MATH 101 Quiz #3 (v.M3)
Last Name:
Friday, February 12
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
1. 1 mark Evaluate
sec4 x tan46 x dx.
tan49 x tan47 x
Answer:
+
+C
49
47
Solution: Use the substitution u = tan x, so that du = sec2 x dx:
Z
Z
Z
4
46
2
46
2
sec x tan x dx = (tan x + 1) tan x sec x dx = (u2 + 1)u46 du
=
tan49 x tan47 x
u49 u47
+
+C =
+
+C
49
47
49
47
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.
x4
Solution: Integrate by parts, using u = log x and dv = x14 dx, so that v = − 3x13 and
du = x1 dx:
Z
Z Z
1
1 1
log x
log x
1
dx = − 3 log x −
− 3
dx = − 3 +
dx
4
x
3x
3x x
3x
3x4
log x
1
= − 3 − 3 +C
3x
9x
Marking scheme:
• 1 mark for integrating by parts with u = log x
• 1 mark for the correct answer
3. 2 marks A 5-meter-long cable of mass 8 kg is used to lift a bucket off the ground. How much
work is needed to raise the entire cable to height 5 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
8 1 − 5 kg. (Note that this linear function equals 8 when y = 0 and 0 when
y = 5.) So,
y
at height y, the cable is subject to a downward gravitational force of 8 1 − 5 · 9.8; to raise
the cable we need to apply a compensating upward force of 8 1 − y5 · 9.8. So the work
required is
5
Z 5 y2 y
· 9.8 dy = 8 y −
· 9.8 = 8 · 2.5 · 9.8 = 196 J.
8 1−
5
10
0
0
Alternatively, the cable has linear density 8 kg/5 m = 1.6 kg/m, and so the work required
to lift a small piece of the cable (of length ∆y) from height y m to height 5 m is 1.6∆y ·
9.8(5 − y). The total work required is therefore
Z 5
1.6 · 9.8(5 − y) dy = 1.6 · 9.8 · 12.5 = 196 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
√ xb2 > 0 be a constant. Let R be the finite region bounded by the graph of
y = −1 + xe , the line y = −1, and the line x = b. 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 b > 0) is
Z
b
2
Z
π[f (x) + 1] dx = π
0
Using the substitution u = 2x2 , so that du = 4x dx:
Z
Volume = π
0
2b2
0
b
2
xe2x dx.
y
y = −1 +
du
π u 2b2
π 2b2
e
= e =
e −1
4
4 0
4
u
√
x
y = −1
x=b
xex
2
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