Part II: The Answers
404
Answers
601–700
Finally, simplify and evaluate this integral:
2 3 2
2 3
3
2 3 3 3 3
3
0 0 0
2
0
3
2
3
0
3
2
3
+
−
(
)
=
+ −
=
+ −
− + −
∫
x x dx
x x
x
( )
(
)
=
+ −
[
]
=
2 9 9 9
18
The following figure shows the region bounded by the given curves:
660.
1
2
Begin by finding the points of intersection on the interval 0 2
, π
by setting the
functions equal to each other:
cos
sin
x
x
=
2
Use an identity on the right-hand side of the equation and factor:
cos
sin cos
sin cos
cos
cos ( sin
)
x
x
x
x
x
x
x
x
=
=
−
=
−
2
0 2
0
2
1
Now set each factor equal to zero and solve for x: cos x = 0 and 2 sin x – 1 = 0, or
sin x = 1
2
. On the interval 0 2
, π
, you have cos x = 0 if x = π
2
and sin x = 1
2
if x = π
6
. On
the interval 0 6
, π
( ) , you have cos x > sin(2x), and on π π
6 2
,
( ) , you have sin(2x) > cos x.
Therefore, the integrals to find the area are
404
Answers
601–700
Finally, simplify and evaluate this integral:
2 3 2
2 3
3
2 3 3 3 3
3
0 0 0
2
0
3
2
3
0
3
2
3
+
−
(
)
=
+ −
=
+ −
− + −
∫
x x dx
x x
x
( )
(
)
=
+ −
[
]
=
2 9 9 9
18
The following figure shows the region bounded by the given curves:
660.
1
2
Begin by finding the points of intersection on the interval 0 2
, π
by setting the
functions equal to each other:
cos
sin
x
x
=
2
Use an identity on the right-hand side of the equation and factor:
cos
sin cos
sin cos
cos
cos ( sin
)
x
x
x
x
x
x
x
x
=
=
−
=
−
2
0 2
0
2
1
Now set each factor equal to zero and solve for x: cos x = 0 and 2 sin x – 1 = 0, or
sin x = 1
2
. On the interval 0 2
, π
, you have cos x = 0 if x = π
2
and sin x = 1
2
if x = π
6
. On
the interval 0 6
, π
( ) , you have cos x > sin(2x), and on π π
6 2
,
( ) , you have sin(2x) > cos x.
Therefore, the integrals to find the area are