MATH 101 Quiz #3 (v.M2)
Last Name:
Friday, February 12
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
1. 1 mark Evaluate
tan3 x sec37 x dx.
Answer:
sec39 x sec37 x
−
+C
39
37
Solution: Use the substitution u = sec x, so that du = sec x tan x dx:
Z
Z
Z
3
37
2
36
tan x sec x dx = (sec x − 1) sec x (sec x tan x) dx = (u2 − 1)u36 du
u39 u37
sec39 x sec37 x
=
−
+C =
−
+C
39
37
39
37
Alternatively, write the original integral in terms of sines and cosines and use the substitution u = cos x, so that du = − sin x dx:
Z
Z
sin3 x
3
37
dx
tan x sec x dx =
cos40 x
Z
Z
1 − cos2 x
1 − u2
=
sin
x
dx
=
−
du
cos40 x
u40
1
1
1
1
=
−
+C =
−
+C
39
37
39
39u
37u
39 cos x 37 cos37 x
Marking scheme: 1 for either correct answer in the box
Short answer questions—you must show your work
Z
2. 2 marks Evaluate
log x
dx.
x2
Solution: Integrate by parts, using u = log x and dv = x12 dx, so that v = − x1 and
du = x1 dx:
Z
Z Z
1
log x
11
log x
1
dx
=
−
log
x
−
−
dx
=
−
+
dx
x2
x
x x
x
x2
log x 1
= −
− +C
x
x
Marking scheme:
• 1 mark for integrating by parts with u = log x
• 1 mark for the correct answer
3. 2 marks A 6-meter-long cable of mass 5 kg is used to lift a bucket off the ground. How much
work is needed to raise the entire cable to height 6 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
5 1 − y6 kg. (Note that this linear function equals 5 when y = 0 and 0 when
y = 6.) So,
y
at height y, the cable is subject to a downward gravitational force of 5 1 − 6 · 9.8; to raise
the cable we need to apply a compensating upward force of 5 1 − y6 · 9.8. So the work
required is
6
Z 6 y
y2 5 1−
· 9.8 dy = 5 y −
· 9.8 = 5 · 3 · 9.8 = 147 J.
6
12
0
0
Alternatively, the cable has linear density 5 kg/6 m = 65 kg/m, and so the work required to
lift a small piece of the cable (of length ∆y) from height y m to height 6 m is 56 ∆y·9.8(6−y).
The total work required is therefore
Z 6
5
5
· 9.8(6 − y) dy = · 9.8 · 18 = 147 J
6
0 6
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 c > 0 be a constant. Let R be the finite region bounded by the graph of
2
y = 2 + xex , the line y = 2, and the line x = c. Using vertical slices, find the volume
generated when R is rotated about the line y = 2.
√ 2
Solution: Let f (x) = 2 + xex . On the vertical slice a distance x from the y-axis,
sketched in the figure below, y runs from 2 to f (x). Upon rotation about the line y = 2,
this slice sweeps out a cylinder of thickess ∆x and radius f (x) − 2 and hence of volume
π[f (x) − 2]2 ∆x. The full volume generated (for any fixed c > 0) is
Z c
Z c
2
2
π[f (x) − 2] dx = π
xe2x dx.
√ 2
0
0
y = 2 + xex
y
Using the substitution u = 2x2 , so that du = 4x dx:
Z
Volume = π
0
2c2
eu
π 2c2
du
π 2c2
= eu =
e −1
4
4 0
4
y=2
x=c 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