 ```MATH 101 Quiz #1 (v.M3)
Last Name:
Friday, January 15
First Name:
Student-No:
Section:
R7
1. 1 mark Estimate −2 x2 dx using three approximating rectangles and left hand end points.
Solution: The distance from −2 to 7 is 9, so for 3 rectangles we want ∆x = 3 and x0 = −2,
x1 = 1, x2 = 4, x3 = 7 and
Z 7
x2 dx ≈ f (x0 ) + f (x1 ) + f (x2 ) ∆x = (−2)2 + 12 + 42 &times; 3 = 63
−2
Marking scheme: 1 for a correct answer in the box
2. 2 marks A car’s brakes are applied at t = 0 seconds and the car decelerates continuously
until it comes to a stop at t = 2 seconds. The car’s speed at half-second intervals is given in
the table below. Find the best possible lower estimate for the distance that the car traveled
t (s)
0
0.5
1.0
1.5
2
v (m/s)
50
30
16
6
0
Solution: The car’s speed decreases with time. So its lowest speed on any time interval
occurs at the right hand end of the interval and the best possible lower 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) &times; 0.5 = 30 + 16 + 6 + 0 &times; 0.5 = 26 m
Marking scheme:
• 1 mark for giving any correct (left, right, midpoint, &middot; &middot; &middot; ) Riemann sum for
• 1 mark for the correct answer (even if the units are missing).
R2
0
v(t) dt.
3. 2 marks The functions f (x) and g(x) obey
Z
−1
Z
f (x) dx = 1
0
Find
2
Z
0
f (x) dx = 2
Z
g(x) dx = 3
−1
0
2
g(x) dx = 4
0
−1
Solution:
Z 2
Z 2
Z 2
f (x) + 2g(x) dx =
f (x) dx + 2
g(x) dx
−1
−1
−1
Z 2
Z 0
Z 2
Z 0
g(x) dx
g(x) dx + 2
f (x) dx + 2
f (x) dx +
=
0
−1
0
−1
Z −1
Z 2
Z 0
Z 2
=−
f (x) dx +
f (x) dx + 2
g(x) dx + 2
g(x) dx
0
−1
0
= −1 + 2 + 2 &times; 3 + 2 &times; 4 = 15
Marking scheme:
• 1 mark for any correct application of one of
Z b
Z b
Z b
[A F (x) + BG(x)] dx = A
F (x) dx + B
G(x) dx
a
a
a
Z b
Z c
Z b
F (x) dx
F (x) dx +
F (x) dx =
a
• 1 mark for the correct answer.
a
c
0
4. 5 marks
(a) Express
n
X
2
lim
n→∞
n
i=1
s
2i
4 − −2 +
n
2
as a definite integal.
(b) Evaluate the integral of part (a).
Solution: (a) Set xi = −2 +
n
X
2
lim
n→∞
n
i=1
s
2i
4 − −2 +
n
2i
.
n
Then a = x0 = −2 and b = xn = 0 and ∆x = n2 . So
2
= lim
n→∞
Z
0
=
n
X
f (xi )∆x
with f (x) =
√
4 − x2 and ∆x =
i=1
√
2
n
4 − x2 dx
−2
√
(b) y = 4 − x2 is equivalent to x2 + y 2 = 4, y ≥ 0. So the integral represents the area
between the upper half of the circle x2 + y 2 = 4 (which has radius 2) and the x-axis with
−2 ≤ x ≤ 0, which is a quarter circle with area 41 π22 = π.
Marking scheme:
• 3 marks for part (a), including 1 mark for a correct choice of f , 1 mark for a correct
choiceRof p
∆x and 1 mark for a correct choice of a and b. A second correct answer for part
2
(a) is 0 4 − (−2 + x)2 dx.
• 2 marks for part (b), including 1 mark for realising that there is a circle involved.
```