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Quiz #1 (v.T2): Page 1 of 4
Thursday, January 19
Short answer question
1. 2 marks Each part is worth 1 mark. Please write your answers in the boxes. Only answers in the boxes will
be marked. “Calculator-ready” answers are acceptable.
Z 2
Z 2
f (x) dx = 2.
f (x) dx = 4 and
(a) The function f (x) satisfies
1
0
Z 1
3f (x) dx.
Find
0
Answer: 6
Solution: We have
Z
1
3f (x) dx = 3
Z
1
Z
f (x) dx = 3
Z
f (x) dx +
0
2
0
0
2
1
f (x) dx
2
Z 2
Z
=3
f (x) dx −
f (x) dx
1
0 = 3 4 − 2 = 6.
Z
−1
(b) Calculate
x3 dx.
−2
Answer:
Solution: One function with derivative x3 is
Z
−1
x3 dx =
−2
x4
4 .
1 − 24
−15
=
4
4
So, by the Fundamental Theorem of Calculus,
−1
(−1)4
(−2)4
−15
x4 =
−
=
.
4 −2
4
4
4
Quiz #1 (v.T2): Page 2 of 4
Thursday, January 19
Long answer question—you must show your work
Z
x
2. 4 marks Part (a) is worth 1 mark and part (b) is worth 3 marks. Define f (x) =
x2
(a) Find f (1). Simplify your answer.
Answer: 0
Solution: Evaluating the definition of f (x) at x = 1 gives
Z
1
f (1) =
1
Z
t2
12
2
et dt = 0
e dt =
1
since the upper and lower endpoints of integration are equal.
Marking scheme:
• 1 mark for the answer 0
(b) Find f 0 (1). Simplify your answer.
Answer: −e
Solution: Write
x
Z
2
et dt
f (x) =
x2
x
Z
Z
t2
x2
e dt −
=
0
2
et dt
0
= G(x) − G(x2 )
where
Z
G(y) =
y
2
et dt
0
By the Fundamental Theorem of Calculus,
G0 (y) = ey
2
Hence, by the chain rule,
d G(x) − G(x2 )
dx
= G0 (x) − 2xG0 (x2 )
f 0 (x) =
2
= ex − 2xex
4
When x = 1,
f 0 (1) = e1 − 2e1 = −e
Marking scheme:
• 1 mark for splitting up the integral
• 1 mark for a correct application of the chain rule
• 1 mark for the correct evaluation at x = 1
Quiz #1 (v.T2): Page 3 of 4
Thursday, January 19
2
et dt.
Long answer question—you must show your work
3. 4 marks Each part is worth 2 marks.
(a) For a certain function f (x) and a certain number b, the following equation holds:
Z b
n
X
4 1 4
f (x) dx.
lim
i−
− 2 =
n→∞
n
2 n
2
i=1
Find f (x) and b.
Answer: f (x) = |x − 4|, b = 6
Solution: The given sum is of the form
n
n
X
X
1 4
1 4
4 4 − 2 = lim
− 4
i−
2 + i −
n→∞
n→∞
n
2 n
n
2 n
i=1
i=1
lim
= lim
n→∞
n
X
n
X
f (x∗i ) ∆x
∆xx∗i − 4 = lim
n→∞
i=1
i=1
with ∆x = n4 , x∗i = 2+ i− 2 n = a+ i− 2 ∆x, a = 2, and f (x) = |x−4|. Since x0 = 2+i n3 i=0 = 2
and xn = 2 + i n4 i=n = 6, the left hand side is the definition (using the midpoint Riemann sum) of
R6
f (x) dx.
2
Marking scheme:
1 4
1
• 1 mark for f (x) = |x − 4|
• 1 mark for b = 6
Z
(b) Evaluate
b
f (x) dx, using the f (x) and b you found in part (a). Simplify your answer.
2
Answer: 4
Solution: The integral
R6
2
|x − 4| dx represents the shaded area in the figure below, which is
1
1
×2×2+ ×2×2=4
2
2
y
2
2
4
6
x
Marking scheme:
• 1 mark for identifying their integral from part (a) as the area of a correct figure
• 1 mark for the getting the area of their figure correct.
Quiz #1 (v.T2): Page 4 of 4
Thursday, January 19