MATH 101 Quiz #4 (v.A2)
Last Name:
Friday, March 4
First Name:
Grade:
Student-No:
Section:
Very short answer question
1
3x3 − 2x2 + 9
in the partial fraction expansion of 2
.
x−1
x (x − 1)(x2 − 2x + 3)
Simplify your answer completely.
1. 1 mark Find the coefficient of
Answer: 5
Solution: The partial fraction expansion has the form
3x3 − 2x2 + 9
A
=
+ various terms;
2
2
x (x − 1)(x − 2x + 3)
x−1
when we multiply through by the original denominator, this becomes
3x3 − 2x2 + 9 = x2 (x2 − 2x + 3)A + (x − 1)(other terms).
Evaluating both sides at x = 1 yields 3 · 13 − 2 · 12 + 9 = 12 (12 − 2 · 1 + 3)A + 0, or A = 5.
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
Z
2. 2 marks Evaluate
(x2
1
dx.
+ 9)3/2
Solution: Let x = 3 tan θ, so that x2 + 9 = 9 tan2 θ + 9 = 9 sec2 θ and dx = 3 sec2 θ dθ.
Then
Z
Z
1
1
dx =
· 3 sec2 θ dθ
2
3/2
(x + 9)
(9 sec2 θ)3/2
Z
3 sec2 θ
=
dθ
27 sec3 θ
Z
9
2 +
1
√
x
=
cos θ dθ
x
9
θ
1
1
x
3
= sin θ + C = √
+ C.
2
9
9 x +9
The fact that sin θ =
√ x
x2 +9
when tan θ =
x
3
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(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 ∈ [a, b].
Solution: Let f (x) = cos(x2 ). Then f 0 (x) = −2x sin(x2 ) and
f 00 (x) = −2 sin(x2 ) − 4x2 cos(x2 ).
Since |x2 | ≤ 1 when |x| ≤ 1, we have |−2 sin(x2 ) − 4x2 cos(x2 )| ≤ 2 + 4 = 6 since |sin θ| ≤ 1
and |cos θ| ≤ 1 for all θ. 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 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 = 8/x3 .
Solution: The area of the region is
Z t
Z ∞
8
8
dx = lim
dx
3
t→∞
x3
2
2 x
t 4 = lim − 2 t→∞
x 2
4
4
= lim − 2 + 2 = 0 + 1.
t→∞
t
2
To find the x-coordinate of the centre of mass, the total moment is
Z t
Z ∞
Z ∞
8
8
8
x · 3 dx =
dx = lim
dx
2
t→∞
x
x2
2
2
2 x
t 8 = lim − t→∞
x 2
8 8
= lim − +
= 0 + 4.
t→∞
t 2
The x-coordinate of the centre of mass is therefore x̄ =
4
1
= 4.
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