MATH 101 Quiz #3 (v.A3)
Last Name:
Friday, February 12
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
1. 1 mark Evaluate
sin36 t cos3 t dt.
Answer:
sin37 t
37
−
sin39 t
39
+C
Solution: Make the substitution u = sin t, so that du = cos t dt:
Z
Z
Z
36
3
36
2
sin t cos t dt = sin t (1 − sin t) cos t dt = u36 (1 − u2 ) du
=
sin37 t sin39 t
u37 u39
−
+C =
−
+C
37
39
37
39
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
Z
2. 2 marks Evaluate
2t arctan(t) dt.
Solution: We integrate by parts, using u = arctan(t), dv = 2t dt so that v = t2 and
dt
du = 1+t
2:
Z
Z
2
2t arctan(t) dx = t arctan(t) −
We now notice that
Z
t2
t2 +1
=
t2 +1
t2 +1
t2
dt =
t2 + 1
−
1
.
t2 +1
Z t2
dt .
t2 + 1
We therefore have
1
1− 2
t +1
dt = t − arctan(t) + C .
The final answer is then
Z
2t arctan(t) dt = t2 arctan(t) − t + arctan(t) + C .
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 30 cm beyond equilibrium, if
its spring constant is k = 20 N/m. Simplify your answer completely.
Answer: 0.9 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
0.3 m
Z 0.3 m
1
1 2 = · 20 · (0.3)2 J = 0.9 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
2
4. 5 marks The finite region between the curves y = 2 cos x and y = 2x2 − π2 is rotated about
2
the line y = − π2 . 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 = ± π2 where they both vanish. The volume is therefore
2 2 2 2 #
Z π/2 "
2
π
π
π
π
2 cos x − −
−
2x2 −
dx
− −
2
2
2
−π/2
Z π/2 π4
2
2
4
=π
4 cos x + 2π cos x +
− 4x dx
4
−π/2
Z π/2 1 + cos(2x)
π4
2
4
=π
4·
+ 2π cos x +
− 4x dx
2
4
−π/2
Z π/2 π4
2
4
= 2π
2 + 2 cos(2x) + 2π cos x +
− 4x dx
4
−π/2
π/2
π4
4 5 2
= 2π 2x + sin(2x) + 2π sin x + x − x 4
5
0
5
5
π
π
= 2π π + 0 + 2π 2 +
−
8
40
6
π
= 2π 2 + 4π 3 + .
5
In the middle line, we used the fact that the integrand is an even function and the interval
of integration [− π2 , π2 ] is symmetric, but one can also compute directly.
− π2
y = −π 2 /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
2