Math 20B (Bender)
Final Exam Solutions
Fall 2000
1. (a) Since this is of the form 0
/
0, we apply l’Hˆ
opital’s rule:
lim
x
→
0
e
x

1
sin
x
= lim
x
→
0
e
x
cos
x
=
1
1
=1
.
(b) You could convert to polar form:
r
=
√
2 and
θ
=
π/
4, giving
r
30
=2
15
and
30
θ
=15
π/
2=8
π

π/
2. Thus (1 +
i
)
30
=

2
15
i
. Alternatively, you can take
powers:
(1 +
i
)
30
= ((1 +
i
)
2
)
15
=(2
i
)
15
=2
15
i
15
=2
15
i
3
=

2
15
i.
2. (a) The denominator factors as
x
(
x
2

1) =
x
(
x

1)(
x
+ 1). Hence
2
x
3

x
=
A
x
+
B
x

1
+
C
x
+1
=
A
(
x
2

1) +
B
(
x
2
+
x
)+
C
(
x
2

x
)
x
(
x

1)(
x
+1)
.
Thus (
A
+
B
+
C
)
x
2
+(
B

C
)
x

A
= 2. From constant and linear terms,
A
=

2
and
B
=
C
. Thus the quadratic gives

2+2
B
= 0 and so
B
= 1. Thus we have
Z
2
x
3

x
dx
=
Z
±

2
x
+
1
x

1
+
1
x
+1
¶
dx
=ln
ﬂ
ﬂ
ﬂ
ﬂ
x
2

1
x
2
ﬂ
ﬂ
ﬂ
ﬂ
+
C
=ln

1

x

2

+
C.
(b) First method: Integrate by parts with
u
=ln
x
and
dv
=
dx
. This gives
Z
e
1
ln
xdx
=
x
ln
x
ﬂ
ﬂ
e
1

Z
e
1
x
dx
x
=
e

(
x
ﬂ
ﬂ
e
1
=
e

(
e

1) = 1
.
Second method:
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 Winter '08
 Justin
 Math, Calculus, dx

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