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Portillo, Tom – Homework 8 – Due: Oct 24 2006, 3:00 am – Inst: David Benzvi
1
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printout
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have
21
questions.
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beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
AFter partitioning the interval [0
,
4] into 4
equal subintervals, use the trapezoidal rule to
estimate the integral
I
=
Z
4
0
7
√
4 +
x
2
dx.
1.
I
≈
10
.
4794
2.
I
≈
10
.
2794
3.
I
≈
9
.
87944
4.
I
≈
10
.
6794
5.
I
≈
10
.
0794
correct
Explanation:
When the interval [0
,
4] is partitioned into
4 equal subintervals the trapezoidal rule esti
mates the integral
I
=
Z
4
0
f
(
x
)
dx
as
I
≈
1
2
‡
f
(0)+2
f
(1)+2
f
(2)+2
f
(3)+
f
(4)
·
.
When
f
(
x
) =
7
√
4 +
x
2
,
thereFore,
I
≈
7
2
‡
1
2
+
2
√
4 + 1
+
2
√
4 + 4
+
2
√
4 + 9
+
1
√
4 + 16
·
.
Consequently,
I
≈
10
.
0794
.
keywords:
partition, trapezoidal rule, inte
gral, estimate
002
(part 1 oF 1) 10 points
A radar gun was used to record the speed
oF a runner during the frst 5 seconds oF a race
as shown in the table
t
(secs)
vel
(m/sec)
0
0
0.5
4
.
6
1.0
7
.
6
1.5
8
.
9
2.0
9
.
1
2.5
10
.
1
3.0
10
.
3
3.5
10
.
7
4.0
10
.
8
4.5
10
.
88
5.0
10
.
99
Use Simpson’s Rule and all the given data
to estimate the distance the runner covered
during those 5 seconds.
1.
dist
≈
44
.
49 meters
2.
dist
≈
44
.
53 meters
3.
dist
≈
44
.
51 meters
4.
dist
≈
44
.
55 meters
correct
5.
dist
≈
44
.
47 meters
Explanation:
The distance covered during those 5 sec
onds is given by the integral
I
=
Z
5
0
v
(
t
)
dt
where
v
(
t
) is the velocity oF the runner at time
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2
t
. Simpson’s Rule estimates this integral as
I
≈
1
6
n
v
(0) + 4
v
‡
1
2
·
+ 2
v
(1) + 4
v
‡
3
2
·
+ 2
v
(2) + 4
v
‡
5
2
·
+ 2
v
(3) + 4
v
‡
7
2
·
+ 2
v
(4) + 4
v
‡
9
2
·
+
v
(5)
o
.
Reading of the values oF
v
(
t
) From the table
we thus see that
I
≈
44
.
55 meters
.
keywords: de±nite integral, table, Simpson’s
rule, velocity, distance
003
(part 1 oF 1) 10 points
²ind the total area under the graph oF
y
=
9
x
4
For
x
≥
1
.
1.
Area = 3
correct
2.
Area =
5
3
3.
Area =
7
3
4.
Area =
∞
5.
Area =
8
3
6.
Area = 2
Explanation:
The total area under the graph For
x
≥
1 is
an improper integral whose value is the limit
lim
t
→∞
Z
t
1
9
x
4
dx.
Now
Z
t
1
9
x

4
dx
=

3
h
x

3
i
t
1
.
Consequently,
Area =
lim
t
→∞
3
h
1

t

3
i
= 3
.
keywords: improper integral, area, limit
004
(part 1 oF 1) 10 points
Determine iF the improper integral
I
=
Z
0
∞
4
2
x

5
dx
converges, and iF it does, ±nd its value.
1.
I
=
5
4
2.
I
= 2
3.
I
=
4
5
4.
I
=
1
2
5.
integral is not convergent
correct
Explanation:
The integral is improper because oF the in
±nite interval oF integration. To test For con
vergence, thereFore, we have to check iF the
limit
lim
t
→∞
Z
0
t
4
2
x

5
dx
exists. But
Z
0
t
4
2
x

5
dx
=
h
2 ln

2
x

5

i
0
t
=
‡
2 ln 5

2 ln

2
t

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 Spring '06
 Benzvi

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