MATH 101 Quiz #1 (v.M2)
Last Name: First Name:
Friday, January 15
Student-No:
Grade:
Section:
1. 1 mark Estimate R
3 x 2 dx using three approximating rectangles and left hand end points.
− 3
Please write your answer in the box. A “calculator-ready” answer is acceptable.
Answer: 22
Solution: The distance from − 3 to 3 is 6, so for 3 rectangles we want ∆ x = 2 and x
0 x
1
= − 1, x
2
= 1, x
3
= 3 and
= − 3,
Z
3
− 3 x
2 dx ≈ f ( x
0
) + f ( x
1
) + f ( x
2
) ∆ x = ( − 3)
2
+ ( − 1)
2
+ 1
2
× 2 = 22
Marking scheme: 1 for a correct answer in the box
2. 2 marks A car’s gas pedal is applied at t = 0 seconds and the car accelerates continuously until t = 2 seconds. The car’s speed at half-second intervals is given in the table below.
Find the best possible upper estimate for the distance that the car traveled during these two seconds. A “calculator-ready” answer is acceptable.
t (s) 0 0 .
5 1 .
0 1 .
5 2 v (m/s) 0 14 22 30 40
Solution: The car’s speed increases with time. So its highest speed on any time interval occurs at the right hand end of the interval and the best possible upper estimate for the distance traveled is given by the right Riemann sum with ∆ x = 0 .
5, which is v (0 .
5) + v (1 .
0) + v (1 .
5) + v (2 .
0) × 0 .
5 = 14 + 22 + 30 + 40 × 0 .
5 = 53 m
Marking scheme:
• 1 mark for giving any correct (left, right, midpoint, · · · ) Riemann sum for
R
2
0 v ( t ) dt .
• 1 mark for the correct answer (even if the units are missing).

3. 2 marks The functions f ( x ) and g ( x ) obey
Z
− 1 f ( x ) dx = 1
0
Z
2 f ( x ) dx = 2
0
Z
0
− 1 g ( x ) dx = 3
Find
R
2
− 1
3 g ( x ) − f ( x ) dx . A “calculator-ready” answer is acceptable.
Z
2 g ( x ) dx = 4
0
Solution:
Z
2
− 1
3 g ( x ) − f ( x ) dx = 3
= 3
= 3
Z
2
− 1
Z
0
− 1
Z
0
− 1 g g g
(
(
( x x x
)
)
) dx dx dx
−
Z
2
+ 3
+ 3
− 1
Z
2 f ( x ) dx
0
Z
2
0 g ( x ) dx − g ( x ) dx +
Z
0
− 1
Z
− 1 f ( x ) dx − f ( x ) dx −
Z
2
0
Z
2 f ( x ) dx f ( x ) dx
0 0
= 3 × 3 + 3 × 4 + 1 − 2 = 20
Marking scheme:
• 1 mark for any correct application of one of
Z b a
[ A F ( x ) + BG ( x )] dx = A
Z b
Z c a
F ( x ) dx =
Z b
F ( x ) dx
F ( x ) dx +
+
Z
B b
Z a b
G (
F ( x ) dx x ) dx a a c
• 1 mark for the correct answer.
Page 2

4. 5 marks (a) Express lim n →∞ n
X i =1
6 s
9 − − 3 + n
6 i n as a definite integal.
(b) Evaluate the integral of part (a).
2
Solution: (a) Set x i
= − 3 + 6 i n
. Then a = x
0
= − 3 and b = x n
= 3 and ∆ x = 6 n
. So lim n →∞ n
X i =1
6 n s
9 − − 3 +
6 n i
2
= lim n →∞ n
X i =1 f ( x i
)∆ x
=
Z
3 √
9 − x 2 dx
− 3 with f ( x ) =
√
9 − x 2 and ∆ x =
6 n
(b) y =
√
9 − x 2 is equivalent to x 2 + y between the upper half of the circle x
− 3 ≤ x ≤ 3, which is
1
2
π 3
2
=
9 π
2
.
2
2 = 9, y ≥ 0. So the integral represents the area
+ y 2 = 9 (which has radius 3) and the x -axis with
Marking scheme:
• 3 marks for part (a), including 1 mark for a correct choice of f , 1 mark for a correct choice of ∆ x and 1 mark for a correct choice of a and b . A second correct answer for part
R
6
(a) is
0 p
9 − ( − 3 + x ) 2 dx .
• 2 marks for part (b), including 1 mark for realising that there is a circle involved.
Page 3