MATH 101 Quiz #4 (v.M2)
Last Name:
Friday, March 4
First Name:
Student-No:
Grade:
Section:
Very short answer question
1. 1 mark Write out the general form of the partial-fractions decomposition of
You need not determine the values of any of the coefficients.
4x3 − 3
.
(x2 − 4)2 (1 + 2x2 )
Solution:
4x3 − 3
4x3 − 3
=
(x2 − 4)2 (1 + 2x2 )
(x − 2)2 (x + 2)2 (1 + 2x2 )
=
A
B
D
C
Ex + F
+
+
+
+
2
2
x − 2 (x − 2)
x + 2 (x + 2)
1 + 2x2
Marking scheme: 1 for a correct answer in the box. Of course the terms can be written in
any order, and the names of the constants don’t have to be the same.
Short answer questions—you must show your work
Z
2
2. 2 marks Determine whether the integral
−2
1
dx is convergent or divergent. If it is
(x + 1)4/3
convergent, find its value. Show all your work.
Solution: The integrand has a singularity at x = −1. So
Z 2
Z t
Z 2
1
1
1
dx = lim−
dx + lim+
dx
4/3
4/3
4/3
t→−1
t→−1
−2 (x + 1)
−2 (x + 1)
t (x + 1)
Since
Z
t
−2
t
1
3
3
3
dx = −
+
=−
4/3
1/3
1/3
(x + 1)
(x + 1)
(t + 1)
(−1)1/3
−2
diverges as t → −1− , the integral diverges . (A similar argument shows that the second
integral diverges, which is also enough to conclude that the original integral diverges.)
Marking scheme:
• 1 mark for observing the singularity at x = −1 (for example, by splitting up the domain
of integration).
• 1 mark for a correct argument giving divergence, using either of the two integrals. No
marks, for the word “divergence” without any justification.
3. 2 marks Express the x–coordinate of the centroid of the triangle with vertices (0, −1), (0, 1),
and (2, 0) in terms of a definite integral. Do not evaluate the integral.
(0, 1)
Solution:
y = 1 − x/2
(2, 0)
y = x/2 − 1
(0, −1)
The equations of the top and bottom of the triangle are
y = T (x) = 1 − x/2
and
y = B(x) = x/2 − 1.
The area of the triangle is A = 12 (2)(2) = 2. Using vertical slices,
1
x̄ =
A
2
Z
Z
1 2 x T (x) − B(x) dx =
x (1 − x/2) − (x/2 − 1) dx
2 0
Z
1 2 =
x 2 − x dx.
2 0
0
Marking scheme:
• 1 mark for a correct equation for either the top or the bottom of the triangle.
• 1 mark for a correct answer.
Long answer question—you must show your work
Z
4. 5 marks Evaluate
4
1
dx. A “calculator-ready” answer is acceptable, but your
(4 + x2 )3/2
answer may not contain inverse trigonometric functions.
0
Solution: Substitute x = 2 tan u, so that dx = 2 sec2 u du:
Z 4
Z arctan 2
1
1
dx =
2 sec2 u du
3/2
3/2
2
2
(4
+
x
)
0
0
(4 + 4 tan u)
Z
1 arctan 2 sec2 u
=
du
√
4 0
sec3 u
5
2
Z arctan 2
u
1
=
cos u du
1
4 0
arctan 2
1
= sin u
4
0
1
1
=
sin(arctan 2) − 0 = √
4
2 5
Marking scheme:
• 2 marks for correct substitution
• 1 mark for correct limits of integration
• 1 mark for evaluating the integral
• 1 mark for sin(arctan)