mao (tm23477) – Hw 12 – gualdani – (57180)
1
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printout
should
have
20
questions.
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before answering.
001
10.0 points
Find all functions
g
such that
g
′
(
x
) =
x
2
+ 3
x
+ 4
√
x
.
1.
g
(
x
) = 2
√
x
parenleftbigg
1
5
x
2
+
x
−
4
parenrightbigg
+
C
2.
g
(
x
) = 2
√
x
(
x
2
+ 3
x
+ 4
)
+
C
3.
g
(
x
)
=
2
√
x
parenleftbigg
1
5
x
2
+
x
+ 4
parenrightbigg
+
C
cor
rect
4.
g
(
x
) =
√
x
parenleftbigg
1
5
x
2
+
x
+ 4
parenrightbigg
+
C
5.
g
(
x
) = 2
√
x
(
x
2
+ 3
x
−
4
)
+
C
6.
g
(
x
) =
√
x
(
x
2
+ 3
x
+ 4
)
+
C
Explanation:
After division
g
′
(
x
) =
x
3
/
2
+ 3
x
1
/
2
+ 4
x
−
1
/
2
,
so we can now find an antiderivative of each
term separately. But
d
dx
parenleftbigg
ax
r
r
parenrightbigg
=
ax
r
−
1
for all
a
and all
r
negationslash
= 0. Thus
2
5
x
5
/
2
+ 2
x
3
/
2
+ 8
x
1
/
2
= 2
√
x
parenleftbigg
1
5
x
2
+
x
+ 4
parenrightbigg
is an antiderivative of
g
′
. Consequently,
g
(
x
) = 2
√
x
parenleftbigg
1
5
x
2
+
x
+ 4
parenrightbigg
+
C
with
C
an arbitrary constant.
002
10.0 points
Determine
f
(
t
) when
f
′′
(
t
) = 6(
t
+ 1)
and
f
′
(1) = 2
,
f
(1) = 4
.
1.
f
(
t
) =
t
3
−
3
t
2
+ 7
t
−
1
2.
f
(
t
) =
t
3
+ 3
t
2
−
7
t
+ 7
correct
3.
f
(
t
) = 3
t
3
+ 3
t
2
−
7
t
+ 5
4.
f
(
t
) =
t
3
−
6
t
2
+ 7
t
+ 2
5.
f
(
t
) = 3
t
3
+ 6
t
2
−
7
t
+ 2
6.
f
(
t
) = 3
t
3
−
6
t
2
+ 7
t
+ 0
Explanation:
The most general antiderivative of
f
′′
has
the form
f
′
(
t
) = 3
t
2
+ 6
t
+
C
where
C
is an arbitrary constant.
But if
f
′
(1) = 2, then
f
′
(1) =
3 + 6 +
C
= 2
,
i.e.,
C
=
−
7
.
From this it follows that
f
′
(
t
) = 3
t
2
+ 6
t
−
7
.
The most general antiderivative of
f
is thus
f
(
t
) =
t
3
+ 3
t
2
−
7
t
+
D ,
where
D
is an arbitrary constant.
But if
f
(1) = 4, then
f
(1) = 1 + 3
−
7 +
D
= 4
,
i.e.,
D
= 7
.
Consequently,
f
(
t
) =
t
3
+ 3
t
2
−
7
t
+ 7
.
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mao (tm23477) – Hw 12 – gualdani – (57180)
2
003
10.0 points
Consider the following functions:
(
A
)
F
1
(
x
) =
−
cos 2
x
4
,
(
B
)
F
2
(
x
) = sin
2
x ,
(
C
)
F
3
(
x
) =
cos
2
x
2
.
Which are antiderivatives of
f
(
x
) = sin
x
cos
x
?
1.
F
3
only
2.
F
2
only
3.
F
1
only
correct
4.
all of them
5.
F
1
and
F
3
only
6.
F
1
and
F
2
only
7.
none of them
8.
F
2
and
F
3
only
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
)
Antiderivative.
(
B
)
Not antiderivative.
(
C
)
Not antiderivative.
004
10.0 points
Find
f
(
t
) when
f
′
(
t
) = 2 cos
1
3
t
−
4 sin
2
3
t
and
f
(
π
2
) = 7.
1.
f
(
t
) = 10 cos
1
3
t
−
6 sin
2
3
t
+ 5
2.
f
(
t
) = 6 cos
1
3
t
+ 6 sin
2
3
t
+ 1
3.
f
(
t
) = 6 sin
1
3
t
+ 6 cos
2
3
t
+ 1
correct
4.
f
(
t
) = 6 sin
1
3
t
+ 2 cos
2
3
t
+ 3
5.
f
(
t
) = 10 sin
1
3
t
−
6 cos
2
3
t
+ 5
6.
f
(
t
) = 6 cos
1
3
t
+ 2 sin
2
3
t
+ 3
Explanation:
The function
f
must have the form
f
(
t
) = 6 sin
1
3
t
+ 6 cos
2
3
t
+
C
where the constant
C
is determined by the
condition
f
parenleftBig
π
2
parenrightBig
= 6 sin
π
6
+ 6 cos
π
3
+
C
= 7
.
But by known trig values
sin
π
6
= cos
π
3
=
1
2
,
so 6 +
C
= 7. Consequently,
f
(
t
) = 6 sin
1
3
t
+ 6 cos
2
3
t
+ 1
.
005
10.0 points
Find the most general antiderivative of the
function
f
(
x
) = 2 cos
x
−
5 sin
x .
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 Fall '08
 schultz
 Cos

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