MATH 101 Quiz #2 (v.M2)
Last Name:
Friday, January 29
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
3
cos x − x) dx. Simplify as far as possible.
1. 1 mark Evaluate
0
Answer: sin 3 −
9
2
Solution: By FTC2,
Z
0
3
cos x − x) dx =
3
x2 9
9
sin x −
−
sin
0
−
0
= sin 3 − .
=
sin
3
−
2 0
2
2
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
Z
2. 2 marks Evaluate
(9 − x2 )x
√
dx.
x2 − 9
Solution: We use the substitution u = x2 − 9, for which du = 2x dx:
Z
Z
9 − x2
1 x2 − 9
√
x dx = − √
(2x) dx
2 x2 − 9
x2 − 9
Z
1
u
√ du
=−
2
u
Z
√
1
u du
=−
2
1 u3/2
=−
+C
2 3/2
1
= − (x2 − 9)3/2 + C
3
Marking scheme:
• 1 mark for substituting u = x2 − 9
• 1 mark for the correct answer (including restoring the x–dependence).
3. 2 marks Find the area to the right of the y–axis and to the left of the curve x = y − 2y 2 . A
“calculator-ready” answer is acceptable.
Solution: A point (x, y) on the curve x = y − 2y 2 = y(1 − 2y) has x = 0 for y = 0, 1/2,
has x > 0 for 0 < y < 1/2 (then both factors y and 1 − 2y are positive) and has x < 0
for y < 0 and y > 1/2 (then one of the factors y and 1 − 2y is positive and the other is
negative). This leads to the figure below. So, using horizontal slices,
y
(0, 1/2)
x = y − 2y 2
1/2
Z
y 2 2y 3
(y − 2y ) dy =
−
2
3
2
area =
0
1/2
1
1
1
−
=
8 12
24
=
0
(0, 0)
x
Marking scheme:
• 1 mark for any correct integral. If they use vertical slices, the resulting integral is
√
√
Z 1/8 Z 1/8 √
1 + 1 − 8x 1 − 1 − 8x
1 − 8x
−
dx
dx =
4
4
2
0
0
• 1 mark for the correct answer.
Long answer question—you must show your work
Z
x2
4. 5 marks Consider the function F (x) =
−t
e
Z
0
0
2
e−t dt.
dt +
−x
0
(a) Find F (x).
(b) Find the value of x for which F (x) takes its minimum value.
Solution: (a) Write
2
F (x) = G(x ) − H(−x)
Z
with G(y) =
y
e
−t
Z
dt, H(y) =
0
y
2
e−t dt
0
By the Fundamental Theorem of Calculus,
G0 (y) = e−y
H 0 (y) = e−y
2
Hence, by the chain rule,
2
2
F 0 (x) = 2xG0 (x2 ) − (−1)H 0 (−x) = 2xe−(x ) + e−(−x) = (2x + 1)e−x
2
(b) Observe that F 0 (x) < 0 for x < −1/2 and F 0 (x) > 0 for x > −1/2. Hence F (x) is
decreasing for x < −1/2 and increasing for x > −1/2, and F (x) must take its minimum
value when x = −1/2
Marking scheme:
• 4 marks for part (a),
– with 2 marks for 2xG0 (x2 ) (including 1 mark for use of the chain rule),
R0
R −x 2
2
– 1 mark for −x e−t dt = − 0 e−t dt and
– 1 mark for −H 0 (−x)
2
2
If the student writes down F 0 (x) = 2xe−x + e−x without any justification, they still get
4 marks.
• 1 mark for part (b), even if they just solve F 0 (x) = 0, without justifying why it is a
minimum.