MATH 101 Quiz #3 (v.T2)
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(5x) + 1 over the interval −π/2 ≤ x ≤ π/2. Your
answer must be fully simplified.
Answer: 1
Solution: By definition, the average value is
1
π
Z
π/2
sin(5x) + 1 dx.
−π/2
We now observe that sin(5x) is an odd function, and hence its integral over the symmetric
interval [− π2 , π2 ] equals zero. So the average value of f (x) on this interval is 1.
Alternatively, the integral equals (by FTC2)
π/2
− cos(5x)
− cos(5π/2) π cos(−5π/2) −π
+ x
=
+ +
−
= π.
5
5
2
5
2
−π/2
Marking scheme: 1 for a correct answer in the box, either by integrating or by observing
that sin(2x) is an odd function
Short answer questions—you must show your work
2. 2 marks A 9-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 9 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 − 9 kg. (Note that this linear function equals 8 when y = 0 and 0 when
y = 9.) So,
at height y, the cable is subject to a downward gravitational force of 8 1 − y9 · 9.8; to raise
the cable we need to apply a compensating upward force of 8 1 − y9 · 9.8. So the work
required is
9
Z 9 y
y2 9
8 1−
· 9.8 dy = 8 y −
· 9.8 = 8 · · 9.8 J.
9
18
2
0
0
Alternatively, the cable has linear density 8 kg/9 m, and so the work required to lift a
small piece of the cable (of length ∆y) from height y m to height 9 m is 89 ∆y · 9.8(9 − y).
The total work required is therefore
Z 9
8
· 9.8(9 − y) dy = 36 · 9.8J
0 9
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
4
√
3 x log x dx.
1
√
Solution: This one cries out for integration by parts, with u(x) = log x and v 0 (x) = 3 x.
Therefore, u0 (x) = 1/x and v(x) = 2x3/2 , and so
Z 4
Z 4 3/2
4
√
x
3/2
3 x log x dx = 2x log x − 2
dx
x
1
1
1
4
4 3/2 x
= 16 log 4 − 2 log 1 −
3
1
32 4
28
= 16 log 4 −
−
= 16 log 4 − .
3
3
3
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) = xe3x and B(x) = x(1 + 2x)
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
√
√
( xe3x )2 − ( x(1 + 2x))2 dx
Z0 3
xe6x − (x + 4x2 + 4x3 ) dx
0
2
3
Z 3
x
4x3
6x
4 xe dx − π
=π
+
+x 2
3
0
2
0
Z 3
3
3
4·3
xe6x dx − π
=π
+
+ 34 .
2
3
0
=π
For the first integral, we use integration by parts with u(x) = x, v 0 (x) = e6x , so that
u0 (x) = 1 and v(x) = 16 e6x :
Z
0
3
3 Z 3
xe6x 1 6x
xe dx =
−
e dx
6 0
0 6
3
18
3e18
e18
1 6x e
1
=
−0− e =
−
−
.
6
36
2
36
36
0
6x
Therefore, the total volume is
18 18
2
e
e
1
3
4 · 33
17e18 − 4373
4
−
−
+
+3 =π
V =π
−π
.
2
36
36
2
3
36
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