Version 045 – EXAM 1 – Radin – (56635)
1
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001
10.0 points
IF the graph oF
f
is
which one oF the Following contains only
graphs oF antiderivatives oF
f
?
1.
2.
3.
4.
5.
cor
rect
6.
Explanation:
IF
F
1
and
F
2
are antiderivatives oF
f
then
F
1
(
x
)

F
2
(
x
) = constant
independently oF
x
; this means that For any
two antiderivatives oF
f
the graph oF one
is just a vertical translation oF the graph oF
the other. But no horizontal translation oF
the graph oF an antiderivative oF
f
will be
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2
the graph of an antiderivative of
f
, nor can
a horizontal and vertical translation be the
graph of an antiderivative. This rules out
two sets of graphs.
Now in each of the the remaining four Fg
ures the dotted and dashed graphs consist of
vertical translations of the graph whose line
style is a continuous line. To decide which of
these Fgures consists of antiderivatives of
f
,
therefore, we have to look more carefully at
the actual graphs. But calculus ensures that
(i) an antiderivative of
f
will have a local
extremum at the
x
intercepts of
f
.
This eliminates two more Fgures since they
contains graphs whose local extrema occur at
points other than the
x
intercepts of
f
.
(ii) An antiderivative of
f
is increasing on
interval where the graph of
f
lies above the
x
axis, and decreasing where the graph of
f
lies below the
x
axis.
Consequently, of the two remaining Fgures
only
consists entirely of graphs of antiderivatives
of
f
.
keywords: antiderivative, graphical, graph,
geometric interpretation
/* If you use any of these, Fx the comment
symbols.
002
10.0 points
±or each
n
the interval [5
,
8] is divided into
n
subintervals [
x
i
−
1
, x
i
] of 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
(4
x
∗
i
sin
x
∗
i
) Δ
x
as a deFnite integral.
1.
limit =
i
8
5
4
x dx
2.
limit =
i
5
8
4 sin
x dx
3.
limit =
i
5
8
4
x dx
4.
limit =
i
5
8
4
x
sin
x dx
5.
limit =
i
8
5
4 sin
x dx
6.
limit =
i
8
5
4
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 [
a, b
] is divided into
n
subinter
vals [
x
i
−
1
, x
i
] of equal length Δ
x
and
x
∗
i
is
some point in [
x
i
−
1
, x
i
].
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 Spring '08
 RAdin
 Sin, lim g, dx, cu. units

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