MATH 101 Quiz #4 (v.T3)
Last Name:
Thursday, March 3
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
∞
1. 1 mark What is the largest value of q for which the integral
1
1
dx diverges?
x5q
Answer: 1/5
Solution: Notice that
Z
t
1

1

(t1−5q − 1) with 1 − 5q > 0, if q < 15 ,

1−5q
1
dx = log t,
if q = 15 ,

x5q
 1
1
(1 − t5q−1
) with 5q − 1 > 0, if q > 15 .
5q−1
Therefore
Z
1
∞
1
dx = lim
t→∞
x5q
t
Z
1

1
1−5q
− 1) = ∞,
if q < 15 ,

 1−5q (limt→∞ t
1
dx = limt→∞ log t = ∞,
if q = 15 ,

x5q
 1
1
1
(1 − limt→∞ t5q−1
) = 5q−1
, if q > 15 ;
5q−1
the first two cases are divergent, and so the largest such value is q = 15 . (Alternatively, we
might recognize this as a “p-integral” with p = 5q, and recall that the p-integral diverges
precisely when p ≤ 1.)
Marking scheme: 1 mark for the correct answer.
Short answer questions—you must show your work
2. 2 marks The total error using the trapezoid rule to integrate f (x) over [a, b] is bounded by
M (b − a)3 /(12n2 ), where |f 00 (x)| ≤ M for all a ≤ x ≤ b.
R1
If the integral −2 2x4 dx is approximated using the trapezoid rule with 90 subintervals, what
is the largest possible error between the approximation T90 and the true value of the integral?
Show your work, but you may leave your answer in calculator-ready form.
Solution: Setting f (x) = 2x4 and b − a = 3, we compute f 00 (x) = 24x2 . The largest value
of 24x2 on the interval [−2, 1] occurs at x = −2, so we can take M = 24 · 22 = 96. Thus
the total error for the trapezoid rule with n = 90 points is bounded by
96 × 33
2
M (b − a)3
=
= .
2
12n
12 × 90 × 90
75
Marking scheme:
• 1 mark for a correct value of M (with enough written work to know that they’re not
guessing a value).
• 1 mark for a correct answer, given their value of M .
Z
3. 2 marks Evaluate
(x2
1
dx.
+ 25)3/2
Solution: Let x = 5 tan θ, so that x2 + 25 = 25 tan2 θ + 25 = 25 sec2 θ and dx = 5 sec2 θ dθ.
Then
Z
Z
1
1
dx =
· 5 sec2 θ dθ
2
3/2
(x + 25)
(25 sec2 θ)3/2
Z
5 sec2 θ
=
dθ
125 sec3 θ
Z
25
1
2 +
√
x
=
cos θ dθ
x
25
θ
1
1
x
5
√
+ C.
=
sin θ + C =
25
25 x2 + 25
The fact that sin θ =
√ x
x2 +25
when tan θ =
x
5
can be deduced from the right triangle above.
Marking scheme:
• 1 mark for a correct trigonometric substitution
• 1 mark for getting the right answer
Long answer question—you must show your work
4. 5 marks Evaluate the following indefinite integral using partial fractions:
Z
3x2 − 4
F (x) =
dx.
(x − 2)(x2 + 4)
Solution: This is of the form N (x)/D(x) with D(x) already factored and N (x) of lower
degree. We immediately look for a partial fractions separation:
3x2 − 4
A
Bx + C
=
+ 2
.
2
(x − 2)(x + 4)
x−2
x +4
Multiplying through and setting x = 2 we find:
8 = 12 − 4 = (4 + 4)A = 8A ⇒ A = 1.
Then we find:
3x2 − 4 − (x2 + 4) = (x − 2)(Bx + C) ⇒ B = 2, C = 4.
Thus, we have:
1
2x + 4
1
2x
4
3x2 − 4
=
+ 2
=
+ 2
+ 2
.
2
(x − 2)(x + 4)
x−2 x +4
x−2 x +4 x +4
The first two of these are directly integrable:
2
F (x) = log |x − 2| + log |x + 4| +
Z
4
dx.
x2 + 4
We substitute: y = x/2, dy = dx/2, and see that:
Z
Z
1
4
dx = 2
dy = 2 arctan y + D = 2 arctan(x/2) + D,
2
2
x +4
y +1
for any constant D. Finally we have:
F (x) = log |x − 2| + log |x2 + 4| + 2 arctan(x/2) + D.
Marking scheme:
• 1 mark for writing down the partial fractions in the correct general form.
• 2 marks for finding A, B, and C.
• 2 marks for the correct evaluation of the different integrals, given the values of A, B,
and C they found. (No marks lost if they forget the arbitrary constant in the final
antiderivative.)