MATH 101 Quiz #3 (v.A2)
Last Name:
Friday, February 12
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
1. 1 mark Evaluate
sin62 x cos3 x dx.
Answer:
sin63 x
63
−
sin65 x
65
+C
Solution: Make the substitution u = sin x, so that du = cos x dx:
Z
Z
Z
62
3
62
2
sin x cos x dx = sin x (1 − sin x) cos x dx = u62 (1 − u2 ) du
=
sin63 x sin65 x
u63 u65
−
+C =
−
+C
63
65
63
65
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
Z
2. 2 marks Evaluate
4y arctan(2y) dy.
Solution: We integrate by parts, using u = arctan(2y), dv = 4y dy so that v = 2y 2 and
2 dy
du = 1+(2y)
2:
Z
Z
2
4y arctan(2y) dy = 2y arctan(2y) −
We now notice that
Z
4y 2
4y 2 +1
=
4y 2 +1
4y 2 +1
4y 2
dy =
4y 2 + 1
−
1
.
4y 2 +1
Z 1−
4y 2
dy.
(2y)2 + 1
We therefore have
1
2
4y + 1
dy = y −
1
arctan(2y) + C.
2
The final answer is then
Z
1
4y arctan(2y) dy = 2y 2 arctan(2y) − y + arctan(2y) + C.
2
Marking scheme:
• 1 mark for correctly integrating by parts
• 1 mark for getting the right answer
3. 2 marks Find the work (in joules) required to stretch a string 20 cm beyond equilibrium, if
its spring constant is k = 30 N/m. Simplify your answer completely.
Answer: 0.6 J
Solution: By Hooke’s Law, the force exerted by the spring (and hence against the spring)
at displacement x m from its natural length is F = kx, where k is the spring constant.
Measuring distance in meters and force in newtons, the total work is
0.2 m
Z 0.2 m
1
1 2 = · 30 · (0.2)2 J = 0.6 J.
kx dx = kx 2
2
0
0
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 The finite region between the curves y = cos( x2 ) and y = x2 − π 2 is rotated about
the line y = −π 2 . Using vertical slices (disks and annuli), find the volume of the resulting
solid. A calculator-ready answer is acceptable. (Hint: to find where the curves intersect, look
at where they both vanish.)
Solution: The curves meet at x = ±π where they both vanish. The volume is therefore
Z πh
i
2 2
2
2
2 2
x
dx
cos( 2 ) − (−π ) − (x − π ) − (−π )
π
−π
Z π
=π
cos2 ( x2 ) + 2π 2 cos( x2 ) + π 4 − x4 dx
Z−π
π 1 + cos(x)
2
4
4
x
+ 2π cos( 2 ) + π − x dx
=π
2
−π
Z π
1 + cos(x)
2
4
4
x
= 2π
+ 2π cos( 2 ) + π − x dx
2
0
π
1
1
1 5 2
4
x
= 2π
x + sin(x) + 4π sin( 2 ) + π x − x 2
2
5
0
5
π
π
= 2π
+ 0 + 4π 2 + π 5 −
2
5
6
8π
= π 2 + 8π 3 +
.
5
In the middle line, we used the fact that the integrand is an even function and the interval
of integration [−π, π] is symmetric, but one can also compute directly.
−π
y = −π 2
Marking scheme:
• 1 mark for correct limits of integration
• 1 mark for correct integrand (including order)
• 1 mark for squaring correctly
• 1 mark for the half-angle formula
• 1 mark for correct evaluation of the integral
π