MATH 101 Quiz #1 (v.T1)
Last Name:
Thursday, January 14
First Name:
Grade:
Student-No:
Section:
Very short answer question
Z
2
Z
0
0
2
Z
3f (x) − 2g(x) dx. Please write
g(x) dx = 3, calculate
f (x) dx = 4 and
1. 1 mark If
2
your answer in the box. Simplify your answer completely.
0
Answer: 6
Marking scheme: 1 for a correct answer in the box
Solution: The operation of integration is linear, so that:
Z 2
Z 2
Z 2
[3f (x) − 2g(x)] dx =
3f (x) dx −
2g(x) dx
0
0
0
Z 2
Z 2
g(x) dx = (3 × 4) − (2 × 3) = 6
f (x) dx − 2
= 3
0
0
Short answer questions—you must show your work
Z
3
f (x) dx, where
2. 2 marks Use elementary geometry to calculate
0
(
x, if x ≤ 1,
f (x) =
1, if x > 1.
Solution: If we sketch the function, we have a trapezoid:R a linear increase from x = 0 to
3
x = 1, followed by a constant. Using the interpretation of 0 f (x) dx as the area under the
function f (x) between 0 and 3, we can break this trapezoid into:
•
R1
•
R3
0
1
f (x) dx: a right-angled triangle of height 1 and base 1. Area = 0.5.
f (x) dx: a rectangle of height 1 and base 2. Area = 2.
Summing up:
R3
0
f (x) dx = 2.5.
Marking scheme:
• 1 mark for splitting the integral into 2, or otherwise identifying simpler geometric
structures, or using the formula for the area of a trapezoid.
• 1 mark for calculating the correct answer.
3. 2 marks The value of the following limit is equal to the area below a graph of y = f (x),
integrated over the interval [0, b]:
2
n
X
4i
4
cos 2 +
.
lim
n→∞
n
n
i=1
Find f (x) and b. (Do not evaluate the limit.) Please write your answers in the boxes.
Answer: b = 4
Answer: f (x) = cos2 (2 + x)
Solution: We identify this as the right Riemann sum, with interval ∆x = 4/n and xi =
4i/n. Therefore, b = 4 and f (x) = cos2 (2 + x).
Marking scheme: 1 mark for each answer
Long answer question—you must show your work
4. 5 marks Consider the integral:
Z
5
(5 + x2 ) dx.
(∗)
0
(a) (1 mark) Approximate this integral using the left Riemann sum with n = 5 intervals.
Solution: The left Riemann sum is defined as
Ln =
n−1
X
f (xi )∆x.
i=0
We subdivide into 5 intervals, (i.e. ∆x = 1), defining for i = 0, . . . , 4, the values
xi = 0, 1, 2, 3, 4. The function f (x) = 5+x2 has the values: f (xi ) = 5, 6, 9, 14, 21,
from which we evaluate L5 = 55. Marking scheme: 1 mark for a correct answer
(b) (4 marks) Write down the expression for the right Riemann sum with n intervals and
calculate the sum. Now take the limit n → ∞ in your expression for the Riemann sum,
to evaluate the integral (∗) exactly. You can leave your answer in “calculator-ready” form.
Hint: you may use the identity
n
X
i=1
2n3 + 3n2 + n
i =
.
6
2
Solution: We divide into n intervals ∆x = 5/n and set xi = 5i/n. The right Riemann
sum is therefore:
n
X
(5i)2
5
5+ 2 .
Rn =
n
n
i=1
To calculate the sum:
Rn =
!
n
25 X
1 +
n i=1
= 25 +
n
125 X 2
i
n3 i=1
!
125
125 2n3 + 3n2 + n
= 25 +
(2 + 3/n + 1/n2 )
×
3
n
6
6
To evaluate the limit exactly, we take n → ∞. The expressions involving 1/n vanish
leaving:
Z 5
125
2
= 66 .
(5 + x2 ) dx = lim Rn = 25 +
n→∞
3
3
0
Marking scheme:
• 1 mark for writing Rn .
• 2 marks for calculating the sum using the formula in the hint, or otherwise.
• 1 mark for taking the limit and getting the answer.