Math 1B Quiz 2 Solutions
Section 107
September 12, 2009
1. (4 points)
R
3
2
√
4
x

x
2
dx
Method 1:
Complete the square to obtain
R
3
2
p
4

(
x

2)
2
dx
. Then substitute
u
=
x

2,
du
=
dx
to obtain
R
1
0
√
4

u
2
du
. (For the limits of integration,
x
= 2 when
u
= 0; and
x
= 3 when
u
= 1). Now use the
trig substitution
u
= 2 sin
θ, du
= 2 cos
θdθ
for

π/
2
≤
θ
≤
π/
2 to obtain
Z
1
0
p
4

u
2
du
=
Z
π/
6
0
p
4

4 sin
2
θ
(2 cos
θ
)
dθ.
For the limits of integration,
u
= 0 when
θ
= 0; and
u
= 1 = 2 sin
θ
when
θ
=
π/
6.
We have
p
4

4 sin
2
θ
= 2

cos
θ

= 2 cos
θ
since cos
θ
is positive for

π/
2
≤
θ
≤
π/
2. Therefore,
Z
π/
6
0
p
4

4 sin
2
θ
(2 cos
θ
)
dθ
=
Z
π/
6
0
4 cos
2
θ dθ
= 2
Z
π/
6
0
1 + cos 2
θ dθ
by the trig identity cos
2
θ
=
1
2
(1 + cos 2
θ
). Now we can evaluate the integral:
Z
3
2
p
4
x

x
2
dx
= 2
Z
π/
6
0
1 + cos 2
θ dθ
= 2
θ
+
sin 2
θ
2

π/
6
0
= 2
π
6
+
sin(
π
3
)
2
=
π
3
+
√
3
2
.
Note:
This problem was a variation of Problem 27 from Section 7.3 of your homework, and the above
solution is the standard way to solve it. Another method (below) uses a substitution not covered in
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 Fall '08
 Reshetiken
 Limits, Sin, Cos

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