Version 010 – EXAM 1 – Odell – (58340)
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
] = [3
,
10]
,
and
x
∗
i
=
x
i
, we see that
area =
lim
n
n
s
i
= 1
p
3 +
7
i
n
P
5
7
n
.
003
10.0 points
Express the limit
lim
n
n
s
i
=1
5
x
i
sin
x
i
Δ
x
as a deFnite integral on the interval [1
,
6].
1.
limit =
i
6
1
5 sin
x dx
2.
limit =
i
1
6
5 sin
x dx
3.
limit =
i
6
1
5
x dx
4.
limit =
i
1
6
5
x dx
5.
limit =
i
1
6
5
x
sin
x dx
6.
limit =
i
6
1
5
x
sin
x dx
correct
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
) = 5
x
sin
x,
[
a, b
] = [1
,
6]
.
Thus
limit =
i
6
1
5
x
sin
x dx
.
004
10.0 points
If
F
(
x
) =
i
x
0
5
e
8 sin
2
θ
dθ ,
Fnd the value of
F
′
(
π/
4).
1.
F
′
(
4) = 4
e
4
2.
F
′
(
4) = 4
e
5
3.
F
′
(
4) = 4
e
8
4.
F
′
(
4) = 5
e
4
correct
5.
F
′
(
4) = 5
e
8
Explanation: