MATH 101 Quiz #2 (v.T1)
Last Name:
Thursday, January 28
First Name:
Grade:
Student-No:
Section:
Very short answer question
2
1. 1 mark Suppose that f (t) is a√function and F (t) = e(t −2) + 6 is an antiderivative of f (t).
Z 3
f (t) dt. Simply your answer as far as possible.
Evaluate the definite integral
1
Answer: e − e−1
Marking scheme: 1 for a correct answer in the box
Solution: FTC2 tells us that
Z √3
√
f (t) dt = F ( 3) − F (1)
1
√ 2
2
= e( 3 −2) + 6 − e(1 −2) + 6
= e3−2 − e1−2 = e − e−1 .
Short answer questions—you must show your work
Z
2. 2 marks Use substitution to evaluate
1
2
2
xex cos(ex ) dx.
0
2
2
2
Solution: We write u(x) = ex and find u0 (x) = 2xex . Note that u(1) = e1 = e when
2
x = 1, and u(0) = e0 = 1 when x = 0. Therefore:
Z 1
Z
1 x=1
x2
x2
xe cos(e ) dx =
cos(u(x))u0 (x) dx
2
0
Zx=0
1 u=e
cos(u) du
=
2 u=1
e 1
1
= sin(u)1 =
sin(e) − sin(1) .
2
2
Marking scheme:
• 1 mark for identifying u(x)
Z correctly and changing from the x-integral to the u1
2
2
2
integral, or for computing xex cos(ex ) dx = sin(ex ) + C
2
• 1 mark for calculating the correct answer.
3. 2 marks Find the area between the curves: y = 2x and y = 6x − 2x2 , by first identifying the
points of intersection and then integrating.
Solution: Equating the two curves, we find that the intersections are when 2x = 6x − 2x2 ,
or equivalently x2 = 2x which has solutions x = 0 and x = 2. We therefore evaluate the
integral:
2 Z 2
Z 2
2x3 8
16
2
2
2
(6x − 2x ) − 2x dx =
[4x − 2x ] dx = 2x −
−0= .
= 8−
3
3
3
0
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 = 3x, 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 3x for 0 ≤ x ≤ 4. (Alternatively, one can algebraically
solve the equation x 25 − x2 = 3x 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 − 3x] dx =
x 25 − x2 dx −
A=
0
0
3
3x 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
3
A2 =
0
Therefore the total area is:
A=
3
27
3x2 3x dx =
= .
2 0
2
61 27
41
−
= .
3
2
6
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 .