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Math 20B (Bender)
Solutions to First Exam
18 October 2000
I’ve noted if the problem or a near miss is in the text.
1. (10 pts.) Evaluate
Z
2
0
p
4

x
2
dx
by interpreting it as an area.
A. (p.383, Ex.4) Squaring and rearranging
y
=
√
4

x
2
gives
x
2
+
y
2
= 4, a circle of
radius 2 centered at the origin. The integral is the area in the ﬁrst quadrant and so
equals (
π
2
2
)
/
4=
π
.
Since the problem did not ask for an exact answer, you will receive credit for a
reasonable numerical evaluation.
2. (30 pts.) Evaluate the following integrals using the tools discussed in the text.
Z
(1

x
)
p
2
x

x
2
dx
Z
2
0

sin
πx

dx.
A. (p.426, #26, #39) The substitution
u
=2
x

x
2
converts the ﬁrst to
R
1
2
u
1
/
2
du
=
u
3
/
2
/
3+
C
and so the answer is (2
x

x
2
)
3
/
2
/
C
.
The second integral equals
R
1
0
sin
πx dx

R
2
1
sin
. The substitution
u
=
gives
R
sin
=(

cos
)
/π
+
C
. Thus the answer is
(

cos
π
+ cos 0)
/π

(

cos
π
+ cos 2
π
)
/π
=4
/π
.
3. (30 pts.) Diﬀerentiate the functions
F
(
x
)=
Z
x
1
p
1+
u
4
du
G
(
x
Z
1
x
2
ln(1

t
3
)
dt.
A. Both rely on the Fundamental Theorem of Calculus.
We have
F
0
(
x
√
x
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This note was uploaded on 06/17/2009 for the course MATH 20B taught by Professor Justin during the Winter '08 term at UCSD.
 Winter '08
 Justin
 Math, Calculus

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