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Mcbiles, Emily – Homework 4 – Due: Sep 26 2006, 3:00 am – Inst: David Fonken
1
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21
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The due time is Central
time.
001
(part 1 o± 1) 10 points
Determine the indefnite integral
I
=
Z
1
θ
2
‰
4 sin
‡
1
θ
·

2
θ
¾
dθ
1.
I
=

1
θ
2
+ 4 sin
‡
1
θ
·
+
C
2.
I
=

1
θ
2
+ 4 cos
‡
1
θ
·
+
C
3.
I
=
1
θ
2

4 cos
‡
1
θ
·
+
C
4.
I
=
1
θ
2
+ 4 sin
‡
1
θ
·
+
C
5.
I
=
1
θ
2
+ 4 cos
‡
1
θ
·
+
C
correct
6.
I
=
1
θ
2

4 sin
‡
1
θ
·
+
C
Explanation:
Set
u
= 1
/θ
. Then
du
=

1
θ
2
dθ ,
so
I
=

Z
(4 sin
u

2
u
)
du
= 4 cos
u
+
u
2
+
C
with
C
an arbitrary constant. Consequently,
I
=
1
θ
2
+ 4 cos
‡
1
θ
·
+
C
.
keywords: integral, trig ±unction, substitution
002
(part 1 o± 1) 10 points
Evaluate the defnite integral
I
=
Z
π/
4
0
{
9 sec
2
x

6 sec
2
x
tan
2
x
}
dx.
1.
I
= 8
2.
I
=
22
3
3.
I
=
25
3
4.
I
=
23
3
5.
I
= 7
correct
Explanation:
Since the integrand is o± the ±orm
sec
2
xf
(tan
x
)
,
f
(
x
) = 9

6
x
2
,
the substitution
u
= tan
x
is needed.
For
then
du
= sec
2
xdx,
while
x
= 0
=
⇒
u
= 0
,
x
=
π
4
=
⇒
u
= 1
.
Thus
I
=
Z
1
0
(9

6
u
2
)
du
=
h
9
u

2
u
3
i
1
0
.
Consequently,
I
= 7
.
keywords:
indefnite integral, trigonometric
substitution
003
(part 1 o± 1) 10 points
Evaluate the defnite integral
I
=
Z
6
3
x
+ 1
√
x

2
dx.
Correct answer: 10
.
6667 .
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View Full Document Mcbiles, Emily – Homework 4 – Due: Sep 26 2006, 3:00 am – Inst: David Fonken
2
Explanation:
Set
u
=
x

2. Then
du
=
dx,
so
Z
6
3
x
+ 1
√
x

2
dx
=
Z
4
1
u
+ 3
√
u
du
=
Z
4
1
‡
u
1
/
2
+ 3
u

1
/
2
·
du.
Consequently,
I
=
Z
4
1
‡
u
1
/
2
+ 3
u

1
/
2
·
du
=
h
2
3
u
3
/
2
+ 6
u
1
/
2
i
4
1
=
32
3
.
keywords: integral, square root, substitution
004
(part 1 of 1) 10 points
Find the area between the graph of
f
and
the
x
axis on the interval [0
,
6] when
f
(
x
) = 3
x

x
2
.
1.
Area = 26 sq.units
2.
Area = 23 sq.units
3.
Area = 25 sq.units
4.
Area = 27 sq.units
correct
5.
Area = 24 sq.units
Explanation:
The graph of
f
is a parabola opening down
wards and crossing the
x
axis at
x
= 0 and
x
= 3.
Thus the required area is similar to
the shaded region in the ±gure below.
graph of
f
In terms of de±nite integrals, therefore, the
required area is given by
Z
3
0
(3
x

x
2
)
dx

Z
6
3
(3
x

x
2
)
dx.
Now
Z
3
0
(3
x

x
2
)
dx
=
h
3
2
x
2

1
3
x
3
i
3
0
=
9
2
,
while
Z
6
3
(3
x

x
2
)
dx
=
h
3
2
x
2

1
3
x
3
i
6
3
=

45
2
.
Consequently,
Area = 27 sq.units.
keywords: integral, graph, area
005
(part 1 of 1) 10 points
Find the area enclosed by the graphs of
f
(
x
) = cos
x,
g
(
x
) = sin
x
on [0
, π
].
1.
area =
√
2 + 1
2.
area = 2(
√
2 + 1)
3.
area =
√
2
4.
area = 4(
√
2 + 1)
5.
area = 2
√
2
correct
6.
area = 4
√
2
Explanation:
Mcbiles, Emily – Homework 4 – Due: Sep 26 2006, 3:00 am – Inst: David Fonken
3
The area between the graphs of
y
=
f
(
x
)
and
y
=
g
(
x
) on the interval [
a, b
] is expressed
by the integral
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This note was uploaded on 03/19/2008 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin

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