MATH 101 Quiz #3 (v.M1)
Last Name:
Friday, February 12
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
1. 1 mark Evaluate
tan62 x sec4 x dx.
tan65 x tan63 x
Answer:
+
+C
65
63
Solution: Use the substitution u = tan x, so that du = sec2 x dx:
Z
Z
Z
62
4
62
2
2
tan x sec x dx = tan x (tan x + 1) sec x dx = u62 (u2 + 1) du
=
tan65 x tan63 x
u65 u63
+
+C =
+
+C
65
63
65
63
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
Z
2. 2 marks Evaluate
log x
dx.
x3
Solution: Integrate by parts, using u = log x and dv = x13 dx, so that v = − 2x12 and
du = x1 dx:
Z
Z Z
1
1 1
log x
log x
1
dx = − 2 log x −
− 2
dx = − 2 +
dx
3
x
2x
2x x
2x
2x3
log x
1
= − 2 − 2 +C
2x
4x
Marking scheme:
• 1 mark for integrating by parts with u = log x
• 1 mark for the correct answer
3. 2 marks A 10-meter-long cable of mass 7 kg is used to lift a bucket off the ground. How
much work is needed to raise the entire cable to height 10 m? Ignore the weight of the bucket
and its contents. Use g = 9.8 m/s2 for the acceleration due to gravity. A calculator-ready
answer is acceptable.
Solution:
When the bucket is at height y, the cable that remains to be lifted has mass
y
7 1 − 10
kg. (Note that this linear function equals 7 when y = 0 and 0 when y = 10.)
y
· 9.8;
So, at height y, the cable is subject to a downward gravitational force of 7 1 − 10
y
to raise the cable we need to apply a compensating upward force of 7 1 − 10
· 9.8. So the
work required is
10
Z 10 y
y2 7 1−
· 9.8 dy = 7 y −
· 9.8 = 7 · 5 · 9.8 = 343 J.
10
20
0
0
Alternatively, the cable has linear density 7 kg/10 m = 0.7 kg/m, and so the work required
to lift a small piece of the cable (of length ∆y) from height y m to height 10 m is 0.7∆y ·
9.8(10 − y). The total work required is therefore
Z 10
0.7 · 9.8(10 − y) dy = 0.7 · 9.8 · 50 = 343 J
0
as before.
Marking scheme: Full marks (2) for a correct integral representing the total work, even if
that integral isn’t evaluated. Partial credit (1 mark) possible for a reasonable attempt with
minor mistakes.
Long answer question—you must show your work
4. 5 marks √Let a > 0 be a constant. Let R be the finite region bounded by the graph of
2
y = 1 + xex , the line y = 1, and the line x = a. Using vertical slices, find the volume
generated when R is rotated about the line y = 1.
√ 2
Solution: Let f (x) = 1 + xex . On the vertical slice a distance x from the y-axis,
sketched in the figure below, y runs from 1 to f (x). Upon rotation about the line y = 1,
this slice sweeps out a cylinder of thickess ∆x and radius f (x) − 1 and hence of volume
π[f (x) − 1]2 ∆x. The full volume generated (for any fixed a > 0) is
Z a
Z a
2
2
π[f (x) − 1] dx = π
xe2x dx.
√ 2
0
0
y = 1 + xex
y
Using the substitution u = 2x2 , so that du = 4x dx:
Z
Volume = π
0
2a2
eu
π 2a2
π 2a2
du
= eu =
e −1
4
4 0
4
y=1
x=a x
Marking scheme:
• 2 marks for the correct slice volume
• 1 mark for an integral giving the full volume, assuming that the slice volume is correct
• 1 mark for a correct substitution that could be used to evaluate their integral
• 1 mark for the correct answer