Mathematics 1A, Fall 2009 — M. Christ — Final Examination Solutions
There were two versions of the exam. They were very similar, so solutions are given here
for only one version.
(1)
Calculate the following.
(1a)
The equation of the line tangent to
f
(
x
) =
x
+
e
x
at
x
= 2.
Solution.
y
= (2 +
e
2
) +
e
2
(
x

2).
(1b)
d
dx
p
3 + ln(ln(
x
)).
Solution.
1
2
(3 + ln(ln(
x
)))

1
/
2
·
1
ln(
x
)
·
1
x
.
(1c)
lim
x
→
π/
2
cos(
x
)
x

π/
2
Solution.
= lim
x
→
π/
2

sin(
x
)
1
=

sin(
π/
2) =

1 by L’Hˆ
opital’s
rule.
(1d)
d
dx
x
cos(
x
)
. (Here
x >
0.)
Solution.
=
d
dx
e
cos(
x
) ln(
x
)
= (

sin(
x
) ln(
x
) + cos(
x
)
x

1
)
e
cos(
x
) ln(
x
)
= (

sin(
x
) ln(
x
) +
cos(
x
)
x

1
)
x
cos(
x
)
.
(1e)
R
d
dx
p

sin(
x
) + cos(
x
)

dx
.
Solution.
p

sin(
x
) + cos(
x
)

+
C
where
C
is an arbitrary
constant.
(1f)
d
dx
R
sin(2
x
)
0
arcsin(
t
)
dt
.
Solution.
= arcsin(sin(2
x
))
·
2 cos(2
x
).
If 2
x
is in the range [

π/
2
, π/
2], then this can be simplified, since then arcsin(sin(2
x
)) =
2
x
.
If 2
x
is not in this range then arcsin(sin(2
x
)) is the unique number
t
in this range
which satisfies sin(
t
) = sin(2
x
).
(1h)
R
sin(
x
) cos(
x
)
dx
Solution.
Substitute
u
= sin(
x
). Then
du
= cos(
u
)
du
, and the
integral is
R
u du
=
1
2
u
2
+
C
=
1
2
sin
2
(
x
) +
C
, where
C
is an arbitrary constant.
(2j)
R
1
0
(1

x
2
)

1
/
2
dx
Solution.
An antiderivative for (1

x
2
)

1
/
2
is arcsin(
x
), so by the
FTC part II, the integral equals arcsin(1)

arcsin(0) =
π
2

0 =
π
2
.
(2k)
∑
4
i
=1
i
2
cos(
πi
)
Solution.
= 1
2
cos(
π
) + 2
2
cos(2
π
) + 3
2
cos(3
π
) + 4
2
cos(4
π
) =

1 +
4

9 + 16 = 10.
(2l)
R
2

2
x
ln(1+
x
4
)
dx
Solution.
We cannot easily evaluate the indefinite integral
R
x
ln(1+
x
4
)
dx
using techniques from this course.
However, the integrand
x
ln(1 +
x
4
) is an odd
function, and the integral is from

a
to
a
where
a
= 2, so the definite integral equals 0.
(2m)
R
(1

x
2
)

3
/
2
dx
(You need not simplify your answer.)
Solution.
Substitute
x
=
sin(
θ
) where
θ
∈
(

π/
2
, π/
2). Then
dx
= cos(
θ
)
dθ
, and cos(
θ
) =
p
cos
2
(
θ
) since cos(
θ
)
≥
0.
Therefore the integral becomes
R
(cos(
θ
))

3
cos(
θ
)
dθ
=
R
1
cos
2
(
θ
)
dθ
=
R
sec
2
(
θ
)
dθ
=
tan(
θ
) +
C
= tan(arcsin(
x
)) +
C
where
C
is an arbitrary constant.
This can be simplified since
sin(arcsin(
x
))
cos(arcsin(
x
))
=
x
√
1

x
2
, but full credit was given for the above
answer.
(2n)
lim
x
→∞
(
x
+
x
1
/
3
)
2
/
3

x
2
/
3
Solution.
This is most easily done using L’Hˆ
opital’s
rule.
= lim
x
→∞
x
2
/
3
(1 +
x

2
/
3
)
2
/
3

1
= lim
x
→∞
(1 +
x

2
/
3
)
2
/
3

1
x

2
/
3
= lim
t
→
0
+
(1 +
t
2
/
3
)
2
/
3

1
t
2
/
3
= lim
t
→
0
+
2
3
(1 +
t
2
/
3
)

1
/
3
·
2
3
t

1
/
3
2
3
t

1
/
3
= lim
t
→
0
+
2
3
(1 +
t
2
/
3
)

1
/
3
=
2
3
·
(1 + 0)

1
/
3
=
2
3
.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2
(2o)
Express an approximation to
R
3
1
e
x
2
dx
as a right endpoint Riemann sum with
n
= 3.
Your answer need not be simplified; it could be expressed as a sum of several numbers.
Solution.
The endpoints are
a
= 1 and
b
= 3, so
b

a
n
=
2
3
.
Thus
x
1
= 1 +
2
3
=
5
3
,
x
2
=
x
1
+
2
3
=
7
3
,
x
3
= 3. The Riemann sum is
b

a
n
3
X
i
=1
e
x
2
i
=
2
3
(
e
25
/
9
+
e
49
/
9
+
e
9
)
.
(2p)
lim
n
→∞
∑
n
i
=1
n
n
2
+
i
2
. (Either use a method taught in this course, or justify your steps
in full detail.)
Solution.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 WILKENING
 Math, Calculus, Derivative, Continuous function, Convex function

Click to edit the document details