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Version 075 – EXAM 1 – Cepparo – (58400)
1
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beFore answering.
001
10.0 points
When
2
4
6
8
10
2
4
−
2
−
4
is the graph oF a Function
f
, use rectangles to
estimate the defnite integral
I
=
i
10
0

f
(
x
)

dx
by subdividing [0
,
10] into 10 equal subin
tervals and taking right endpoints oF these
subintervals.
1.
I
≈
24
2.
I
≈
25
3.
I
≈
21
4.
I
≈
23
5.
I
≈
22
correct
Explanation:
The defnite integral
I
=
i
10
0

f
(
x
)

dx
is the area between the graph oF
f
and the
interval [0
,
10]. The area is estimated using
the grayshaded rectangles in
2
4
6
8
10
2
4
−
2
−
4
Each rectangle has baselength 1; and it’s
height can be read o± From the graph. Thus
Area = 4 + 3 + 2 + 1 + 1 + 3 + 4 + 3 + 1
.
Consequently,
I
≈
22
.
002
10.0 points
²ind an expression For the area oF the region
under the graph oF
f
(
x
) =
x
5
on the interval [4
,
8].
1.
area =
lim
n
→∞
n
s
i
= 1
p
4 +
5
i
n
P
5
5
n
2.
area =
lim
n
→∞
n
s
i
= 1
p
4 +
7
i
n
P
5
4
n
3.
area =
lim
n
→∞
n
s
i
= 1
p
4 +
7
i
n
P
5
5
n
4.
area =
lim
n
→∞
n
s
i
= 1
p
4 +
4
i
n
P
5
5
n
5.
area =
lim
n
→∞
n
s
i
= 1
p
4 +
5
i
n
P
5
4
n
6.
area =
lim
n
→∞
n
s
i
= 1
p
4 +
4
i
n
P
5
4
n
correct
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View Full Document Version 075 – EXAM 1 – Cepparo – (58400)
2
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
s
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
].
Consequently, when
f
(
x
) =
x
5
,
[
a, b
] = [4
,
8]
,
and
x
∗
i
=
x
i
, we see that
area =
lim
n
→∞
n
s
i
= 1
p
4 +
4
i
n
P
5
4
n
.
003
10.0 points
Express the limit
lim
n
→∞
n
s
i
=1
2
x
i
sin
x
i
Δ
x
as a deFnite integral on the interval [3
,
7].
1.
limit =
i
3
7
2
x
sin
x dx
2.
limit =
i
3
7
2 sin
x dx
3.
limit =
i
7
3
2
x dx
4.
limit =
i
7
3
2
x
sin
x dx
correct
5.
limit =
i
7
3
2 sin
x dx
6.
limit =
i
3
7
2
x dx
Explanation:
By deFnition, the deFnite integral
I
=
i
b
a
f
(
x
)
dx
of a continuous function
f
on an interval [
a, b
]
is the limit
I
=
lim
n
→∞
n
s
i
= 1
f
(
x
∗
i
) Δ
x
of the Riemann sum
n
s
i
= 1
f
(
x
∗
i
) Δ
x
formed when the interval [
a, b
] is divided into
n
subintervals of equal width Δ
x
and
x
∗
i
is any
sample point in the
i
th
subinterval [
x
i
−
1
, x
i
].
In the given example,
f
(
x
) = 2
x
sin
x,
[
a, b
] = [3
,
7]
.
Thus
limit =
i
7
3
2
x
sin
x dx
.
004
10.0 points
If
F
(
x
) =
i
x
0
5
e
12 sin
2
θ
dθ ,
Fnd the value of
F
′
(
π/
4).
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This note was uploaded on 03/24/2008 for the course M 408L taught by Professor Radin during the Spring '08 term at University of Texas at Austin.
 Spring '08
 RAdin

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