MATH 101 Quiz #2 (v.A2)
Last Name:
Friday, January 29
First Name:
Grade:
Student-No:
Section:
Very short answer question
d
1. 1 mark Find
dx
5
Z
x2
dt
.
1 + t4
Answer: −
Z
Solution: Define g(x) =
x
5
dt
=−
1 + t4
Z
5
x
2x
1 + x8
1
1
0
dt,
so
that
g
(x)
=
−
by FTC1.
1 + t4
1 + x4
Then by the Chain Rule,
Z 5
dt
d
d
1
d
=
g(x2 ) = g 0 (x2 ) · x2 = −
· 2x.
4
dx
dx
dx
1 + x8
x2 1 + t
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
π/2
Z
cos x · (1 + sin2 x) dx. Simplify your answer as much as possible.
2. 2 marks Evaluate
0
Answer: 4/3
Solution: Setting u = sin x, we have du = cos x dx and so
Z
π/2
2
Z
x=π/2
cos x · (1 + sin x)dx =
0
2
Z
u=1
(1 + sin x) · cos x dx =
x=0
(1 + u2 ) du,
u=0
since u = sin 0 = 0 when x = 0 and u = sin(π/2) = 1 when x = π/2. Then, by FTC2,
Z 1
1
(1 + u2 ) du = (u + u3 /3)0 = (1 + 1/3) − 0 = 4/3.
0
Marking scheme:
• 1 mark Z
for correctly using substitution, either for converting to the u-integral or for
getting cos x · (1 + sin2 x) dx = sin x + (sin3 x)/3 + C
• 1 mark for getting the right answer.
3. 2 marks A car traveling at 16 m/s applies its brakes at time t = 0, its velocity (in m/s)
decreasing according to the formula v(t) = 16 − 4t. How far does the car go before it stops?
Simplify your answer completely.
Answer: 32 m
Solution: The car stops when v(t) = 0, which occurs at time t = 4. The distance covered
up to that time is
Z 4
4
v(t) dt = (16t − 2t2 )0 = (64 − 32) − 0 = 32 m.
0
Marking scheme:
• 1 mark for recognizing that distance is the the integral of velocity
• 1 mark for getting the right answer (even without units)
Long answer question—you must show your work
√
4. 5 marks Find the area of the finite region that is bounded by the graphs of f (x) = x2 x3 + 1
and g(x) = 3x2 . Your answer may be left in calculator-ready form.
√
Solution: For that computation, we will need an antiderivative of x2 x3 + 1, which can
be found using substitution using u = x3 + 1, so that 13 du = x2 dx:
Z
Z
Z
√
√
1
2
1
1 u3/2
1/2
3
x x + 1 dx =
u · du =
u du =
+ C = (x3 + 1)3/2 + C.
3
3
3 3/2
9
2
The two functions f (x) and g(x) are clearly equal at x = 0. If x 6= 0, then the functions
are equal when
√
3x2 = x2 x3 + 1
√
3 = x3 + 1
9 = x3 + 1
8 = x3
2 = x.
The function g(x) = 3x2 is the larger of the two on the interval [0, 2], as can be seen either
algebraically or by plugging in x = 1, say.
12
y = 3x2
√
y = x2 x3 + 1
2
The area in question is therefore
Z
2
2
3x − x
0
2
√
x3
2
2 3
3
3/2 + 1 dx = x − (x + 1)
9
0 2 3
2 3
3/2
3
3/2
3
− 0 − (0 + 1)
= 2 − (2 + 1)
9
9
2
20
= 8−6 − 0−
= .
9
9
Marking scheme:
• 1 mark for correct integrand (including order)
• 1 mark for correct limits of integration
R2
• 1 mark for correct evaluation of 0 3x2 dx
R2 √
• 2 marks for correct evaluation of 0 x2 x3 + 1 dx