MATH 101 Quiz #3 (v.T3) Thursday, February 11 Grade:
Last Name: First Name: Student-No: Section:
1. 1 mark Find the average value of f ( x ) = sin(7 x ) over the interval − π/ 2 ≤ x ≤ π/ 2. Your answer must be fully simplified.
Answer: 0
Solution: By definition, the average value is
1 Z
π/ 2 sin(7 x ) dx.
π
− π/ 2
We now observe that sin(7 x ) is an odd function, and hence the integral over the symmetric interval [ − π
2
, π
2
] equals zero.
Alternatively, the integral equals (by FTC2)
− cos(7 x )
7 π
π/ 2
− π/ 2
=
− cos(7 π/ 2)
7 π
+ cos( − 7 π/ 2)
7 π
= 0 − 0 = 0 .
Marking scheme: 1 for a correct answer in the box, either by integrating or by observing that sin(7 x ) is an odd function
2. 2 marks An 11-meter-long cable of mass 6 kg is used to lift a bucket off the ground. How much work is needed to raise the entire cable to height 11 m? Ignore the weight of the bucket and its contents. Use g = 9 .
8 m / s 2 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
6 1 − y
11 kg. (Note that this linear function equals 6 when y = 0 and 0 when y = 11.)
So, at height y , the cable is subject to a downward gravitational force of 6 1 to raise the cable we need to apply a compensating upward force of 6 1 − y
− y
· 9 .
8;
11
· 9 .
8. So the
11 work required is
Z
11
6 1 −
0 y
11 y 2
· 9 .
8 dy = 6 y −
22
· 9 .
8
0
11
= 6 ·
11
· 9 .
8 J .
2
Alternatively, the cable has linear density 6 kg / 11 m, and so the work required to lift a small piece of the cable (of length ∆ y ) from height y m to height 11 m is
6
11
∆ y · 9 .
8(11 − y ).
The total work required is therefore
Z
11
0
6
11
· 9 .
8(11 − y ) dy = 33 · 9 .
8J 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
9
4 log x
√ x dx.
Solution: This one cries out for integration by parts, with u ( x ) = log x and v
Therefore, u
0
( x ) = 1 /x and v ( x ) = 2 x , and so
0
( x ) = 1 /
√ x .
Z
9
4 log x
√ x dx = 2
√ x log x
9
− 2
Z
9
4 4
= 6 log 9 − 4 log 4 − 4
√ x
√ x dx x
9
4
= 6 log 9 − 4 log 4 − (12 − 8) = 6 log 9 − 4 log 4 − 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.
4. 5 marks Let R be the region between the curves T ( x ) =
√ xe
2 x and B ( x ) =
√ x (1 + 2 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 :
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V =
Z
3
π [ T ( x )
2 − B ( x )
2
] dx
= π
0
Z
3
(
√ xe
2 x
)
2
0
Z
3
= π xe
4 x
− (
√ x (1 + 2
− ( x + 4 x
2
+ 4 x x
3
))
)
2 dx dx
= π
0
Z
3 xe
4 x dx − π x
2
+
4 x
3
+ x
4
3
= π
0
Z
3
0 xe
4 x dx − π
3
2
2
2
+
4
3
·
3
3
3
+ 3
4
0
.
For the first integral, we use integration by parts with u ( x ) = x , v
0
( x ) = e 4 x , so that u
0
( x ) = 1 and v ( x ) =
1
4 e 4 x :
Z
3 xe
4 x dx =
0
= xe
4 x
3 e
4
4
12
3
0
−
Z
3
0
1
4 e
4 x dx
3
− 0 −
1
16 e
4 x
0
=
3 e 12
−
4
Therefore, the total volume is e 12
16
1
−
16
.
V = π
3 e 12
−
4 e 12
16
1
−
16
− π
3 2
2
+
4 · 3 3
3
+ 3
4
= π
11 e 12 − 1943
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
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