Mcbiles, Emily – Homework 2 – Due: Sep 12 2006, 3:00 am – Inst: David Fonken
1
This
printout
should
have
20
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
Rewrite the sum
n
4+
‡
1
9
·
2
o
+
n
8+
‡
2
9
·
2
o
+
. . .
+
n
28+
‡
7
9
·
2
o
using sigma notation.
1.
9
X
i
= 1
n
4
i
+
‡
i
9
·
2
o
2.
7
X
i
= 1
4
n
i
+
‡
i
9
·
2
o
3.
9
X
i
= 1
4
n
i
+
‡
i
9
·
2
o
4.
7
X
i
= 1
n
i
+
‡
4
i
9
·
2
o
5.
9
X
i
= 1
4
n
i
+
‡
4
i
9
·
2
o
6.
7
X
i
= 1
n
4
i
+
‡
i
9
·
2
o
correct
Explanation:
The terms are of the form
n
4
i
+
‡
i
9
·
2
o
,
with
i
= 1
,
2
, . . . ,
7. Consequently, in sigma
notation the sum becomes
7
X
i
= 1
n
4
i
+
‡
i
9
·
2
o
.
keywords: Stewart5e, summation notation,
002
(part 1 of 1) 10 points
Estimate the area,
A
, under the graph of
f
(
x
) =
4
x
on [1
,
5] by dividing [1
,
5] into four equal
subintervals and using right endpoints.
Correct answer: 5
.
133 .
Explanation:
With four equal subintervals and right end
points as sample points,
A
≈
n
f
(2) +
f
(3) +
f
(4) +
f
(5)
o
1
since
x
i
=
x
*
i
=
i
+ 1. Consequently,
A
≈
5
.
133
.
keywords: Stewart5e, area, rational function,
Riemann sum,
003
(part 1 of 1) 10 points
Decide which of the following regions has
area =
lim
n
→ ∞
n
X
i
= 1
π
5
n
sin
iπ
5
n
without evaluating the limit.
1.
n
(
x, y
) : 0
≤
y
≤
sin 3
x,
0
≤
x
≤
π
5
o
2.
n
(
x, y
) : 0
≤
y
≤
sin
x,
0
≤
x
≤
π
10
o
3.
n
(
x, y
) : 0
≤
y
≤
sin
x,
0
≤
x
≤
π
5
o
correct
4.
n
(
x, y
) : 0
≤
y
≤
sin 3
x,
0
≤
x
≤
π
10
o
5.
n
(
x, y
) : 0
≤
y
≤
sin 2
x,
0
≤
x
≤
π
10
o
6.
n
(
x, y
) : 0
≤
y
≤
sin 2
x,
0
≤
x
≤
π
5
o
Explanation:
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Mcbiles, Emily – Homework 2 – Due: Sep 12 2006, 3:00 am – Inst: David Fonken
2
The area under the graph of
y
=
f
(
x
) on
an interval [
a, b
] is given by the limit
lim
n
→ ∞
n
X
i
= 1
f
(
x
i
)Δ
x
when [
a, b
] is partitioned into
n
equal subin
tervals
[
a, x
1
]
,
[
x
1
, x
2
]
,
. . . ,
[
x
n

1
, x
n
]
each of length Δ
x
= (
b

a
)
/n
.
When the area is given by
A
=
lim
n
→ ∞
n
X
i
= 1
π
5
n
sin
iπ
5
n
,
therefore, we see that
f
(
x
i
) = sin
iπ
5
n
,
Δ
x
=
π
5
n
,
where in this case
x
i
=
iπ
5
n
,
f
(
x
) = sin
x,
[
a, b
] =
h
0
,
π
5
i
.
Consequently, the area is that of the region
under the graph of
y
= sin
x
on the interval
[0
, π/
5].
In setbuilder notation this is the
region
n
(
x, y
) : 0
≤
y
≤
sin
x,
0
≤
x
≤
π
5
o
.
keywords: limit Riemann sum, area, trig func
tion
004
(part 1 of 1) 10 points
Find an expression for the area of the region
under the graph of
f
(
x
) =
x
3
on the interval [3
,
10].
1.
area =
lim
n
→ ∞
n
X
i
= 1
‡
3 +
10
i
n
·
3
8
n
2.
area =
lim
n
→ ∞
n
X
i
= 1
‡
3 +
8
i
n
·
3
7
n
3.
area =
lim
n
→ ∞
n
X
i
= 1
‡
3 +
7
i
n
·
3
7
n
correct
4.
area =
lim
n
→ ∞
n
X
i
= 1
‡
3 +
10
i
n
·
3
7
n
5.
area =
lim
n
→ ∞
n
X
i
= 1
‡
3 +
8
i
n
·
3
8
n
6.
area =
lim
n
→ ∞
n
X
i
= 1
‡
3 +
7
i
n
·
3
8
n
Explanation:
The area of the region under the graph of
f
on an interval [
a, b
] is given by the limit
A
=
lim
n
→ ∞
n
X
i
= 1
f
(
x
*
i
) Δ
x
when [
a, b
] is partitioned into
n
equal subin
tervals
[
a, x
1
]
,
[
x
1
, x
2
]
, . . . ,
[
x
n

1
, b
]
each of length Δ
x
= (
b

a
)
/n
and
x
*
i
is an
arbitrary sample point in [
x
i

1
, x
i
].
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 Fall '08
 RAdin
 Derivative, dx, Riemann sum, David Fonken, Mcbiles

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