Math 105 - Practice Midterm 2
1 Compute the following:
a
d
dx
R
2
x
1
1+
t
3
dt
Solution:
We rewrite this as
-
d
dx
R
x
2
1
1+
t
3
dt
and apply the Funda-
mental Theorem of Calculus to yield
-
1
1+
x
3
.
b
R
∞
-∞
xdx
Solution:
This needs to be split into two improper integrals because
it ranges over (
-∞
,
∞
)
.
If the integral converges,
I
=
R
0
-∞
xdx
+
R
∞
0
xdx
. Both must converge for the integral to converge.
Note that
R
∞
0
xdx
= lim
R
→∞
R
R
0
xdx
= lim
R
→∞
x
2
2
|
R
0
=
∞
.
Thus
I
diverges and we don’t even need to consider the other integral.
c
R
∞
0
dx
x
1
/
2
+
x
3
/
2
Solution:
This integral is improper for two reasons:
the vertical
asymptote at
x
= 0 and the infinite range of integration. We split it
into
R
1
0
dx
x
1
/
2
+
x
3
/
2
+
R
∞
1
dx
x
1
/
2
+
x
3
/
2
.
Both must converge for the integral
to converge.
Making a substitution
u
=
x
1
/
2
with
du
=
1
2
x
1
/
2
dx
so
dx
= 2
udu
in
both integrals, we have
R
dx
x
1
/
2
+
x
3
/
2
=
R
2
1+
u
2
du
= 2 arctan
u
+
C
=
2 arctan
√
x
+
C
.
Thus,
R
∞
0
dx
x
1
/
2
+
x
3
/
2
= lim
t
→
0
+
2 arctan
√
x
|
1
t
+lim
R
→∞
2 arctan
√
x
|
R
1
=
2(
π/
4
-
0) + 2(
π/
2
-
π/
4) =
π.
d
R
dy
1+cos(4
y
)
Solution:
We write 1 + cos(4
y
) = 2 cos
2
(2
y
)
.
Thus,
R
dy
1+cos(4
y
)
=
R
1
2
sec
2
(2
y
) =
1
4
tan(2
y
) +
C.
e
R
(ln(
x
8
))
5
dx
Solution:
First of all we pull out the 8 to find
R
(8 ln
x
)
5
dx
= 8
5
R
ln
5
xdx.
This will involve multiple integration-by-parts. We’ll first note that
R
ln
n
xdx
=
R
ln
n
x
1
dx
so that if
f
= ln
n
x
with
df
=
n
(ln
n
-
1
x
)
1
x
and
dg
=
dx
with
g
=
x
then
R
ln
n
xdx
=
x
ln
n
x
-
n
R
ln
n
-
1
xdx.
Life is good now:
8
5
R
ln
5
xdx
= 8
5
(
x
ln
5
x
-
5
R
ln
4
xdx
) = 8
5
(
x
ln
5
x
-
5(
x
ln
4
x
-
R
4 ln
3
xdx
)) = 8
5
(
x
ln
5
x
-
5
x
ln
4
x
+20(
R
ln
3
xdx
))
=
|{z}
guess pattern
8
5
(
x
ln
5
x
-
5
x
ln
4
x
+5
×
4
x
ln
3
x
-
5
×
4
×
3
x
ln
2
x
+5
×
4
×
3
×
2 ln
x
-
5
×
4
×
3
×
2
×
1
x
)
= 8
5
(
x
ln
5
x
-
5
x
ln
4
x
+ 20
x
ln
3
x
-
60
x
ln
2
x
+ 120
x
ln
x
-
120
x
) +
C
.
This result can be verified by taking a derivative.
1
f
R
(tan
2
t
sec
t
-
tan
3
t
sec
t
)
dt
Solution:
We split the integral into two, as they require different
strategies.
•
I
=
R
tan
2
t
sec
tdt
=
R
(sec
2
t
-
1) sec
tdt
=
R
sec
3
t
-
R
sec
tdt.
