MATH 101 Quiz #4 (v.A3)
Last Name:
Friday, March 4
First Name:
Grade:
Student-No:
Section:
Very short answer question
1. 1 mark Find the coefficient of
1
x
Simplify your answer completely.
1
in the partial fraction expansion of
3x3
x2 (x
2x2 + 11
.
1)(x2 + 3)
Answer: 3
Solution: The partial fraction expansion has the form
3x3
x2 (x
2x2 + 11
A
=
+ various terms;
2
1)(x + 3)
x 1
when we multiply through by the original denominator, this becomes
3x3
2x2 + 11 = x2 (x2 + 3)A + (x
Evaluating both sides at x = 1 yields 3 · 13
1)(other terms).
2 · 12 + 11 = 12 (12 + 3)A + 0, or A = 3.
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
2. 2 marks Evaluate
Z
(x2
1
dx.
+ 25)3/2
Solution: Let x = 5 tan ✓, so that x2 + 25 = 25 tan2 ✓ + 25 = 25 sec2 ✓ and dx = 5 sec2 ✓ d✓.
Then
Z
Z
1
1
dx =
· 5 sec2 ✓ d✓
2
3/2
(x + 25)
(25 sec2 ✓)3/2
Z
5 sec2 ✓
=
d✓
125 sec3 ✓
Z
25
1
2 +
p
x
=
cos ✓ d✓
x
25
✓
1
1
x
5
p
=
sin ✓ + C =
+ C.
2
25
25 x + 25
The fact that sin ✓ =
p x
x2 +25
when tan ✓ =
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
R1
3. 2 marks The integral 1 sin(x2 ) dx is estimated using the Midpoint Rule with 1000 points.
Show that the error in this approximation is at most 2 · 10 6 in absolute value.
Rb
You may use the fact that when approximating a f (x) dx with the Midpoint Rule using n
points, the absolute value of the error is at most K(b a)3 /24n2 where |f 00 (x)|  K for all
x 2 [a, b].
Solution: Let f (x) = sin(x2 ). Then f 0 (x) = 2x cos(x2 ) and
f 00 (x) = 2 cos(x2 )
4x2 sin(x2 ).
Since |x2 |  1 when |x|  1, we have |2 cos(x2 ) 4x2 sin(x2 )|  2 + 4 = 6 since |sin ✓|  1
and |cos ✓|  1 for all ✓. We can therefore choose K = 6, and it follows that the error is at
most
6 · 23
2
= 6 = 2 · 10 6 .
2
24 · 1000
10
Marking scheme:
• 1 point for correctly estimating f 00 (x) on the interval
• 1 point for arithmetic in the end
Long answer question—you must show your work
4. 5 marks Find the y-coordinate of the centre of mass of the (infinite) region lying to the right
of the line x = 1, above the x-axis, and below the graph of y = 10/x3 .
Solution: The area of the region is
✓Z t
◆
Z 1
10
10
dx = lim
dx
3
t!1
x3
1
1 x
✓
t◆
5
= lim
t!1
x2 1
✓
◆
5
5
= lim
+
= 0 + 5.
t!1
t2 12
To find the y-coordinate of the centre of mass, we compute
✓Z t
◆
Z 1 ✓ ◆2
1 10
50
dx = lim
dx
6
t!1
2 x3
1
1 x
✓
t◆
10
= lim
t!1
x5 1
✓
◆
10 10
= lim
+ 5 = 0 + 10.
t!1
t5
1
The y-coordinate of the centre of mass is therefore ȳ = 10/5 = 2.
Marking scheme:
• 1 mark for knowing the centre of mass’s coordinate is moment/area
• 1 mark for correct handling of an improper integral
• 1 mark for correct area of the region
• 1 mark for integrating by parts
• 1 mark for correct moment