M ATH 31B S PRING 2017 M IDTERM 1 P RACTICE
Joseph Breen
Contents
1
Problems
1
2
Solutions
2
1
Problems
1. Let f (x) = sin (xx ) for x > 0. Compute f 0 (x).
2. As in the previous problem, let f (x) = sin (xx ) for x > 0. Does f (x) have an inverse function
f −1 (x)? If so, compute f −1 (x). If not, explain why.
3. Evaluate the following limit:
sinh(x)
.
x→0 sin(x)
lim
4. Evaluate the following limit:
x2
lim cos e−x
.
x→∞
5. Evaluate the following limit:
lim 2 + e−x
arctan(x)
x→∞
6. Evaluate the following antiderivative:
Z
2x arctan(x) dx.
7. Evaluate the following antiderivative:
Z
√
x
dx.
4 − 9x4
1
.
2
Solutions
(I’ll keep updating this document with solutions)
1. Let f (x) = sin (xx ) for x > 0. Compute f 0 (x).
Solution.
2. As in the previous problem, let f (x) = sin (xx ) for x > 0. Does f (x) have an inverse function
f −1 (x)? If so, compute f −1 (x). If not, explain why.
Solution.
3. Evaluate the following limit:
lim
x→0
sinh(x)
.
sin(x)
Solution.
4. Evaluate the following limit:
x2
lim cos e−x
.
x→∞
Solution.
5. Evaluate the following limit:
lim 2 + e−x
arctan(x)
x→∞
.
Solution.
6. Evaluate the following antiderivative:
Z
2x arctan(x) dx.
Solution. Because the function we are antideriving is a product of functions (2x and arctan(x)) and
because a u-substitution doesn’t seem helpful, this looks like an integration by parts problem.
R
R
Towards the goal of using the integration by parts formula u dv = uv − v du, we need to decide
what u and dv are in the given integral. Finding the antiderivative of arctan(x) seems more complicated than finding the antiderivative for 2x, so let’s try:
u = arctan(x)
Then:
du =
1
dx
1 + x2
dv = 2x dx.
v = x2 .
The integration by parts formula then gives:
Z
2
2x arctan(x) dx = arctan(x) · x −
2
Z
x2
dx.
1 + x2
R x2
It remains to compute 1+x
2 dx. This is not an integral that we know immediately, nor does it look
like a u-substitution will be helpful. We could probably try doing integration by parts again, but
instead we can use some clever algebra:
x2 + 1 − 1
1 + x2
1
1
x2
=
=
−
=1−
.
2
2
1+x
1+x
1 + x2
1 + x2
1 + x2
Therefore,
Z
x2
dx =
1 + x2
Z
1−
1
dx =
1 + x2
Z
Z
1 dx −
1
dx
1 + x2
= x + arctan(x) + C.
Putting all of this together, we have:
Z
Z
2
2x arctan(x) dx = arctan(x) · x −
x2
dx
1 + x2
= x2 arctan(x) − x − arctan(x) + C.
7. Evaluate the following antiderivative:
Z
√
x
dx.
4 − 9x4
R
1
1
x
looks sort of like √1−x
, and we know that √1−x
dx = arcsin(x) + C.
Solution. The function √4−9x
4
2
2
We can probably make a clever u-substitution to get what we want. Our first step is to decide what
u-substitution to make.
Ultimately, we want the inside of the square root to look like 1 − u2 . So the first thing we can do is
factor out the 4 from the inside of the square root:
9
4 − 9x4 = 4 1 − x4 .
4
This means we want a u-substitution that gives 49 x4 = u2 . Taking the square root of both sides of this
equation suggests the following substitution: u = 32 x2 .
Let’s try it. We have:
u=
3 2
x
2
du = 3x dx
1
3
so that x dx =
Z
du. Therefore, our integral is:
Z
Z
Z
1
1
x
x
1
1
√
q
√
√
dx =
dx
=
du
=
du.
4
2
3
6
4 − 9x
2 1−u
1 − u2
4 1 − 94 x4
We know that
√ 1
1−u2
du = arcsin(u) + C, so
Z
√
1
x
dx = arcsin
4
6
4 − 9x
3
3 2
x
2
+ C.