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Quiz #4 (v.T3): Page 1 of 4
Thursday, March 9
Short answer question
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Only your
answers in the boxes will be marked (no partial credit).
(a) A variable force moves an object along the x–axis. When the object has x-coordinate x,
the force is F (x) = kx newtons. If the work done in moving the object from x = 2 meters
to x = 4 meters is 24 joules, what is the value of k? Simplify your answer completely, but
don’t worry about the units of k.
Answer: k = 4
Solution: By definition, the work done in moving the object from x = 2 meters to
x = 4 meters by the force F (x) is
4
Z 4
Z 4
k
k 2 F (x) dx =
kx dx = x = (42 − 22 ) = 6k.
W =
2
2
2
2
2
To have W = 24, we need k = 4.
(b) Express as a definite integral the x–coordinate of the centroid of the region to the right
of the y–axis and to the left of x2 + y 2 = 4. Do not evaluate the integral.
Z
1 2 √
Answer:
x 4 − x2 dx
π 0
Solution: We use vertical strips, as in the sketch below.
x2 + y 2 = 4
(2, 0)
The equations of the top and bottom of the specified region are
√
√
y = T (x) = 4 − x2
and
y = B(x) = − 4 − x2 .
The area of the half disk is A = 21 π22 = 2π. Using vertical slices,
Z
Z 2 √
√
1 2 1
x T (x) − B(x) dx =
x 4 − x2 − (− 4 − x2 ) dx
x̄ =
A 0
2π 0
Z 2 √
1
=
x 4 − x2 dx
π 0
Quiz #4 (v.T3): Page 2 of 4
Thursday, March 9
Long answer question—you must show your work
Z
2. 4 marks Determine whether the integral
0
verges, evaluate it.
∞
8 −1/x2
e
dx converges or diverges; if it conx3
Answer: converges to 4
Solution: This is an improper integral because of a potential singularity at x = 0 and
also because of an infinite domain. So we split it into two integrals.
Z ∞
Z 1
Z ∞
8 −1/x2
8 −1/x2
8 −1/x2
e
dx
=
e
dx
+
e
dx
3
x3
x3
0
0 x
1
Z 1
Z R
8 −1/x2
8 −1/x2
e
e
= lim
dx + lim
dx.
3
t→0+ t x
R→∞ 1 x3
To evaluate both integrals we substitute u = −
Z
8 −1/x2
e
dx =
x3
Z
1
2
, so that du = 3 dx. Since
2
x
x
2
4eu du = 4eu + C = 4e−1/x + C,
we have
Z
0
∞
h
i1
h
iR
8 −1/x2
−1/x2
−1/x2
e
dx
=
lim
4e
+
lim
4e
t→0+
R→∞
x3
t
1
−1
2
2
= 4 lim e − e−1/t + 4 lim e−1/R − e−1
t→0+
R→∞
−1
−1
=4 e −0 +4 1−e
= 4.
Marking scheme:
• 1 mark for splitting up the integral at some point a > 0. (It is not necessary to
choose a = 1.)
• 1 mark for correctly formulating the improper integrals as limits of proper integrals.
• 1 mark for evaluating the proper integrals.
• 1 mark for the final answer.
Quiz #4 (v.T3): Page 3 of 4
Thursday, March 9
Long answer question—you must show your work
2
ey y 0
x
1
3. 4 marks Find the solution to
− = that obeys y(−1) = −3. Solve completely
2
4(x + 1)
y
y
for y as a function of x.
q Answer: y = − log e9 + 2(x + 1)4
Solution: This is a separable differential equation and we solve it accordingly.
2
2
ey y 0
1
x
ey y 0
x+1
y2 0
−
=
⇐⇒
=
⇐⇒
ye
y = 4(x + 1)3
4(x + 1)2 y
y
4(x + 1)2
y
Z
Z
1 2
y2
⇐⇒
e y dy = 4(x + 1)3 dx ⇐⇒ ey = (x + 1)4 + C
2
To determine C, we substitute in that y = −3 when x = −1.
e9
1 (−3)2
e
= (−1 + 1)4 + C = C ⇐⇒ C =
2
2
So
1 y2
e9
2
e = (x + 1)4 +
⇐⇒ ey = 2(x + 1)4 + e9 ⇐⇒ y 2 = log e9 + 2(x + 1)4
2
2
q ⇐⇒ y = − log e9 + 2(x + 1)4
We need the negative sign so that y(−1) = −3.
Marking scheme:
•
•
•
•
1 mark for separating the variables
1 mark for integrating both sides of the equation
1 mark for finding the value of the constant, assuming that they integrated correctly
1 mark for solving for y, assuming that their previous work was correct
Quiz #4 (v.T3): Page 4 of 4
Thursday, March 9