MATH 101 Quiz #2 (v.M3)
Last Name:
Friday, January 29
First Name:
Grade:
Student-No:
Section:
Very short answer question
1
Z
x4 + ex ) dx. Simplify as far as possible.
1. 1 mark Evaluate
0
Answer: e −
4
5
Solution: By FTC2,
Z
1
4
x
x + e ) dx =
0
1
1
x5
4
x +e =
+e − 0+1 =e− .
5
5
5
0
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
Z
2. 2 marks Evaluate
(x3 − 1)x2
√
dx.
1 − x3
Solution: We use the substitution u = 1 − x3 , for which du = −3x2 dx:
Z
Z
x3 − 1 2
1 1 − x3
√
√
x dx =
(−3x2 ) dx
3 1 − x3
1 − x3
Z
1
u
√ du
=
3
u
Z
√
1
u du
=
3
1 u3/2
=
+C
3 3/2
=
2
(1 − x3 )3/2 + C
9
Marking scheme:
• 1 mark for substituting u = 1 − x3
• 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 − y 2 . A
“calculator-ready” answer is acceptable.
Solution: A point (x, y) on the curve x = y − y 2 = y(1 − y) has x = 0 for y = 0, 1, has
x > 0 for 0 < y < 1 (then both factors y and 1 − y are positive) and has x < 0 for y < 0
and y > 1 (then one of the factors y and 1 − y is positive and the other is negative). This
leads to the figure below. So, using horizontal slices,
y
(0, 1)
x = y − y2
Z
area =
0
1
y2 y3
−
(y − y ) dy =
2
3
2
1
1 1
1
− =
2 3
6
=
0
(0, 0)
x
Marking scheme:
• 1 mark for any correct integral. If they use vertical slices, the resulting integral is
√
√
Z 1/4 Z 1/4
√
1 + 1 − 4x 1 − 1 − 4x
−
1 − 4x dx
dx =
2
2
0
0
• 1 mark for the correct answer.
Long answer question—you must show your work
x2
Z
t3/2
e
4. 5 marks Consider the function F (x) =
0
Z
0
3
et 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
t3/2
e
Z
dt, H(x) =
0
x
3
et dt
0
By the Fundamental Theorem of Calculus,
G0 (y) = ey
3/2
H 0 (x) = ex
3
Hence, by the chain rule,
3
3
F 0 (x) = 2xG0 (x2 ) − H 0 (x) = 2xex − ex = (2x − 1)ex
3
(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 3
Rx 3
– 1 mark for x et dt = − 0 et dt and
– 1 mark for H 0 (x)
3
3
If the student writes down F 0 (x) = 2xex − ex 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.