MATH 101 Quiz #3 (v.T1)
Last Name:
Thursday, February 11
First Name:
Grade:
Student-No:
Section:
Very short answer question
1. 1 mark Find the average value of f (x) = sin(3x) over the interval −π/2 ≤ x ≤ π/2. Your
answer must be fully simplified.
Answer: 0
Solution: By definition, the average value is
1
π
Z
π/2
sin(3x) dx.
−π/2
We now observe that sin(3x) is an odd function, and hence the integral over the symmetric
interval [− π2 , π2 ] equals zero.
Alternatively, the integral equals (by FTC2)
π/2
− cos(3x) − cos(3π/2) cos(−3π/2)
=
+
= 0 − 0 = 0.
3
3
3
−π/2
Marking scheme: 1 for a correct answer in the box, either by integrating or by observing
that sin(3x) is an odd function
Short answer questions—you must show your work
2. 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
So, at height y, the cable is subject to a downward gravitational force of 7 1 − 10
· 9.8;
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.
3. 2 marks Evaluate the following integral, leaving your answer in calculator-ready form.
Z
1
4
log x
dx.
x3/2
Solution: This one cries out for integration
by parts, with u(x) = log x and v 0 (x) = 1/x3/2 .
√
Therefore, u0 (x) = 1/x and v(x) = −2/ x, and so
4 Z 4
Z 4
log x
−2
−2
−
√
√
dx
=
log
x
dx
3/2
x
x·x
1 x
1
1
4
−2 = − log 4 + 2 log 1 + 2 √
x 1
= − log 4 + 2(−1 − (−2)) = 2 − log 4.
Marking scheme:
• 1 mark for correctly using integration by parts. (Solutions that begin with a substitution will need to use integration by parts at some point.)
• 1 mark for the correct answer.
Long answer question—you must show your work
√
√
4. 5 marks Let R be the region between the curves T (x) = xe2x and B(x) = x(1 + x) on the
interval 0 ≤ x ≤ 3. (It is true that T (x) ≥ B(x) for all 0 ≤ x ≤ 3.) Compute the volume of the
solid formed by rotating R about the x-axis. You may leave your answer in calculator-ready
form.
Solution: For a fixed value of x, if we rotate about the x-axis, we form an annulus of
area π[T (x)2 − B(x)2 ]. We integrate this function from x = 0 to x = 3 to find the total
volume V :
3
Z
π[T (x)2 − B(x)2 ] dx
V =
0
Z
3
=π
√
√
( xe2x )2 − ( x(1 + x))2 dx
Z0 3
xe4x − (x + 2x2 + x3 ) dx
0
2
3
Z 3
x
2x3 x4 4x
xe dx − π
=π
+
+
2
3
4 0
0
2
Z 3
3
2 · 33 34
4x
xe dx − π
+
+
.
=π
2
3
4
0
=π
For the first integral, we use integration by parts with u(x) = x, v 0 (x) = e4x , so that
u0 (x) = 1 and v(x) = 14 e4x :
Z
3
0
3 Z 3
xe4x 1 4x
xe dx =
−
e dx
4 0
0 4
3
12
3e12
3e12
1 4x e
1
=
−0− e =
−
−
.
4
16
4
16
16
0
4x
Therefore, the total volume is
12 12
2
e
3e
1
3
2 · 33 34
11e12 − 683
−
−
+
+
V =π
−π
=π
.
4
16
16
2
3
4
16
Marking scheme:
• 2 marks for writing down the integral for V
• 1 mark for correct evaluation of the polynomial part of the integral
• 2 marks for the correct evaluation of the rest of the integral