MATH 101 Quiz #1 (v.A1)
Last Name:
Friday, January 15
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
7
f (x) dx as a limit of Riemann
1. 1 mark Let f be a function on the whole real line. Express
−1
sums, using the right endpoints. Write your answer in the box.
Answer:P
limn→∞ ni=1 f −1 +
8i
n
8
n
Solution: We have ∆x = (7 − (−1))/n = 8/n and hence xi = −1 + 8i/n.
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
2. 4 marks Each part is worth 2 marks.
√
R3
(a) Evaluate −3 2 + 9 − x2 dx. Simplify your answer completely.
Answer: 12 +
Solution: We first use additivity:
Z
Z 3
√
2
2 + 9 − x dx =
3
−3
−3
Z
3
2 dx +
√
9π
2
9 − x2 dx.
−3
The first integral represents the area of a rectangle of height 2 and width 6. The
second integral represents the area of a semicircle of radius 3.
Marking scheme:
• 1 mark for using additivity or for intepreting an integral as an area (or for
correctly using integration techniques all the way to the final answer)
• 1 mark for writing the correct answer in the box
n
X
2 −i/n
(b) Express lim
e
cos(2i/n) as a definite integral. Write your answer in the box.
n→∞
n
i=1
R2
Answer: 0 e−x/2 cos(x) dx
Solution: As i ranges from 1 to n, 2i/n range from 2/n to 2 with jumps of 2/n, so
this is
n
X
lim
f (xi )∆x
n→∞
where xi = 2i/n, f (x) = e
−x/2
cos(x).
i=1
R1
We also accept 0 2e−x cos(2x) dx.
Marking scheme: 2pts correct answer, 1 if minor error (endpoints or integrand)
Long answer question—you must show your work
Z
3
(x + x3 ) dx using the definition of the definite integral. You may use the
0
P
P
2
4
3
2
summation formulas ni=1 i = n 2+n and ni=1 i3 = n +2n4 +n . You may leave your answer in
“calculator-ready” form. (No credit will be given for using integration rules, but you may use
them to check your answer.)
3. 5 marks Find
Solution: We have ∆x = 3/n and breakpoints xi = 3i/n. Using right-hand points, we
need to evaluate
3 !
n
n X
3i
3 X 3i 27i3
3i
3
+
= lim
+ 3
lim
n→∞
n
n
n n→∞ n i=1 n
n
i=1
!
n
n
81 X 3
9 X
i+ 4
i
= lim
n→∞
n2 i=1
n i=1
9(n2 + n) 81(n4 + 2n3 + n2 )
= lim
+
n→∞
n2 · 2
n4 · 4
9
1
1
81
2
= lim
1+
+
1+ + 2
n→∞
2
n
4
n n
99
3
9 81
=
= 24 .
= +
2
4
4
4
Marking scheme:
• 1 mark setting up the limit correctly.
• 2 marks for manipulating the sums
• 1 mark for correctly using at least one summation formula
• 1 mark for evaluting the limit correctly (no need to simplify fully)