Version 065 – Homework 01 – Gompf – (58370)
1
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001
10.0 points
±ind the value of
f
(0) when
f
′′
(
t
) = 2(3
t

1)
and
f
′
(1) = 5
,
f
(1) = 2
.
1.
f
(0) =

2
correct
2.
f
(0) = 2
3.
f
(0) = 0
4.
f
(0) =

1
5.
f
(0) = 1
Explanation:
The most general antiderivative of
f
′′
has
the form
f
′
(
t
) = 3
t
2

2
t
+
C
where
C
is an arbitrary constant.
But if
f
′
(1) = 5, then
f
′
(1) = 3

2 +
C
= 5
,
i.e.,
C
= 4
.
±rom this it follows that
f
′
(
t
) = 3
t
2

2
t
+ 4
,
and the most general antiderivative of the
latter is
f
(
t
) =
t
3

t
2
+ 4
t
+
D ,
where
D
is an arbitrary constant.
But if
f
(1) = 2, then
f
(1) = 1

1 + 4 +
D
= 2
,
i.e.,
D
=

2
.
Consequently,
f
(
t
) =
t
3

t
2
+ 4
t

2
.
At
x
= 0, therefore,
f
(0) =

2
.
002
10.0 points
Consider the following functions:
(
A
)
F
1
(
x
) = sin
2
x ,
(
B
)
F
2
(
x
) =
cos
2
x
2
,
(
C
)
F
3
(
x
) =
cos 2
x
4
.
Which are antiderivatives of
f
(
x
) = sin
x
cos
x
?
1.
F
1
only
2.
F
2
and
F
3
only
3.
none of them
correct
4.
F
2
only
5.
F
1
and
F
2
only
6.
F
3
only
7.
F
1
and
F
3
only
8.
all of them
Explanation:
By trig identities,
cos 2
x
= 2 cos
2
x

1 = 1

2 sin
2
x ,
while
sin 2
x
= 2 sin
x
cos
x .
But
d
dx
sin
x
= cos
x,
d
dx
cos
x
=

sin
x .
Consequently, by the Chain Rule,
(
A
) Not antiderivative.
(
B
) Not antiderivative.
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2
(
C
) Not antiderivative.
003
10.0 points
Find
f
(
x
) on
(

π
2
,
π
2
)
when
f
′
(
x
) =
√
2 sin
x
+ 2 sec
2
x
and
f
(
π
4
)
= 5.
1.
f
(
x
) = 2 tan
x

√
2 cos
x
+ 4
correct
2.
f
(
x
) = 2 tan
x
+
√
2 sin
x
+ 4
3.
f
(
x
) = 8

2 tan
x

√
2 cos
x
4.
f
(
x
) = 6

2 tan
x
+
√
2 sin
x
5.
f
(
x
) = 2 tan
x
+
√
2 cos
x
+ 2
Explanation:
The most general antiderivative of
f
′
(
x
) =
√
2 sin
x
+ 2 sec
2
x
is
f
(
x
) =

√
2 cos
x
+ 2 tan
x
+
C
with
C
an arbitrary constant.
But if
f
p
π
4
P
= 5, then
f
p
π
4
P
=

1 + 2 +
C
= 5
,
so
C
= 4
.
Consequently,
f
(
x
) = 2 tan
x

√
2 cos
x
+ 4
.
004
10.0 points
Find the unique antiderivative
F
of
f
(
x
) =
3
e
3
x
+ 2
e
2
x
+ 5
e
−
x
e
2
x
for which
F
(0) = 0.
1.
F
(
x
) = 3
e
x
+ 2
x

5
3
e
−
3
x

4
3
correct
2.
F
(
x
) =
e
3
x

2
x
+ 3
e
−
x

2
3
3.
F
(
x
) = 3
e
x

2
x
+
5
3
e
−
3
x
+
14
3
4.
F
(
x
) = 3
e
x

2
x
+
5
3
e
−
x
+
4
3
5.
F
(
x
) = 3
e
x
+ 2
x

3
e
−
x
6.
F
(
x
) =
e
3
x
+ 2
x

5
3
e
−
3
x

2
3
Explanation:
After division,
3
e
3
x
+ 2
e
2
x
+ 5
e
−
x
e
2
x
= 3
e
x
+ 2 + 5
e
−
3
x
.
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 Spring '08
 RAdin
 Calculus, Sin, lim

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