University of Toronto
Faculty of Arts and Science
MAT137Y Test 4
X. Cui, V. Dimitrov, A. Malusà, B. Khesin, Z.Qian
March 17, 2022
Duration: 110 minutes
No Aids Permitted
Last Name:
First Name:
UToronto email:
@mail.utoronto.ca
UTORid:
Student Number:
This exam contains 12 double-sided pages (including this cover page) and is printed on 6 sheets of paper.
There are 7 problems. If you booklet is missing one page, please raise your hand to notify an
invigilator as soon as possible.
Instructions
• Keep this booklet closed, until an invigilator announces that the test has begun. However, you
may fill out your information in this page before the test starts.
• Please place your student ID in a location on your desk that is easy for an invigilator to check without
disturbing you during the test.
• Turn your cell phone completely off (not just in vibrate mode) and store it with your other
belongs.
• No aids are permitted on this examination. Examples of illegal aids include but are not limited
to textbooks, notes, calculators, cellphones, or any electronic device.
• Write clearly and concisely in a line fashion by using dark pencil or pen. Work scattered all over
the page without a clear ordering will receive very little credit.
• For questions with a boxed area, ensure your answer is completely inside the box.
• If you need more space, use the blank pages at the end of the exam and clearly indicate on the
question page when you have done this. Do not tear any pages off this exam.
Question:
1
2
3
4
5
6
7
Total
Points:
4
5
6
9
6
4
6
40
Score:
Do not tear this page off. This page is a formula sheet and can be used for
rough work only. It will not be graded under any circumstances.
tan θ =
sin θ
cos θ
sec θ =
sin2 θ + cos2 θ = 1
cos 2θ = cos2 θ − sin2 θ
1
cos θ
cot θ =
cos θ
1
=
sin θ
tan θ
1 + tan2 θ = sec2 θ
sin 2θ = 2 sin θ cos θ
csc θ =
1
sin θ
1 + cot2 θ = csc2 θ
tan 2θ =
2 tan θ
1 − tan2 θ
= 2 cos2 θ − 1
= 1 − 2 sin2 θ
cos2 θ =
1 + cos 2θ
2
sin2 θ =
1
cos(A − B) − cos(A + B)
2
1
cos A cos B =
cos(A − B) + cos(A + B)
2
1
sin A cos B =
sin(A + B) + sin(A − B)
2
sin A sin B =
c2 = a2 + b2 − 2ab cos C
1 − cos 2θ
2
tan2 θ =
1 − cos 2θ
1 + cos 2θ
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B − sin A sin B
tan(A + B) =
tan A + tan B
1 − tan A tan B
sin A
sin B
sin C
=
=
a
b
c
You may use the rest of this formula sheet page for rough work. It will not be graded. If you
need more space for a solution, please use one of the blank pages at the back of the test.
1. For each question below, only your final answer will be graded. No justification is necessary.
Z
(1a) (1 point) For which values of q is the integral
0
1
1
x2q−1
dx convergent?
Final Answer
q<1
Remark:The integral is improper when 2q − 1 > 0, i.e. q > 1/2. For 2q − 1 ≤ 0, the integral
Z 1
1
dx is a definite integral and we could compute it by using FTC. So both q < 1 and
2q−1
x
0
1/2 < q < 1 are marked correct in this problem.
ln n + 2n + 6n!
.
(1b) (1 point) Evaluate lim n
n→∞ 3 − 2n! + n5
Final Answer
-3
Z
π
2
(1c) (1 point) Compute
sin5 x cos x dx.
0
Final Answer
1
6
Z
(1d) (1 point) Compute
π
cos2 x dx.
0
Final Answer
π
2
2. For each question below, only your final answer will be graded. No justification is necessary.
∞
Let {an }n=0 be a sequence.
∞
(2a) (2 points) Write the formal definition of “The sequence {an }n=0 is divergent to ∞.”
Final Answer
∀M > 0, ∃n0 ∈ N, s.t. ∀n ∈ N, n ≥ n0 =⇒ an > M
Or
∀M ∈ R, ∃n0 ∈ N, s.t. ∀n ∈ N, n ≥ n0 =⇒ an > M
∞
(2b) (1 point) Write the formal definition of “The sequence {an }n=0 is (strictly) increasing.”
Final Answer
∀n ∈ N, an < an+1
Or
∀n, m ∈ N, n < m =⇒ an < am
(2c) (2 points) State the Monotone Convergence Theorem for Sequences. Make sure to specify all assumptions.
Final Answer
If a sequence is (eventually) monotonic and bounded, then it is convergent.
3. (6 points) For each statement, determine whether it is TRUE or FALSE. Circle the correct answer.
Again, no justification is necessary.
(3a)
∞
• IF {an }n=0 diverges and is unbounded above THEN lim an = ∞.
TRUE
FALSE
∞
TRUE
FALSE
∞
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
TRUE
FALSE
n→∞
• If {an }n=0 converges THEN it is bounded.
• If {an }n=0 converges THEN it is eventually monotonic.
Let f and g be positive, continuous functions on [1, ∞).
Z ∞
f (x)
= ∞ and
f (x)dx diverges
(3b) • IF lim
x→∞ g(x)
1
Z ∞
THEN
g(x)dx diverges.
1
f (x)
= 0 and
• IF lim
x→∞ g(x)
Z
∞
THEN
Z
∞
g(x)dx converges
1
f (x)dx converges.
1
• IF
Z
∞
f (x)dx converges
Z ∞
f (x) cos2 x dx converges.
THEN
1
1
4. (9 points) Please show all of your work to get full marks.
Z
ln x
(4a) Compute
dx. We use integration by parts to compute this question.
x3
Let u = ln x and dv = x13 dx = x−3 dx. Thus, du = x1 dx and v = − 21 x−2 .
Z
1 −2
1
1
ln x
dx = ln x · − x
− − x−2 · dx
3
x
2
2
x
Z
ln x 1
=− 2 +
x−3 dx
2x
2
ln x 1
= − 2 − x−2 + C
2x
4
Z
Z
(4b) Let f be a twice-differentiable function with domain R. We know
3
f (x)dx = 1. We also know
1
the following table:
3
Z
f 00 (x)x2 dx. We use integration by parts twice to solve this problem.
Compute
1
Let u = x2 and dv = f 00 (x)dx. Thus, du = 2xdx and v = f 0 (x).
Z
3
3
00
0
2
f (x)x dx = f (x)x
1
2
3
Z
0
−
0
0
3
0
Z
2xf (x)dx = −9+4−
2xf (x)dx = 9f (3)−f (1)−
1
1
Z
1
3
2xf 0 (x)dx
1
To compute the definite integral on the right hand side above, we use IBP again. Let u1 = 2x and
dv1 = f ”(x)dx. Thus, du1 = 2dx and v1 = f (x).
Z
3
3
2xf 0 (x)dx = 2xf (x)
1
Z
3
−
Z
2f (x)dx = 6f (3) − 2f (1) − 2
1
1
3
f (x)dx = 18 − 2 − 2 · 1 = 14
1
Therefore,
Z
3
00
Z
2
f (x)x dx = −9 + 4 −
1
Z
(4c) Compute
x2
3
2xf 0 (x)dx = −5 − 14 = −19
1
3x
dx. First, we use partial fraction decomposition to rewrite the integrand
−x−2
x2
3x
3x
A
B
=
=
+
.
−x−2
(x − 2)(x + 1)
x−2 x+1
Thus,
3x = A(x + 1) + B(x − 2) =⇒ A + B = 3, A − 2B = 0 =⇒ A = 2, B = 1
Therefore,
Z
3x
dx =
2
x −x−2
Z
2
1
+
dx = 2 ln |x − 2| + ln |x + 1| + C
x−2 x+1
5. (6 points) Let R be the bounded region delimited by the curves of equations y = 2x and y =
√
4x.
√
(5a) Find the area of this region. First, we find the points of intersection between y = 2x and y = 4x.
These occurs when
√
2x = 4x =⇒ 4x2 = 4x =⇒ 4x2 − 4x = 0 =⇒ 4x(x − 1) = 0 =⇒ x = 0, 1
√
Thus, the points of intersection are (0, 0), (1, 2). Notice the curve y = 4x is above the straight
line y = 2x, so the area of this region is given by
Z 1√
[ 4x − 2x] dx
A=
0
Z
=2
1
√
[ x − x] dx
0
2
1
= 2 x3/2 − x2
3
2
2 1
=2
−
−0
3 2
1
=
3
1
0
(5b) Write out a definite integral with respect to x (don’t compute) which represents the volume of
solid of revolution obtained by rotating R around y = 2.
We use disk/washer method here. The
√ cross section is a washer. The outer radius is 2−yout =√2−2x
and the inner radius is 2−yin = 2− 4x. The area of the cross section at x is π(2−2x)2 −π(2− 4x)2 .
Therefore, the volume of this solid is
Z 1
√
[π(2 − 2x)2 − π(2 − 4x)2 ] dx
0
(5c) Write out a definite integral with respect to y (don’t compute) which represents the volume of
solid of revolution obtained by rotating R around y = 0.
We use the cylindrical shells method here. The height of the shell is xout − xin = xright − xlef t =
√
2
y
y2
y
4x =⇒ x = y4 .The
2 − 4 since the right curve is y = 2x =⇒ x = 2 and the left curve is y =
radius of the base of the shell is y. Therefore, the volume of the solid is
Z 2
y y2
2πy
−
dy
2
4
0
Z
∞
1
dx. Justify your answer.
x(1 + x)
1
First, by the definition of the improper integral, we have
Z b
Z ∞
1
1
√
√
dx = lim
dx
b→∞
x(1 + x)
x(1 + x)
1
1
6. (4 points) Evaluate the given improper integral
√
b
√
1
√
Second, we now compute the definite integral
dx by using substitution. Let u = x.
x(1
+
x)
1
√
√
√
1
Then x = u2 , du = 2√
dx.Thus, dx = 2 x du = 2u du. Since x ∈ [1, b], we have u = x ∈ [1, b].
x
Z
√
Z √b
b
√
1
1
√
·
2
du
=
2
arctan(u)
dx =
= 2 arctan( b) − 2 arctan(1)
2
1+u
x(1 + x)
1
1
1
√
Third, since lim arctan( b) = π2 and arctan(1) = π4 , we have
Z
b
b→∞
Z
1
∞
√
√
π
π
1
π
dx = lim [2 arctan( b) − 2 arctan(1)] = 2 · − 2 · =
b→∞
2
4
2
x(1 + x)
∞
7. (7a) (1 point) Write the formal definition of “The sequence {an }n=0 is convergent.”
∃L ∈ R, s.t. ∀ > 0, ∃n0 ∈ N s.t. ∀n ∈ N, n ≥ n0 =⇒ |an − L| < (7b) (5 points) Prove the following theorem:
∞
∞
Let{an }n=0 and {bn }n=0 be two sequences. In this problem, N = {0, 1, 2, · · · }. IF
1. lim an = L
n→∞
2. ∀n ∈ N, L ≤ bn ≤ an .
THEN lim bn = L.
n→∞
Note: You need to write a proof directly from the definition of limit of a sequence, without using
any other theorems.
Proof: We want to show: ∀ > 0, ∃n0 ∈ N such that
∀n ∈ N, n0 ≥ N =⇒ |bn − L| < ⇔ L − < bn < L + (1)
• Let > 0 be fixed.
• I use this same value of in the definition of lim an = L.
n→∞
By the definition, we know there exists n0 ∈ N such that
∀n ∈ N, n0 ≥ N =⇒ |an − L| < ⇔ L − < an < L + • I am going to take this same value of n0 and I will show it satisfies (1).
• More specifically, let n ∈ N. Assume n ≥ n0 . I need to show L − < bn < L + .
– We have L − < L because > 0.
– We have L ≤ bn ≤ an by hypothesis.
– We have an < L + from (2)
Putting them all together
L − < 7 ≤ bn ≤ an < L + so I have shown L − < bn < L + , as needed.
(2)
Do not tear this page off. This page is for additional work and will not be
marked, unless you clearly indicate it on the original question page.
Do not tear this page off. This page is for additional work and will not be
marked, unless you clearly indicate it on the original question page.
Do not tear this page off. This page is for additional work and will not be
marked, unless you clearly indicate it on the original question page.