Exam #2 study guide (solutions)
Z
1.
MATH 161
cos4 θ dθ
We will apply the reduction formula cos2 θ = 12 (1 + cos 2θ).
Z
Z
Z
1
1
2
2
2
(cos θ) dθ =
(1 + cos 2θ) dθ =
1 + 2 cos 2θ + cos2 2θ dθ
4
4
Apply it again:
1
4
Z
1 + 2 cos 2θ + cos2 2θ dθ =
1
4
Z
1 + 2 cos θ dθ +
1
8
Z
1 + cos 4θ dθ = · · ·
Integrate:
··· =
Z
2.
1
1
1
3
1
θ + sin 2θ + θ +
sin 4θ + C = θ + sin 2θ +
sin 4θ + C
4
8
32
8
32
∞
x2 e−x dx
This is an improper integral over an unbounded interval.
0
Our first step is to find the antiderivative using integration by parts. We can use the tabular
method or repeated integration by parts to see that
Z
x2 e−x dx = −x2 e−x − 2xe−x − 2e−x + C.
Our improper integral is thus
b
lim −x2 e−x − 2xe−x − 2e−x 0 = 2 − lim −b2 e−b − 2be−b − 2e−b .
b→∞
b→∞
We use L’Hôpital’s rule to evaluate this limit.
2b
2
= lim b = 0.
b
b→∞ e
b→∞ e
lim 2be−b = lim
b→∞
Likewise,
b2
2b
= lim b
b
b→∞ e
b→∞ e
lim b2 e−b = lim
b→∞
which we just saw equals 0. Hence,
2 − lim −b2 e−b − 2be−b − 2e−b = 2.
b→∞
http://douglasweathers.nfshost.com/s17math161.html
Z
3.
x3
dx
4 + x2
There are two totally different approaches to this integral.
One option is to start with long division:
x3
4x
=x−
4 + x2
4 + x2
The integral becomes
Z
Z
x dx − 2
2x
dx.
4 + x2
Substituting u = 4 + x2 in the second integral and antidifferentiating, this becomes
1 2
x − 2 ln(4 + x2 ) + C.
2
The second option is to use the trigonometric substitution x = 2 tan θ. The integral becomes
Z
Z
Z
8 tan3 θ
tan3 θ
2
2
sec
θ
dθ
=
4
tan3 θ dθ.
2
sec
θ
dθ
=
4
sec2 θ
4 + 4 tan2 θ
Next, use the identity tan2 θ = sec2 θ − 1:
Z
Z
Z
3
2
4 tan θ dθ = 4 tan θ sec θ dθ − 4 tan θ dθ
To evaluate the first integral, use the substitution u = tan θ. To evaluate the second, write
tan θ as the quotient of sin θ and cos θ and put u = cos θ. The antiderivative is
2 tan2 θ − 4 ln | sec θ| + C.
After applying the reverse substitutions tan θ = x/2 and sec θ =
is the same as the one from the long division approach.
√
4 + x2 , this antiderivative
4.
R 2x4 + x2 − 1
dx
x5 − x3
This is a rational function with no obvious trigonometric substitution available, so we’ll use
partial fraction decomposition. Start by factoring the denominator.
2x4 + x2 − 1
C
D
2x4 + x2 − 1
A
B
E
=
= + 2+ 3+
+
5
3
3
x −x
x (x + 1)(x − 1)
x
x
x
x+1 x−1
Multiply both sides of the equation by the least common denominator.
2x4 + x2 − 1 = Ax2 (x2 − 1) + Bx(x2 − 1) + C(x2 − 1) + Dx3 (x + 1) + Ex3 (x − 1)
With this many letters, we’ll want to cover some up.
If we suppose x = 0, we’ll see that C = 1;
if we suppose x = 1, we’ll see that D = 1; and
if we suppose x = −1, we’ll see that E = 1.
We’ll have to compare coefficients to find A and B. Since we’re solving for two variables,
we’ll only need two equations, but we should use the remaining four equations to check our
work. Group the right-hand side of the equation according to the power of x:
2x4 + x2 − 1 = (A + D + E)x4 + (B + D − E)x3 + · · ·
We’ve learned A + D + E = 2. Since D = E = 1, this means A = 0. Likewise, since
B + D − E = 0, we know B = 0.
At long last our integral becomes
Z
1
1
1
1
1
+
+
dx = − 2 + ln |x + 1| + ln |x − 1| + C = ln |x2 − 1| − 2 + C.
x3
x−1 x+1
2x
2x
Z
6
5.
p
6x − x2 dx
3
Quadratics suggest trigonometric substitution, but we’ll have to complete the square first.
We could do this either by inspection or by writing x2 − 6x + B 2 = (Ax + B)2 and figuring
out what A and B are.
6x − x2 = −(x2 − 6x) = −(x2 − 6x + 9) + 9 = 9 − (x − 3)2
Our integral becomes
Z
6
6
Z
p
6x − x2 dx =
p
9 − (x − 3)2 dx
3
3
which we can solve with the substitution x − 3 = 3 sin θ. In this substitution, the bound
x = 3 corresponds to θ = 0, and x = 6 corresponds to θ = π/2.
Z
π/2
Z
p
2
9 − 9 sin θ(3 cos θ) dθ = 9
0
π/2
√
cos2
Z
θ(cos θ) dθ = 9
0
π/2
cos2 θ dθ
0
Use the reduction formula again.
9
2
Z
6.
1
4
Z
0
π/2
π/2
9π
9
9
=
1 + cos 2θ dθ =
θ + sin 2θ
.
2
4
4
0
x2 − 2x + 11
dx
x2 − 2x + 10
Long divide and complete the square to see that the integral is
Z
1
4
1
1+
dx = 3 +
(x − 1)2 + 9
Z
1
4
1
dx
(x − 1)2 + 9
The remaining integral may be evaluated with the substitution x − 1 = 3 tan θ, where the
bounds x = 1 and x = 4 become θ = 0 and θ = π/4, respectively.
Z
3+
0
π/4
3 sec2 θ
1
dθ = 3 +
3
9 + 9 tan2 θ
Z
0
π/4
sec2 θ
1
dθ = 3 +
sec2 θ
3
Z
π/4
1 dθ = 3 +
0
π
.
12
Z
7.
√
x2
dx
1 − x2
Hopefully, we see this integral definitely involves a trigonometric substitution. Put x = sin θ.
Z
Z
Z
sin2 θ
sin2 θ
p
cos θ dθ =
cos θ dθ = sin2 θ dθ
cos θ
1 − sin2 θ
Use the reduction formula sin2 θ = 1 − cos 2θ.
Z
1
1
1
1 − cos 2θ dθ = θ − sin 2θ + C
2
2
4
Next, use the double-angle
formula sin 2θ = 2 sin θ cos θ and the reverse substitutions sin θ =
√
x and cos θ = 1 − x2 .
1
1
1
1 p
θ − sin θ cos θ = sin−1 x − x 1 − x2 + C.
2
2
2
2
Z
∞
8.
−∞
x2
1
dx
+ 25
This is an improper integral over an unbounded interval. Let’s start by finding the antiderivative, which definitely looks like an arctangent function. We can verify this with the
substitution x = 5 tan θ.
Z
1
1
x
dx = tan−1 + C
2
x + 25
5
5
Next we take the limits as our upper and lower bounds separately tend to infinity.
Z
lim
lim
b→∞ a→−∞
a
b
b
1
1
1
b
1
a
−1 x dx
=
lim
lim
tan
= lim tan−1 − lim
tan−1
b→∞ a→−∞ 5
x2 + 25
5 a b→∞ 5
5 a→−∞ 5
5
We have seen elsewhere that as y = tan x → ∞, x = tan−1 y → π/2 (because this is the
angle where the denominator, cos x, vanishes). Likewise, tan x → −∞ as x → −π/2. Thus
lim
b→∞
b
a
π
π
π
1
1
tan−1 − lim
tan−1 =
− (− ) = .
5
5 a→−∞ 5
5
10
10
5
Z
sin6 θ cos3 θ dθ
9.
This is one of the integrals where we use the Pythagorean identity sin2 θ +cos2 θ = 1 and the
fact that cosine and sine are one another’s derivatives to evaluate the integral. Since cos θ
is the one with the odd power, we will peel off a cosine and use the Pythagorean identity:
Z
Z
sin6 θ cos2 θ cos θ dθ = sin6 θ(1 − sin2 θ) cos θ
Next, we use the substitution u = sin θ.
Z
6
Z
2
u (1 − u ) du =
Z
∞
u6 − u8 du =
1
1
1 7 1 9
u − u + C = sin7 θ − sin9 θ + C
7
9
7
9
3x2 + 6x + 2
dx
x3 + 3x2 + 2x
10.
1
If we are not mindful, we will find ourselves doing partial fraction decomposition when the
integral is just a simple u-sub! Put u = x3 + 3x2 + 2x.
Z
1
∞
3x2 + 6x + 2
dx = lim
b→∞
x3 + 3x2 + 2x
Z
1
b
b
3x2 + 6x + 2
2
dx
=
lim
ln
|3x
+
6x
+
2|
1
b→∞
x3 + 3x2 + 2x
The integral diverges because lim ln |3b2 + 6b + 2| does not exist.
b→∞
Z
11.
tan θ
dθ
sec4 θ
We would like to put u = sec θ since this function is trapped in the denominator, but du is
not present. We can put it there by multiplying and dividing by another power of sec θ.
Z
tan θ
dθ =
sec4 θ
Z
tan θ sec θ
dθ =
sec5 θ
Z
1
1
1
1
du = C − 4 = C −
= C − cos4 θ
u5
4u
4 sec4 θ
4
Z
12.
3x3 − 14x2 + 29x − 54
dx
(x2 + 9)(x − 3)(x − 4)
If you’ve been paying attention, you know that the exams are more focused on the concepts;
so, a problem like this one (with a lot of simple but tedious algebra and arithmetic) is
unlikely to appear on the exam. It’s still important to know the concepts, so let’s work on
those.
We have a rational function with a fully-factored denominator, so we will decompose the
fraction.
Ax + B
C
D
Ax
B
C
D
3x2 − 14x2 + 29x − 54
= 2
+
+
= 2
+ 2
+
+
2
(x + 9)(x − 3)(x − 4)
x +9
x−3 x−4
x +9 x +9 x−3 x−4
Let’s integrate this without worrying about the constants. The antiderivative is
x
A
B
ln |x2 + 9| + tan−1
+ C ln |x − 4| + D ln |x − 3| + E,
2
3
3
where E is the arbitrary constant of integration.
The partial fraction decomposition gives us
3x2 − 14x2 + 29x − 54 = (Ax + B)(x − 3)(x − 4) + C(x2 + 9)(x − 4) + D(x2 + 9)(x − 3),
which after either comparing coefficients or covering up with x = 3 or x = 4 tells us that
A = 17/15, B = 1/5, C = 6/5, and D = 2/3.
Z
13.
π/2
p
sin3 θ cos3 θ dθ
0
We see that sin θ is trapped under the square root, forcing us to choose u = sin θ. Don’t
forget to peel off the cosine!
Z
π/2
Z
p
sin3 θ(1 − sin2 θ) cos θ dθ =
0
1
u3/2 (1 − u2 ) du
0
Distribute, integrate, and evaluate at the bounds.
Z
0
1
u3/2 − u7/2 du =
1
2 5/2 2 9/2 8
u − u =
5
9
45
0
Z
6
14.
−2
1
p
dx
|x − 2|
Start by noticing a few things about the integrand: first, that it is discontinuous at x = 2;
second, that to the left of the discontinuity |x−2| = 2−x, and to the right of the discontinuity
|x − 2| = x − 2. We correctly write the integral as
Z
b
−1/2
(2 − x)
lim
b→2−
Z
dx + lim
a→2+
−2
a
6
6
b
√
√
(x − 2)−1/2 dx = lim 2 x − 2a − lim− 2 − x−2
a→2+
b→2
The antiderivatives are continuous at x = 2, so we just substitute a = b = 2 to evaluate the
limit.
√
√
2 4−0−0+2 4=8
The integral, hence, converges to 8.
Z
15.
π
tan θ dθ
0
The integrand is discontinuous at θ = π/2, so split up the integral and take limits:
Z
lim
π
b→ 2 −
b
0
Z
tan θ dθ + lim
π
a→ 2 +
π
tan θ dθ
a
R
You may remember that tan θ dθ = ln | sec θ| + C. If you don’t, no worries: write tan θ =
sin θ/ cos θ and put u = cos θ.
b
π
ln | sec θ||a = lim
ln | sec b| + lim
ln | sec a|
lim ln | sec θ||0 + lim
π
π
π
−
b→ π
2
a→ 2 +
b→ 2 −
a→ 2 +
It is enough to see that one of these limits doesn’t exist. As b → π/2 from the left, sec b → ∞.
(Remember, sec b = 1/ cos b, and cos b is approaching 0.) Putting u = sec b, ln u → ∞ as
u → ∞. Since therefore ln | sec b| → ∞ as b → π/2 from the right, one limit doesn’t exist,
and the integral diverges.
16. There are 1,000 fish in a certain fishery when it opens. A year later there are 5,000 fish.
The differential equation corresponding to the population’s growth is y 0 = ky, where y(t)
is the number of fish t years after the fishery opens and k is the proportionality constant.
What is the function y(t)?
Start by solving the differential equation y 0 = ky. First, separate the variables.
1 0
y =k
y
Next, integrate both sides with respect to t. Remember that y 0 dt = dy, so we can treat y
like a variable!
Z
Z
1
dy = k dt
⇒
ln |y| = kt + C
y
Exponentiate both sides. It is customary to write eC as a simpler-looking expression, usually
A.
y = Aekt
This is the general solution to the differential equation. We can use the knowledge that
y(0) = 1000 and y(1) = 5000 to solve for A and k.
y(0) = 1000 = Ae0
y(1) = 5000 = 1000ek
The function is thus y(t) = 1000e
necessary.)
(ln 5)t
⇒
5 = ek
⇒
A = 1000
⇒
k = ln 5
t
= 1000 · 5 . (This extra simplification is nice but not
17. Solve the differential equation y 0 = 2xe−y where y(0) = 0.
As before, separate the variables and integrate both sides with respect to x.
Z
Z
ey dy = 2x dx
⇒
ey = x2 + C
⇒
y = ln(x2 + C)
Use the initial condition y(0) = 0 to solve for C.
y(0) = 0 = ln C
⇒
C=1
Hence, the solution to the initial-value problem is y(x) = ln(x2 + 1).
18. The spread of a rumor is modeled by the differential equation y 0 = ky(1 − y), where k
is some proportionality constant and 0 ≤ y(t) ≤ 1 is the percentage of students who know
about the rumor t hours after it starts. The rumor is initially known by 5% of students,
and an hour later, 10% of students know it. What is the function y(t)?
Start by separating the variables.
1
y0 = k
y(1 − y)
To integrate the left side, we will have to use the method of partial fraction decomposition
to see that
1
1
1
= +
.
y(1 − y)
y 1−y
(This is as easy as partial fraction decompositions get, so it’s left as an exercise to you.)
Now we integrate each side with respect to t:
Z
Z
1
1
+
dy = k dt
⇒
ln |y| − ln |1 − y| = kt + C
y 1−y
Combine logarithms and exponentiate.
y = kt + C
ln 1 − y
⇒
y
= Aekt
1−y
Finally, rearrange and solve for y.
y = Aekt (1 − y)
(1 + Aekt )y = Aekt
⇒
⇒
y=
Aekt
1 + Aekt
Now we use the initial conditions to nail down k and A.
y(0) =
1
A
=
20
1+A
⇒
1 + A = 20A
1
19
⇒
A=
⇒
−k = ln
We can rewrite
y=
1
1 kt
19 e
1 kt
+ 19
e
=
1
,
19e−kt + 1
which is helpful, I promise. Put t = 1 to solve for k:
y(1) =
1
1
=
10
19e−k + 1
⇒
10 = 19e−k + 1
The function that models the propogation of the rumor is
y(t) =
1
=
19eln(9/19)t + 1
19
where, again, the final simplification is not necessary.
1
9 t
19
+1
9
19
19. Write the first five terms of the sequence {an }∞
n=0 whose explicit formula is an = 4 + 2n.
Does this sequence converge (to what?) or diverge?
The first five terms are 4, 6, 8, 10, 12. As we continue adding 2 indefinitely, the sequence will
diverge.
2 n
20. Write the first five terms of the sequence {bn }∞
n=1 where bn = ((n + 1)/n ) . Does this
sequence converge (to what?) or diverge?
2
3 5 4 6 5
The first five terms are 2, 43 , 49 , 16
, 25 . These numbers are getting smaller with
no signs of slowing down, so we see the sequence is converging to 0.
21. Write the first seven terms of the sequence {fn }∞
n=0 given by the recurrence relation
f0 = 1, f1 = 1, and fn = fn−1 + fn−2 . Does this sequence converge (to what?) or diverge?
The first seven terms are 1, 1, 2, 3, 5, 8, 13. As we are continuing to add larger and large are
numbers, the sequence diverges.
22. Evaluate the limit of the sequence whose n-th term is given by the function sin n.
As n → ∞, sin n oscillates between −1 and 1. It never gets and stays close to a value, so
the sequence diverges.
23. Evaluate the limit of the sequence whose n-th term is given by the function
n3 + 30n + 1
.
5n3 + 400n2 − 7
We can divide through the highest power of n:
1 + 30/n2 + 1/n3
5 + 400/n − 7/n3
As n gets large, the terms with n in the denominator vanish, and we are left with 1/5. The
sequence converges to this number.
24. Evaluate the limit of the sequence {xn }∞
n=0 whose first term is x0 = 1, and whose
subsequent terms are given by the recurrence xn = −xn−1 /2.
When in doubt, list some terms of the sequence and see what is happening. The sequence
is 1, −1/2, 1/4, −1/8, 1/16, −1/32, . . . Though the numbers are oscillating from positive to
negative and back, they are getting closer and closer to zero. The sequence converges to 0,
because it is a geometric sequence with |r| < 1.