# Very short answer question

```MATH 101 Quiz #2 (v.A3)
Last Name:
Friday, January 29
First Name:
Student-No:
Section:
Very short answer question
d
1. 1 mark Find
du
Z
0
cos u
dt
.
1 + t3
Z
Solution: Define g(u) =
u
0
dt
=−
1 + t3
u
Z
0
sin u
1 + cos3 u
1
1
0
dt,
so
that
g
(u)
=
−
by FTC1.
1 + t3
1 + u3
Then by the Chain Rule,
Z 0
dt
d
1
d
d
g(cos u) = g 0 (cos x) &middot;
cos u = −
&middot; (− sin u).
=
3
du
du
du
1 + cos3 u
cos u 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
e4
dx
. Simplify your answer as much as possible.
x log x
Solution: Setting u = log x, we have du =
Z
e
e4
dx
=
x log x
Z
x=e4
x=e
1
x
dx and so
1
1
&middot; dx =
log x x
Z
u=4
u=1
1
du,
u
since u = log(e) = 1 when x = e and u = log(e4 ) = 4 when x = e4 . Then, by FTC2,
Z 4
4
1
du = (log |u|)1 = log 4 − log 1 = log 4.
1 u
Marking scheme:
• 1 mark Z
for correctly using substitution, either for converting to the u-integral or for
1
dx = log(log x) + C
getting
x log x
• 1 mark for getting the right answer.
3. 2 marks A car traveling at 40 m/s applies its brakes at time t = 0, its velocity (in m/s)
decreasing according to the formula v(t) = 40 − 8t. How far does the car go before it stops?
Solution: The car stops when v(t) = 0, which occurs at time t = 5. The distance covered
up to that time is
Z 5
5
v(t) dt = (40t − 4t2 )0 = (200 − 100) − 0 = 100 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) = 2x x2 + 7 and g(x) = 8x. Your answer may be left in calculator-ready form.
√
Solution: For that computation, we will need an antiderivative of 2x x2 + 7, which can
be found using substitution using u = x2 + 7, so that du = 2x dx:
Z
Z
Z
√
√
u3/2
2
2
2x x + 7 dx =
u du = u1/2 du =
+ C = (x2 + 7)3/2 + C.
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
√
8x = 2x x2 + 7
√
4 = x2 + 7
16 = x2 + 7
9 = x2
&plusmn;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).
24
y = 8x
√
y = 2x x2 + 7
3
The area in question is therefore
Z
3
8x − 2x
√
0
x2
3
2 2
2
3/2 + 7 dx = 4x − (x + 7)
3
0 2 2
2 2
3/2
2
3/2
2
− 4 &middot; 0 − (0 + 7)
= 4 &middot; 3 − (3 + 7)
3
3
128
20
2
2
= 36 −
− 0 − 73/2 = 73/2 − .
3
3
3
3
Marking scheme:
• 1 mark for correct integrand (including order)
• 1 mark for correct limits of integration
R3
• 1 mark for correct evaluation of 0 8x dx
R3 √
• 2 marks for correct evaluation of 0 2x x2 + 7 dx
```