MATH 101 Quiz #2 (v.A1)
Last Name:
Friday, January 29
First Name:
Grade:
Student-No:
Section:
Very short answer question
d
1. 1 mark Find
dx
Z
1
sin x
dt
.
1 + t4
Answer: −
Z
Solution: Define g(x) =
x
1
dt
=−
1 + t4
Z
x
1
cos x
1 + sin4 x
1
1
dt, so that g 0 (x) = −
by FTC1.
4
1+t
1 + x4
Then by the Chain Rule,
Z 1
dt
d
1
d
d
· cos x.
g(sin x) = g 0 (sin x) ·
sin x = −
=
4
dx
dx
dx
1 + sin4 x
sin x 1 + t
Marking scheme: 1 for a correct answer in the box
Short answer questions—you must show your work
Z
2. 2 marks Evaluate
e
√
e
dx
. Simplify your answer as much as possible.
x log x
Answer: log 2
Solution: Setting u = log x, we have du =
Z
e
√
√
since u = log( e) =
1
2
e
dx
=
x log x
when x =
1
Z
1
2
√
Z
x=e
√
e
1
x
dx and so
1
1
· dx =
log x x
Z
u=1
u= 21
1
du,
u
e and u = log(e) = 1 when x = e. Then, by FTC2,
1
1
1
du = (log |u|) 1 = log 1 − log = log 2.
2
u
2
Marking scheme:
• 1 mark Z
for correctly using substitution, either for converting to the u-integral or for
1
getting
dx = log(log x) + C
x log x
• 1 mark for getting the right answer.
3. 2 marks A car traveling at 30 m/s applies its brakes at time t = 0, its velocity (in m/s)
decreasing according to the formula v(t) = 30 − 10t. How far does the car go before it stops?
Simplify your answer completely.
Answer: 45 m
Solution: The car stops when v(t) = 0, which occurs at time t = 3. The distance covered
up to that time is
Z 3
3
v(t) dt = (30t − 5t2 )0 = (90 − 45) − 0 = 45 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, to the right of the y-axis, that is bounded by the
graphs of f (x) = x x2 + 16 and g(x) = 5x. Your answer may be left in calculator-ready form.
√
Solution: For that computation, we will need an antiderivative of x x2 + 16, which can
be found using substitution using u = x2 + 16, so that 12 du = x dx:
Z
x
√
Z
x2
+ 16 dx =
√ 1
1
u · du =
2
2
Z
u
1/2
1
1 u3/2
du =
+ C = (x2 + 16)3/2 + C.
2 3/2
3
The two functions f (x) and g(x) are clearly equal at x = 0. If x 6= 0, then the functions
are equal when
√
5x = x x2 + 16
√
5 = x2 + 16
25 = x2 + 16
9 = x2
±3 = x.
Since we care about the region to the right of the y-axis, the other point of intersection is
when x = 3. The linear function is the larger of the two on the interval [0, 3], as can be
seen either algebraically or by plugging in x = 1, say (or deduced from concavity).
15
y = 5x
√
y = x x2 + 16
3
The area in question is therefore
2
3
Z 3
√
5x
1 2
3/2 2
− (x + 16)
5x − x x + 16 dx =
2
3
0
0
2
5·3
1 2
5 · 02 1 2
3/2
3/2
=
− (3 + 16)
− (0 + 16)
−
2
3
2
3
64
45 61
13
45 125
−
− 0−
=
−
= .
=
2
3
3
2
3
6
Marking scheme:
• 1 mark for correct integrand (including order)
• 1 mark for correct limits of integration
R3
• 1 mark for correct evaluation of 0 5x dx
R3 √
• 2 marks for correct evaluation of 0 x x2 + 16 dx