 ```MATH 101 Quiz #2 (v.T3)
Last Name:
Thursday, January 28
First Name:
Student-No:
Section:
2
1. 1 mark Suppose that f (y) is a function and F (y) = e(y +1) − 4 is an antiderivative of f (y).
Z 2
Evaluate the definite integral √ f (y) dy. Simply your answer as far as possible.
2
Marking scheme: 1 for a correct answer in the box
Solution: FTC2 tells us that
Z 2
√
f
(y)
dy
=
F
(2)
−
F
(
2)
√
2
√ 2
2
= e(2 +1) − 4 − e( 2 +1) − 4
= e4+1 − e2+1 = e5 − e3 .
Z
2. 2 marks Use substitution to evaluate
1
2
2
tet sin(et ) dt.
0
2
2
2
Solution: We write u(t) = et and find u0 (t) = 2tet . Note that u(1) = e1 = e when
2
t = 1, and u(0) = e0 = 1 when t = 0. Therefore:
Z
Z 1
1 t=1
t2
t2
te sin(e ) dt =
sin(u(t))u0 (t) dt
2 t=0
0
Z
1 u=e
=
sin(u) du
2 u=1
e 1
1
= − cos(u)1 =
cos(1) − cos(e) .
2
2
Marking scheme:
• 1 mark for identifying
Z u(t) correctly and changing from the t-integral to the u-integral,
−1
2
2
2
or for computing tet sin(et ) dt =
cos(et ) + C
2
• 1 mark for calculating the correct answer.
3. 2 marks Find the area between the curves: y = 2x and y = 6x − 4x2 , by first identifying the
points of intersection and then integrating.
Solution: Equating the two curves, we find that the intersections are when 2x = 6x − 4x2 ,
or equivalently x2 = x which has solutions x = 0 and x = 1. We therefore evaluate the
integral:
1 Z 1
Z 1
4x3 2
4
2
2
2
(6x − 4x ) − 2x dx =
[4x − 4x ] dx = 2x −
−0= .
= 2−
3
3
3
0
0
0
Marking scheme:
• 1 mark for writing down the correct definite integral
• 1 mark for the correct answer.
√
4. 5 marks Find the total area between the curves y = x 25 − x2 and y = 3x, on the interval
0 ≤ x ≤ 4.
√
√
√
2 is at least
2 =
Solution:
Note
that
for
x
≤
4,
the
expression
25
−
x
25
−
4
16 = 3,
√
and so x 25 − x2 is√larger than 3x for 0 ≤ x ≤ 3. (Alternatively, one can algebraically
solve the equation x 25 − x2 = 3x to find that the graphs intersect for x = 0 and x = &plusmn;4.)
The area we need to calculate is therefore:
Z 4 √
Z 4 √
Z
[x 25 − x2 − 3x] dx =
x 25 − x2 dx −
A=
0
0
4
3x dx = A1 − A2 .
0
To evaluate A1 , we use the substitution u(x) = 25 − x2 , for which u0 (x) = −2x; and
u(4) = 25 − 42 = 9 when x = 4, while u(0) = 25 − 02 = 25 when x = 0. Therefore
Z 4 √
Z
1 x=4 0 p
2
A1 =
x 25 − x dx = −
u (x) u(x) dx
2 x=0
0
9
Z
1 u=9 √
1 3/2 125 − 27
98
=−
u du = − u =
= .
2 u=25
3
3
3
25
For A2 we use the antiderivative directly:
Z
4
A2 =
0
Therefore the total area is:
A=
4
3x2 3x dx =
= 24
2 0
98
26
− 24 = .
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 .
```