This preview shows pages 1–3. Sign up to view the full content.
Version 084 – M408L FINAL EXAM – Radin – (57410)
1
This printout should have 25 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
be±ore answering.
001
10.0 points
For each
n
the interval [3
,
9] is divided into
n
subintervals [
x
i
−
1
, x
i
] o± equal length Δ
x
,
and a point
x
∗
i
is chosen in [
x
i
−
1
, x
i
].
Express the limit
lim
n
→∞
n
s
i
= 1
(5
x
∗
i
sin
x
∗
i
) Δ
x
as a defnite integral.
1.
limit =
i
3
9
5
x dx
2.
limit =
i
9
3
5
x dx
3.
limit =
i
3
9
5 sin
x dx
4.
limit =
i
9
3
5
x
sin
x dx
correct
5.
limit =
i
3
9
5
x
sin
x dx
6.
limit =
i
9
3
5 sin
x dx
Explanation:
By defnition, the defnite integral
I
=
i
b
a
f
(
x
)
dx
o± a continuous ±unction
f
on an interval [
a, b
]
is the limit
I
=
lim
n
→∞
n
s
i
= 1
f
(
x
∗
i
) Δ
x
o± the Riemann sum
n
s
i
= 1
f
(
x
∗
i
) Δ
x
±ormed when [
a, b
] is divided into
n
subinter
vals [
x
i
−
1
, x
i
] o± equal length Δ
x
and
x
∗
i
is
some point in [
x
i
−
1
, x
i
].
In the given example,
f
(
x
) = 5
x
sin
x,
[
a, b
] = [3
,
9]
.
Thus
limit =
i
9
3
5
x
sin
x dx
.
002
10.0 points
Determine i±
lim
x
→
0
sin
−
1
(4
x
)
tan
−
1
(3
x
)
exists, and i± it does, fnd its value.
1.
limit = 3
2.
limit = 0
3.
limit =
4
3
correct
4.
limit =
3
4
5.
limit does not exist
6.
limit = 4
Explanation:
Since the limit has the ±orm 0
/
0, it is in
determinate and L’Hospital’s Rule can be ap
plied:
lim
x
→
0
f
(
x
)
g
(
x
)
= lim
x
→
0
f
′
(
x
)
g
′
(
x
)
with
f
(
x
) = sin
−
1
(4
x
)
,
g
(
x
) = tan
−
1
(3
x
)
,
and
f
′
(
x
) =
4
√
1
−
16
x
2
,
g
′
(
x
) =
3
1 + 9
x
2
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document Version 084 – M408L FINAL EXAM – Radin – (57410)
2
In this case,
f
′
(
x
)
g
′
(
x
)
=
4
3
p
1 + 9
x
2
√
1
−
16
x
2
P
−→
4
3
as
x
→
0. Consequently, the limit exists and
limit =
4
3
.
003
10.0 points
Determine
g
′
(
x
) when
g
(
x
) =
i
x
0
r
5
−
6
t
2
dt .
1.
g
′
(
x
) =
6
x
√
5
−
6
x
2
2.
g
′
(
x
) =
r
5
−
6
x
2
correct
3.
g
′
(
x
) =
−
r
5
−
6
x
2
4.
g
′
(
x
) =
x
√
5
−
6
x
2
5.
g
′
(
x
) =
−
6
x
r
5
−
6
x
2
6.
g
′
(
x
) =
−
x
√
5
−
6
x
2
Explanation:
By the Fundamental Theorem of Calculus,
d
dx
p
i
x
a
f
(
t
)
dt
P
=
f
(
x
)
.
When
g
(
x
) =
i
x
0
f
(
t
)
dt ,
f
(
t
) =
r
5
−
6
t
2
,
therefore,
g
′
(
x
) =
r
5
−
6
x
2
.
004
10.0 points
Find
f
(1) when
f
′′
(
x
) = 9
x
and
f
(
−
1) = 5
,
f
′
(
−
1) = 4
.
1.
f
(1) = 4
2.
f
(1) = 7
correct
3.
f
(1) = 6
4.
f
(1) = 8
5.
f
(1) = 5
Explanation:
When
f
′′
(
x
) = 9
x
the most general anti
derivative of
f
′′
is
f
′
(
x
) =
9
2
x
2
+
C
where the arbitrary constant
C
is determined
by the condition
f
′
(
−
1) = 4. For
f
′
(
−
1) = 4
=
⇒
f
′
(
x
) =
9
2
x
2
−
1
2
.
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 12/23/2009 for the course M 408L taught by Professor Radin during the Fall '08 term at University of Texas at Austin.
 Fall '08
 RAdin

Click to edit the document details