MATH 101 Quiz #3 (v.A1)
Last Name:
Friday, February 12
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
sin84 x cos3 x dx.
1. 1 mark Evaluate
Answer:
sin85 x
85
−
sin87 x
87
+C
Solution: Make the substitution u = sin x, so that du = cos x dx:
Z
Z
Z
84
3
84
2
sin x cos x dx = sin x (1 − sin x) cos x dx = u84 (1 − u2 ) du
=
sin85 x sin87 x
u85 u87
−
+C =
−
+C
85
87
85
87
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
Z
2. 2 marks Evaluate
2x arctan(x) dx.
Solution: We integrate by parts, using u = arctan(x), dv = 2x dx so that v = x2 and
dx
du = 1+x
2:
Z
Z
2
2x arctan(x) dx = x arctan(x) −
We now notice that
Z
x2
x2 +1
=
x2 +1
x2 +1
x2
dx =
x2 + 1
−
1
.
x2 +1
Z x2
dx .
x2 + 1
We therefore have
1
1− 2
x +1
dx = x − arctan(x) + C .
The final answer is then
Z
2x arctan(x) dx = x2 arctan(x) − x + arctan(x) + 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 10 cm beyond equilibrium, if
its spring constant is k = 50 N/m. Simplify your answer completely.
Answer: 0.25 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
0.1 m
Z 0.1 m
1
1
1 2 = · 50 · (0.1)2 J = J.
kx dx = kx 2
2
4
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 = cos(x) and y = x2 − π4 is rotated about
2
the line y = − π4 . 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
cos(x) − −
−
x2 −
dx
− −
4
4
4
−π/2
Z π/2 π2
π4
2
4
=π
cos x +
cos x +
− x dx
2
16
−π/2
Z π/2 π4
1 + cos(2x) π 2
4
+
cos x +
− x dx
=π
2
2
16
−π/2
Z π/2 1 + cos(2x) π 2
π4
4
= 2π
+
cos x +
− x dx
2
2
16
0
π/2
1
1
π2
π4
1 5 = 2π
x + sin(2x) +
sin x + x − x 2
4
2
16
5
0
2
5
5
π
π
π
π
= 2π
+0+
+
−
4
2
32 160
2
6
π
π
=
+ π3 + .
2
20
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 /4
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