MATH 101 Quiz #1 (v.A3)
Last Name:
Friday, January 15
First Name:
Grade:
Student-No:
Section:
Very short answer question
10
Z
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 +
9i
n
9
n
Solution: We have ∆x = (10 − 1)/n = 9/n and hence xi = 1 + 9i/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.
√
R2
(a) Evaluate −2 5 + 4 − x2 dx. Simplify your answer completely.
Answer: 20 + 2π
Solution: We first use additivity:
Z 2
Z
√
5 + 4 − x2 dx =
−2
2
−2
Z
2
5 dx +
√
4 − x2 dx.
−2
The first integral represents the area of a rectangle of height 5 and width 4. The
second integral represents the area of a semicircle of radius 2.
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 −4i/n
e
sin(2i/n) as a definite integral. Write your answer in the box.
n→∞
n
i=1
R2
Answer: 0 e−2x sin(x) dx
(b) Express lim
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→∞
−2x
i=1
where xi = 2i/n, f (x) = e
sin(x).
R 4 1 −x
We also accept 0 2 e sin(x/2) dx.
Marking scheme: 2pts correct answer, 1 if minor error (endpoints or integrand)
Long answer question—you must show your work
Z
5
(x + x2 ) dx using the definition of the definite integral. You may use the
0
P
P
2
3
2 +n
summation formulas ni=1 i = n 2+n and ni=1 i2 = 2n +3n
. You may leave your answer in
6
“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 = 5/n and breakpoints xi = 5i/n. Using right-hand points, we
need to evaluate
2 !
n n
X
5i
5
5 X 5i 25i2
5i
+
= lim
+ 2
lim
n→∞ n
n→∞
n
n
n
n
n
i=1
i=1
!
n
n
25 X
125 X 2
= lim
i+ 3
i
n→∞
n2 i=1
n i=1
25(n2 + n) 125(2n3 + 3n2 + n)
+
= lim
n→∞
n2 · 2
n3 · 6
1
25
1
125
3
= lim
1+
+
2+ + 2
n→∞
2
n
6
n n
325
25 125
+
=
.
=
2
3
6
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)