 ```MATH 101 Quiz #1 (v.T3)
Last Name:
Thursday, January 14
First Name:
Student-No:
Section:
Z
1. 1 mark If
2
2
Z
f (x) dx = 5 and
0
2
Z
2f (x) − 4g(x) dx. Please write
g(x) dx = 3, calculate
0
0
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) − 4g(x)] dx =
2f (x) dx −
4g(x) dx
0
0
0
Z 2
Z 2
= 2
f (x) dx − 4
g(x) dx = (2 &times; 5) − (4 &times; 3) = −2
0
0
Z
2. 2 marks Use elementary geometry to calculate
3
f (x) dx, where
0
(
1,
if x ≤ −1,
f (x) =
−x, if x &gt; −1.
Solution: On the interval of integration [0, 3], the function f (x) is simply equal to −x,
whose graph on that interval is a straight line segment below the x-axis. [The intention
was for both parts of the definition of f to be relevant, but a typo
R 0 in the statement of
the problem made that not happen.] Using the interpretation of −3 f (x) dx as the signed
area bounded by the function f (x) between 0 and −3, we see from its graph that the
value of the integral is the negative of the area of the triangle with vertices (0, 0), (3, 0),
and
triangle has base 3 and height 3, and so the value of the integral is
R 0 (3, −3). This
1
f (x) dx = − 2 &middot; 3 &middot; 3 = − 29 .
−3
Marking scheme:
• 1 mark for calculating the correct answer.
• 1 mark for recognizing that the integral involves the area of a triangle, or recognizing
that the result must be negative due to the graph lying below the x-axis.
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
sin 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) = sin2 (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) = sin2 (2 + x).
Marking scheme: 1 mark for each answer
4. 5 marks Consider the integral:
Z
4
(6 + x2 ) dx.
(∗)
0
(a) (1 mark) Approximate this integral using the left Riemann sum with n = 4 intervals.
Solution: The left Riemann sum is defined as
Ln =
n−1
X
f (xi )∆x.
i=0
We subdivide into 4 intervals, (i.e. ∆x = 1), defining for i = 0, . . . , 3, the values
xi = 0, 1, 2, 3. The function f (x) = 6 + x2 has the values: f (xi ) = 6, 7, 10, 15, from
which we evaluate L4 = 38. 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
i2 =
2n3 + 3n2 + n
.
6
Solution: We divide into n intervals ∆x = 4/n and set xi = 4i/n. The right Riemann
sum is therefore:
n
X
(4i)2
4
6+ 2 .
Rn =
n
n
i=1
To calculate the sum:
Rn =
!
n
24 X
1 +
n i=1
= 24 +
n
64 X 2
i
n3 i=1
!
64
64 2n3 + 3n2 + n
= 24 + (2 + 3/n + 1/n2 )
&times;
3
n
6
6
To evaluate the limit exactly, we take n → ∞. The expressions involving 1/n vanish
leaving:
Z 4
64
1
= 45 .
(6 + x2 ) dx = lim Rn = 24 +
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.
```