MATH 101 Quiz #4 (v.A1)
Last Name:
Friday, March 4
First Name:
Grade:
Student-No:
Section:
Very short answer question
1. 1 mark Find the coefficient of
x
Simplify your answer completely.
1
1
in the partial fraction expansion of
4x3 3x2 + 7
.
x2 (x 1)(3x2 2x + 1)
Answer: 4
Solution: The partial fraction expansion has the form
4x3 3x2 + 7
A
=
+ various terms;
2
2
x (x 1)(3x
2x + 1)
x 1
when we multiply through by the original denominator, this becomes
4x3
3x2 + 7 = x2 (3x2
2x + 1)A + (x
Evaluating both sides at x = 1 yields 4 · 13
1)(other terms).
3 · 12 + 7 = 12 (3 · 12
2 · 1 + 1)A + 0, or A = 4.
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.
+ 4)3/2
Solution: Let x = 2 tan ✓, so that x2 + 4 = 4 tan2 ✓ + 4 = 4 sec2 ✓ and dx = 2 sec2 ✓ d✓.
Then
Z
Z
1
1
dx =
· 2 sec2 ✓ d✓
2
3/2
(x + 4)
(4 sec2 ✓)3/2
Z
2 sec2 ✓
=
d✓
8 sec3 ✓
Z
4
2 +
1
p
x
=
cos ✓ d✓
x
4
✓
1
1
x
2
= sin ✓ + C = p
+ C.
2
4
4 x +4
The fact that sin ✓ =
p x
x2 +4
when tan ✓ =
x
2
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 cos(x3 ) dx is estimated using the Midpoint Rule with 1000 points.
Show that the error in this approximation is at most 5 · 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) = cos(x3 ). Then f 0 (x) =
f 00 (x) =
3x2 sin(x3 ) and
6x sin(x3 )
9x4 cos(x3 ).
Since |x4 |  1 when |x|  1, we have | 6x sin(x3 ) 9x4 cos(x3 )|  6 + 9 = 15 since
|sin ✓|  1 and |cos ✓|  1 for all ✓. We can therefore choose K = 15, and it follows that
the error is at most
15 · 23
5
= 6 = 5 · 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 x-coordinate of the centre of mass of the (infinite) region lying to the right
of the line x = 2, above the x-axis, and below the graph of y = 18/x4 .
Solution: The area of the region is
✓Z t
◆
Z 1
18
18
dx = lim
dx
4
t!1
x4
2
2 x
✓
t◆
6
= lim
t!1
x3 2
✓
◆
6
6
3
= lim
+ 3 =0+ .
3
t!1
t
2
4
To find the x-coordinate of the centre of mass, the total moment is
✓Z t
◆
Z 1
Z 1
18
18
18
x · 4 dx =
dx = lim
dx
3
t!1
x
x3
2
2
2 x
✓
t◆
9
= lim
t!1
x2 2
✓
◆
9
9
9
= lim
+ 2 =0+ .
2
t!1
t
2
4
The x-coordinate of the centre of mass is therefore x̄ =
9/4
3/4
= 3.
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