 ```MATH 101 Quiz #2 (v.M1)
Last Name:
Friday, January 29
First Name:
Student-No:
Section:
Z
2
x3 + sin x) dx. Simplify as far as possible.
1. 1 mark Evaluate
0
Solution: By FTC2,
Z
2
3
x + sin x) dx =
0
2
24
x4
− cos x =
− cos 2 + cos 0 = 4 − cos 2 + 1 = 5 − cos 2.
4
4
0
Marking scheme: 1 for a correct answer in the box
Z
2. 2 marks Evaluate
(x2 − 4)x
√
dx.
4 − x2
Solution: We use the substitution u = 4 − x2 , for which du = −2x dx:
Z
Z
x2 − 4
1 4 − x2
√
√
x dx =
(−2x) dx
2 4 − x2
4 − x2
Z
1
u
√ du
=
2
u
Z
1 √
=
u du
2
1 u3/2
=
+C
2 3/2
=
1
(4 − x2 )3/2 + C
3
Marking scheme:
• 1 mark for substituting u = 4 − x2
• 1 mark for the correct answer (including restoring the x–dependence).
3. 2 marks Find the area to the left of the y–axis and to the right of the curve x = y 2 + y. A
Solution: A point (x, y) on the curve x = y 2 + y = y(y + 1) has x = 0 for y = −1, 0, has
x &lt; 0 for −1 &lt; y &lt; 0 (the factors y and y + 1 have opposite signs) and has x &gt; 0 for
y &lt; −1 and y &gt; 0 (the factors y and y + 1 are either both positive or both negative). This
y
leads to the figure below. So, using horizontal slices,
(0, 0) x
Z
0
3
0
y
1
y2
1 1
0 − (y + y) dy = −
+
=− + =
3
2 −1
3 2
6
2
area =
−1
(0, −1)
x = y + y2
Marking scheme:
• 1 mark for any correct integral. If they use vertical slices, the resulting integral is
√
√
Z 0 Z 0
√
−1 + 1 + 4x −1 − 1 + 4x
−
1 + 4x dx
dx =
2
2
−1/4
−1/4
• 1 mark for the correct answer.
Z
x2
4. 5 marks Consider the function F (x) =
0
Z
t2
0
4
et dt.
e 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
Z
t2
e dt, H(x) =
0
x
4
et dt
0
By the Fundamental Theorem of Calculus,
G0 (y) = ey
2
H 0 (x) = ex
4
Hence, by the chain rule,
4
4
F 0 (x) = 2xG0 (x2 ) − H 0 (x) = 2xex − ex = (2x − 1)ex
4
(b) Observe that F 0 (x) &lt; 0 for x &lt; 1/2 and F 0 (x) &gt; 0 for x &gt; 1/2. Hence F (x) is
decreasing for x &lt; 1/2 and increasing for x &gt; 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 4
Rx 4
– 1 mark for x et dt = − 0 et dt and
– 1 mark for H 0 (x)
4
4
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.
```