Math 1210
Quiz 9 Solutions
March 28th, 2014
Answer the following three (3) questions. The value of every question is
indicated at the beginning of it. You may use scratch paper, but you can
only turn in this sheet. Please write your answer in the space provided. You
have 20 minutes.
Name:
UID:
1. (10 points) Consider the function f (x) =
Rx
0
cos(u) du over the interval [0, 2π].
(i) (5 points) Find f 0 (x) and f 00 (x).
Solution:
d
f (x) =
dx
0
Z
x
f 00 (x) =
cos u du = cos x,
0
d
cos x = − sin x
dx
where the first derivative follows from the First Fundamental Theorem of Calculus.
(ii) (5 points) Over which intervals is f (x) is increasing (resp. decreasing)?
Solution: Recall that a differentiable function is increasing over an interval if its
derivative is positive at every x-valuein that interval. Since f 0 (x) = cos x is positive
, 2π , we conclude that f is increasing over these
over the intervals 0, π2 and 3π
2
, we conclude that f is
intervals, and since f 0 (x) = cos x is negative over π2 , 3π
2
decreasing over this interval.
R cos x
2. (10 points) Consider the function f (x) = sin x t3 dt
(i) (4 points) Compute f π4 and f (0).
Solution:
f
π 4
Z
cos
π
4
=
sin
Z
t3 dt =
√
2
2
π
4
cos 0
3
√
2
2
Z
Z
t dt =
f (0) =
sin 0
0
1
t3 dt = 0
1
1 4 1
t dt = t =
4 0 4
3
(ii) (6 points) Find f 0 (x).
Ra
R cos x
Hint: Write f (x) = sin x t3 dt + a t3 dt.
Solution:
Z a
Z cos x
Z sin x
Z cos x
d
d
d
d
3
3
3
f (x) =
t dt +
t dt = −
t dt +
t3 dt
dx sin x
dx a
dx a
dx a
= − sin3 x cos x − cos3 x(− sin x) = − sin3 x cos x − cos3 x sin x
0
3.
(i) (6 points) Compute the indefinite integral
√
Z
x sin x2 + 4
√
dx
x2 + 4
√
Hint: Use the change of variables u = x2 + 4.
√
Solution: If u = u(x) = x2 + 4, then du
= √xx2 +4 , so dx =
dx
thus rewrite out integral as
Z
√
√
x2 +4
du.
x
We may
dx
z√
}|
{ Z
2
2
x sin x + 4
x sin u
x +4
√
√
du = sin u du = − cos u + C
dx =
2
2
x
x +4
x +4
√
= − cos
x2 + 4 + C
Z
(ii) (4 points) Compute the definite integral
√
Z 1
x sin x2 + 4
√
dx
x2 + 4
0
Solution:
We may proceed as above, using the same change of variables u =
√
x2 + 4, but bearing in mind that the limits of integration
√ with respect to the new
2
variable u are different.
When
√ x = 0, we have u(0) = 0 + 4 = 2 and when x = 1
√
we have u(1) =√ 12 + 4 = 5. Therefore, as x varies between 0 and 1, u varies
between 2 and 5 and hence our integral reads
√
Z 1
Z u(1)
Z √5
√
x sin x2 + 4
√
dx =
sin u du =
sin u du = − cos u|2 5
x2 + 4
2
0
u(0)
√
= − cos 5 + cos 2
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