MATH 101 Quiz #2 (v.T2)
Last Name:
Thursday, January 28
First Name:
Grade:
Student-No:
Section:
Very short answer question
2
1. 1 mark Suppose that f (x) is a√function and F (x) = e(x −3) + 1 is an antiderivative of f (x).
Z 5
f (x) dx. Simply your answer as far as possible.
Evaluate the definite integral
1
Answer: e2 − e−2
Marking scheme: 1 for a correct answer in the box
Solution: FTC2 tells us that
Z √5
√
f (x) dx = F ( 5) − F (1)
1
√ 2
2
= e( 5 −3 + 1 − e(1 −3) + 1
= e5−3 − e1−3 = e2 − e−2 .
Short answer questions—you must show your work
Z
2. 2 marks Use substitution to evaluate
2
2
2
yey cos(ey ) dy.
0
2
2
2
Solution: We write u(y) = ey and find u0 (y) = 2yey . Note that u(2) = e2 = e4 when
2
y = 2, and u(0) = e0 = 1 when y = 0. Therefore:
Z 2
Z
1 y=2
y2
y2
ye cos(e ) dy =
cos(u(y))u0 (y) dy
2
0
y=0
Z u=e4
1
=
cos(u) du
2 u=1
e4
1
1
= sin(u)1 =
sin(e4 ) − sin(1) .
2
2
Marking scheme:
• 1 mark for identifying u(y)
Z correctly and changing from the y-integral to the u1
2
2
2
integral, or for computing yey cos(ey ) dy = sin(ey ) + C
2
• 1 mark for calculating the correct answer.
3. 2 marks Find the area between the curves: y = x and y = 3x − x2 , by first identifying the
points of intersection and then integrating.
Solution: Equating the two curves, we find that the intersections are when x = 3x − x2 ,
or equivalently x2 = 2x which has solutions x = 0 and x = 2. We therefore evaluate the
integral:
2 Z 2
Z 2
x3 4
8
2
2
2
(3x − x ) − x dx =
[2x − x ] dx = x −
−0= .
= 4−
3 0
3
3
0
0
Marking scheme:
• 1 mark for writing down the correct definite integral
• 1 mark for the correct answer.
Long answer question—you must show your work
√
4. 5 marks Find the total area between the curves y = x 25 − x2 and y = 4x, on the interval
0 ≤ x ≤ 3.
√
√
√
2 is at least
2 =
Solution:
Note
that
for
x
≤
3,
the
expression
25
−
x
25
−
3
16 = 4,
√
and so x 25 − x2 is√larger than 4x for 0 ≤ x ≤ 3. (Alternatively, one can algebraically
solve the equation x 25 − x2 = 4x to find that the graphs intersect for x = 0 and x = ±4.)
The area we need to calculate is therefore:
Z 3 √
Z 3 √
Z
[x 25 − x2 − 4x] dx =
x 25 − x2 dx −
A=
0
0
3
4x dx = A1 − A2 .
0
To evaluate A1 , we use the substitution u(x) = 25 − x2 , for which u0 (x) = −2x; and
u(3) = 25 − 32 = 16 when x = 3, while u(0) = 25 − 02 = 25 when x = 0. Therefore
Z 3 √
Z
1 x=3 0 p
2
A1 =
x 25 − x dx = −
u (x) u(x) dx
2 x=0
0
16
Z
1 u=16 √
1 3/2 125 − 64
61
=−
u du = − u =
= .
2 u=25
3
3
3
25
For A2 we use the antiderivative directly:
Z
A2 =
0
3
3
4x dx = 2x = 18.
Therefore the total area is:
A=
2
0
61
7
− 18 = .
3
3
Marking scheme:
• 2 marks for correctly identifying A, either via a sketch or algebraically, and writing
down the correct integral with the limits.
• 2 marks for evaluating A1 via substitution.
• 1 mark for evaluating A2 .