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Cheung, Anthony – Exam 2 – Due: Nov 7 2006, 11:00 pm – Inst: David Benzvi
1
This
printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
IF
f
is the Function whose graph on [0
,
10]
is given by
2
4
6
8
2
4
6
8
use the Trapezoidal Rule with
n
= 5 to esti
mate the defnite integral
I
=
Z
7
2
f
(
x
)
dx.
1.
I
≈
28
2.
I
≈
53
2
correct
3.
I
≈
27
4.
I
≈
57
2
5.
I
≈
55
2
Explanation:
The Trapezoidal Rule estimates the defnite
integral
I
=
Z
7
2
f
(
x
)
dx
by
I
≈
1
2
h
f
(2) + 2
{
f
(3)+
···
+
f
(6)
}
+
f
(7)
i
when
n
= 5. ±or the given
f
, thereFore,
I
≈
1
2
h
8 + 2
{
7 + 5 + 5 + 4
}
+ 3
i
=
53
2
,
reading o² the values oF
f
From the graph.
keywords: trapezoidal rule, integral, graph
002
(part 1 oF 1) 10 points
±ind
Z
e
8
x
81 +
e
16
x
dx.
1.
arcsin
e
8
x
+
C
2.
None oF these.
3.
1
72
arctan
µ
1
9
e
8
x
¶
+
C
correct
4.
1
9
arcsin
e
8
x
+
C
5.
1
9 +
e
8
x
+
C
6.
1
9
arcsec
e
8
x
+
e
8
x
+
C
Explanation:
Z
e
8
x
81 +
e
16
x
dx
=
1
8
Z
8
e
8
x
dx
(9)
2
+ (
e
8
x
)
2
=
1
8
Z
d
(
e
8
x
)
(9)
2
+ (
e
8
x
)
2
=
1
8
·
1
9
arctan
µ
1
9
e
8
x
¶
+
C
=
1
72
arctan
µ
1
9
e
8
x
¶
+
C
keywords: exponential Function, inverse trig
Function
003
(part 1 oF 1) 10 points
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2
Evaluate the defnite integral
I
=
Z
2
0
4
x

8
x
2

2
x

3
dx.
1.
I
= 2 ln 3
correct
2.
I
=

2 ln 3
3.
I
=

4 ln 3
4.
I
= 2 ln 4
5.
I
= 4 ln 4
6.
I
=

2 ln 4
7.
I
= 4 ln 3
8.
I
=

4 ln 4
Explanation:
AFter Factorization
x
2

2
x

3 = (
x
+ 1)(
x

3)
.
But then by partial Fractions,
4
x

8
x
2

2
x

3
=
3
x
+ 1
+
1
x

3
.
Now
Z
2
0
3
x
+ 1
dx
=
h
3 ln

(
x
+ 1)

i
2
0
= 3 ln 3
,
while
Z
2
0
1
x

3
dx
=
h
ln

(
x

3)

i
2
0
=

ln 3
.
Consequently,
I
= 2 ln 3
.
keywords: defnite integral, rational Function,
partial Fractions, natural log
004
(part 1 oF 1) 10 points
Evaluate the defnite integral
I
=
Z
π/
4
0
3 cos
x
+ 6 sin
x
cos
3
x
dx.
1.
I
= 6
correct
2.
I
=
9
2
3.
I
= 9
4.
I
=
3
2
5.
I
= 0
Explanation:
AFter division we see that
3 cos
x
+ 6 sin
x
cos
3
x
= 3 sec
2
x
+ 6 tan
x
sec
2
x
= (3 + 6 tan
x
) sec
2
x.
Thus
I
=
Z
π/
4
0
(3 + 6 tan
x
) sec
2
xdx.
Let
u
= tan
x
; then
du
= sec
2
xdx,
while
x
= 0
=
⇒
u
= 0
,
x
=
π
4
=
⇒
u
= 1
.
In this case
I
=
Z
1
0
(3 + 6
u
)
du
=
£
3
u
+ 3
u
2
/
1
0
.
Consequently,
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 RAdin
 Calculus

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