MATH 101 Quiz #1 (v.A2)
Last Name:
Friday, January 15
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
4
f (x) dx as a limit of Riemann
1. 1 mark Let f be a function on the whole real line. Express
−3
sums, using the right endpoints. Write your answer in the box.
Answer:P
limn→∞ ni=1 f −3 +
7i
n
7
n
Solution: We have ∆x = (4 − (−3))/n = 7/n and hence xi = −3 + 7i/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.
√
R0
(a) Evaluate −3 6 + 9 − x2 dx. Simplify your answer completely.
Answer: 18 +
Solution: We first use additivity:
Z
Z 0
√
2
6 + 9 − x dx =
0
−3
−3
Z
0
6 dx +
√
9π
4
9 − x2 dx.
−3
The first integral represents the area of a rectangle of height 6 and width 3. The
second integral represents the area of a quarter circle 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
3 −i/n
(b) Express lim
e
cos(3i/n) as a definite integral. Write your answer in the box.
n→∞
n
i=1
R3
Answer: 0 e−x/3 cos(x) dx
Solution: As i ranges from 1 to n, 3i/n range from 3/n to 3 with jumps of 3/n, so
this is
n
X
lim
f (xi )∆x
n→∞
where xi = 3i/n, f (x) = e
−x/3
cos(x).
i=1
R1
We also accept 0 3e−x cos(3x) dx.
Marking scheme: 2pts correct answer, 1 if minor error (endpoints or integrand)
Long answer question—you must show your work
Z
2
(x3 + x) dx using the definition of the definite integral. You may use the
0
P
P
4
3
2
2
summation formulas ni=1 i3 = n +2n4 +n and ni=1 i = n 2+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 = 2/n and breakpoints xi = 2i/n. Using right-hand points, we
need to evaluate
!
3
n
n X
2 X 8i3 2i
2i
2i 2
= lim
+
lim
+
n→∞
n
n n n→∞ n i=1 n3
n
i=1
!
n
n
4 X
16 X 3
i + 2
i
= lim
n→∞
n4 i=1
n i=1
16(n4 + 2n3 + n2 ) 4(n2 + n)
= lim
+
n→∞
n4 · 4
n2 · 2
16
1
2
4
1
= lim
1+ + 2 +
1+
n→∞
4
n n
2
n
16 4
+ = 6.
=
4
2
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)