 MATH 101 Quiz #1 (v.M1)

Last Name: First Name:

Friday, January 15

Student-No:

Section:

1. 1 mark Estimate R

5 x 3 dx using three approximating rectangles and left hand end points.

− 1

Solution: The distance from − 1 to 5 is 6, so for 3 rectangles we want ∆ x = 2 and x

0 x

1

= 1, x

2

= 3, x

3

= 5 and

= − 1,

Z

5

− 1 x

3 dx ≈ f ( x

0

) + f ( x

1

) + f ( x

2

) ∆ x = ( − 1)

3

+ 1

3

+ 3

3

× 2 = 54

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 during these two seconds. A “calculator-ready” answer is acceptable.

t (s) 0 0 .

5 1 .

0 1 .

5 2 v (m/s) 40 30 24 10 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) × 0 .

5 = 30 + 24 + 10 + 0 × 0 .

5 = 32 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 f ( x ) − g ( x ) dx . A “calculator-ready” answer is acceptable.

Z

2 g ( x ) dx = 4

0

Solution:

Z

2

− 1

3 f ( x ) − g ( x ) dx = 3

= 3

Z

− 1

Z

0

= − 3

2

0 f f

(

(

− 1

Z

− 1 x x f

)

)

( dx dx x )

Z

2

− 1

Z

2 g ( x ) dx

+ 3 dx

0

+ 3

Z

0

Z

2

Z f (

2 x ) dx −

− 1

Z

0 f ( x ) dx − g ( x ) dx − g ( x ) dx

0

0 − 1 g

Z

2

0

( x g

)

( dx x ) dx

= − 3 × 1 + 3 × 2 − 3 − 4 = − 4

Marking scheme:

• 1 mark for any correct application of one of

Z b a

Z b

Z b

[ A F ( x ) + BG ( x )] dx = A

Z b

F ( x ) dx =

Z c a

F ( x ) dx

F ( x ) dx +

+

Z

B b a

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

4 s

4 − − 2 + n

4 i n as a definite integal.

(b) Evaluate the integral of part (a).

2

Solution: (a) Set x i

= − 2 + 4 i n

. Then a = x

0

= − 2 and b = x n

= 2 and ∆ x = 4 n

. So lim n →∞ n

X i =1

4 n s

4 − − 2 +

4 n i

2

= lim n →∞ n

X i =1 f ( x i

)∆ x

=

Z

2 √

4 − x 2 dx

− 2 with f ( x ) =

4 − x 2 and ∆ x =

4 n

(b) y =

4 − x 2 is equivalent to x 2 + y between the upper half of the circle x

− 2 ≤ x ≤ 2, which is

1

2

π 2

2

= 2 π .

2

2 = 4, y ≥ 0. So the integral represents the area

+ y 2 = 4 (which has radius 2) 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

4

(a) is

0 p

4 − ( − 2 + x ) 2 dx .

• 2 marks for part (b), including 1 mark for realising that there is a circle involved.

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