∫
∫ −
( )
( )
x x dx
x
x
x
x
3
0
3
3
0
4
2
3
0
2
4
0
3
4
3
2
3
2
4
0 0
9
4
(
)
=
−
+
−
= − −
∫
∫ −
−
(
)
− −
( ) + −
( )
=
−
=
=
9
2
9
2
9
4
18
2
18
4
18
4
9
2