 ```MATH 101 Quiz #1 (v.T2)
Last Name:
Thursday, January 14
First Name:
Student-No:
Section:
Z
2
Z
Z
2f (x) + 3g(x) dx. Please
0
0
0
2
g(x) dx = −4, calculate
f (x) dx = 3 and
1. 1 mark If
2
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
[2f (x) + 3g(x)] dx =
2f (x) dx +
3g(x) dx
0
0
0
Z 2
Z 2
g(x) dx = (2 &times; 3) + (3 &times; (−4)) = −6
f (x) dx + 3
= 2
0
0
Z
3
f (x) dx, where
2. 2 marks Use elementary geometry to calculate
0
(
x, if x ≤ 2,
f (x) =
2, if x &gt; 2.
Solution: If we sketch the function, we have a trapezoid:R a linear increase from x = 0 to
3
x = 2, 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:
•
R2
•
R3
0
2
f (x) dx: a right-angled triangle of height 2 and base 2. Area = 2.
f (x) dx: a rectangle of height 2 and base 1. Area = 2.
Summing up:
R3
0
f (x) dx = 4.
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
log 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: f (x) = log2 (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) = log2 (2 + x).
Marking scheme: 1 mark for each answer
4. 5 marks Consider the integral:
Z
3
(7 + x3 ) dx.
(∗)
0
(a) (1 mark) Approximate this integral using the left Riemann sum with n = 3 intervals.
Solution: The left Riemann sum is defined as
Ln =
n−1
X
f (xi )∆x.
i=0
We subdivide into 3 intervals, (i.e. ∆x = 1), defining for i = 0, . . . , 2, the values
xi = 0, 1, 2. The function f (x) = 7 + x3 has the values: f (xi ) = 7, 8, 15, from which
we evaluate L3 = 30. 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,
Hint: you may use the identity
n
X
i=1
i3 =
n4 + 2n3 + n2
.
4
Solution: We divide into n intervals ∆x = 3/n and set xi = 3i/n. The right Riemann
sum is therefore:
n
X
(3i)3
3
7+ 3 .
Rn =
n
n
i=1
To calculate the sum:
Rn =
!
n
21 X
1 +
n i=1
= 21 +
n
81 X 3
i
n4 i=1
!
81
81 n4 + 2n3 + n2
= 21 + (1 + 2/n + 1/n2 )
&times;
4
n
4
4
To evaluate the limit exactly, we take n → ∞. The expressions involving 1/n vanish
leaving:
Z 3
81
1
= 41 .
(7 + x3 ) dx = lim Rn = 21 +
n→∞
4
4
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.
```