405
Answers
601–700
Answers and Explanations
cos
sin
sin
c os
sin
cos
x
x dx
x
x dx
x
x
−
( )
 
  +
( )−
 
 
=
+
(
∫
∫
2
2
2
0
6
6
2
π
π
π
) )


 


 
+
−
( ) −


 


 
=
+
( ) −
2
2
2
6
3
2
0
6
6
2
π
π
π
π
π
cos
sin
sin
cos
s
x
x
i in
cos
cos
sin
cos
sin
0
0
2
2
2
3
2
+
( )


 


 










+ −
−
−
−
( ) −
π
π
π
π
6 6
1
2
1
4
0 1
2
1
2
1 1
4
1
2
1




















=
+ − −

 

 
+
− + +

 

 
= 4 4
1
4
1
2
+ =
The following figure shows the region bounded by the given curves:
661.
13
4
1
8
+ ( )
ln
Begin by finding the point of intersection by setting the functions y = 2e
2x
and y = 3 – 5e
x
equal to each other and solving for x:
2
3 5
2
5
3 0
2
1
3 0
1
2
3
2
2
e
e
e
e
e
e
e
x
x
x
x
x
x
x
= −
+
− =
−
(
) +
(
) =
=
−
,
Because e
x
= –3 has no solution, the only solution is x = ( )
ln 1
2
. Note that ln 1
2
0
( ) < , so the
line x = 0 bounds the region on the right so that the upper limit of integration is b = 0.
To determine which function is larger on the interval ln
,
1
2
0
( )

 

  , pick a point inside the
interval and substitute it into each equation. On the interval ln
,
1
2
0
( )

 

  , notice that
2e
2x
> 5e
x
. Therefore, the integral to find the area is