Miller, Kierste – Exam 1 – Due: Feb 20 2007, 11:00 pm – Inst: Gary Berg
1
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printout
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have
16
questions.
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
When
2
1
0
1
2
3
4
5
6
7
8
9
10
2
4
6
8
10
2
4

2

4
is the graph of a function
f
, use rectangles to
estimate the definite integral
I
=
Z
10
0

f
(
x
)

dx
by subdividing [0
,
10] into 10 equal subin
tervals and taking right endpoints of these
subintervals.
1.
I
≈
19
2.
I
≈
22
correct
3.
I
≈
21
4.
I
≈
23
5.
I
≈
20
Explanation:
The definite integral
I
=
Z
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
1
0
1
2
3
4
5
6
7
8
9
10
2
4
6
8
10
2
4

2

4
Each rectangle has baselength 1;
and it’s
height can be read off from the graph. Thus
Area = 5 + 3 + 2 + 1 + 1 + 3 + 3 + 3 + 1
.
Consequently,
I
≈
22
.
keywords: definite integrals, graph, absolute
value
002
(part 1 of 1) 10 points
For which integral,
I
, is the expression
1
5
ˆ
r
1
5
+
r
2
5
+
r
3
5
+
. . .
+
r
5
5
!
a Riemann sum approximation.
1.
I
=
Z
1
0
√
x dx
correct
2.
I
=
1
5
Z
1
0
√
x dx
3.
I
=
Z
1
0
r
x
5
dx
4.
I
=
1
5
Z
5
0
√
x dx
5.
I
=
1
5
Z
1
0
r
x
5
dx
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Miller, Kierste – Exam 1 – Due: Feb 20 2007, 11:00 pm – Inst: Gary Berg
2
Explanation:
When the interval [
a, b
] is divided into
n
equals intervals, then
n
X
i
= 1
f
a
+
i
(
b

a
)
n
¶
b

a
n
is a Riemann sum approximation for the inte
gral
Z
b
a
f
(
x
)
dx
of
f
over [
a, b
] using right endpoints as sample
points. Comparing this with
1
5
ˆ
r
1
5
+
r
2
5
+
r
3
5
+
. . .
+
r
5
5
!
we see that [
a, b
] = [0
,
1],
n
= 5, and
[
a, b
] = [0
,
1]
,
n
= 5
,
f
(
x
) =
√
x .
Consequently, the given expression is a Rie
mann sum for
I
=
Z
1
0
√
x dx
.
keywords:
integral,
Riemann sum,
square
root
003
(part 1 of 1) 10 points
If
F
(
x
) =
Z
x
0
3
e
8 sin
2
θ
dθ ,
find the value of
F
0
(
π/
4).
1.
F
0
(
π/
4) = 4
e
8
2.
F
0
(
π/
4) = 3
e
4
correct
3.
F
0
(
π/
4) = 4
e
3
4.
F
0
(
π/
4) = 3
e
8
5.
F
0
(
π/
4) = 4
e
4
Explanation:
By the Fundamental theorem of calculus,
F
0
(
x
) = 3
e
8 sin
2
x
.
At
x
=
π/
4, therefore,
F
0
(
π/
4) = 3
e
4
since sin(
π
4
) =
1
√
2
.
keywords: integral, FTC
004
(part 1 of 1) 10 points
Evaluate the definite integral
I
=
Z
π
8
0
(5 cos 4
x

3 sin 4
x
)
dx .
1.
I
=
1
2
correct
2.
I
=

1
4
3.
I
=
3
4
4.
I
=
1
4
5.
I
= 0
Explanation:
To reduce the integral to one involving just
sin
u
and cos
u
set
u
= 4
x
. Then
du
= 4
dx
,
and so
I
=
1
4
Z
π
2
0
(5 cos
u

3 sin
u
)
du
=
1
4
h
5 sin
u
+ 3 cos
u
i
π
2
0
.
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