Pham, Quoc – Homework 5 – Due: Feb 20 2007, 3:00 am – Inst: Eric Katerman
1
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printout
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have
17
questions.
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be±ore answering.
The due time is Central
time.
001
(part 1 o± 1) 10 points
Find the most general ±unction
f
such that
f
0
(
x
) = 6 +
5
√
4

x
2
.
1.
f
(
x
) = 3

5
2
tan

1
x
2
+
C
2.
f
(
x
) = 6
x
+
5
2
tan

1
x
2
+
C
3.
f
(
x
) = 6

5
2
tan

1
x
+
C
4.
f
(
x
) = 6
x
+ 5 sin

1
x
2
+
C
correct
5.
f
(
x
) = 6
x

5 sin

1
x
2
+
C
6.
f
(
x
) = 3
x
+ 5 sin

1
x
+
C
Explanation:
Since
d
dx
‡
sin

1
x
2
·
=
1
√
4

x
2
,
we see that
f
(
x
) = 6
x
+ 5 sin

1
x
2
+
C
with
C
an arbitrary constant.
keywords:
002
(part 1 o± 1) 10 points
Find the value o± the integral
I
=
Z
4
1
4
9 + (
x

1)
2
dx.
1.
I
=
2
3
π
2.
I
= 3
π
3.
I
=
2
3
4.
I
= 3
5.
I
=
1
3
π
correct
6.
I
=
4
3
π
Explanation:
Set 3 tan
u
=
x

1. Then
9 + (
x

1)
2
= 9 + (3 tan
u
)
2
= 9(1 + tan
2
u
) = 9 sec
2
u,
while
3 sec
2
udu
=
Also
x
= 1
=
⇒
u
= 0
,
and
x
= 4
=
⇒
u
=
π
4
.
In this case
I
=
Z
π/
4
0
12 sec
2
u
9 sec
2
u
du
=
4
3
Z
π/
4
0
du.
Consequently,
I
=
4
3
h
u
i
π/
4
0
=
1
3
π
.
keywords:
003
(part 1 o± 1) 10 points
Evaluate the defnite integral
I
=
Z
1
4
0
1
√
1

4
x
2
Correct answer: 0
.
261799 .
Explanation:
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View Full DocumentPham, Quoc – Homework 5 – Due: Feb 20 2007, 3:00 am – Inst: Eric Katerman
2
Since
Z
1
√
1

x
2
dx
= sin

1
x
+
C ,
a change of variable
x
is needed to reduce
I
to
this form. Set
u
= 2
x
. Then
du
= 2
dx
, and
x
= 0
=
⇒
u
= 0
,
while
x
=
1
4
=
⇒
u
=
1
2
.
In this case
I
=
1
2
Z
1
2
0
1
√
1

u
2
du
=
•
1
2
sin

1
u
‚
1
2
0
.
Consequently,
I
=
1
2
arcsin
µ
1
2
¶
= 0
.
261799
.
keywords:
004
(part 1 of 1) 10 points
Find the value of the de±nite integral
I
=
Z
√
3
2
0
8 sin

1
x
√
1

x
2
dx.
1.
I
=
4
25
π
2
2.
I
=
1
3
π
2
3.
I
=
1
9
π
2
4.
I
=
4
9
π
2
correct
5.
I
=
1
4
π
2
Explanation:
Since
Z
1
√
1

x
2
dx
= sin

1
x
+
C ,
this suggests the substitution
u
= sin

1
x
, for
then
du
=
1
√
1

x
2
dx,
while
x
= 0
=
⇒
u
= 0
,
x
=
√
3
2
=
⇒
u
=
π
3
.
Thus
I
= 8
Z
π
3
0
udu
= 4
h
u
2
i
π
3
0
.
Consequently,
I
=
4
9
π
2
.
keywords:
005
(part 1 of 1) 10 points
Evaluate the de±nite integral
I
=
Z
π
π/
4
3 sin 2
x
1 + cos
2
2
x
Correct answer:

1
.
1781 .
Explanation:
Set
u
= cos 2
x
. Then
du
=

2 sin 2
xdx,
while
x
=
π
4
=
⇒
u
= 0
,
x
=
π
=
⇒
u
= 1
.
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