nunez (djn358) – HW02 – Radin – (54915)
1
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beFore answering.
001
10.0 points
±ind all Functions
g
such that
g
′
(
x
) =
2
x
2
+ 5
x
+ 1
√
x
.
1.
g
(
x
) = 2
√
x
(
2
x
2
+ 5
x
−
1
)
+
C
2.
g
(
x
) = 2
√
x
p
2
5
x
2
+
5
3
x
+ 1
P
+
C
cor
rect
3.
g
(
x
) =
√
x
(
2
x
2
+ 5
x
+ 1
)
+
C
4.
g
(
x
) = 2
√
x
p
2
5
x
2
+
5
3
x
−
1
P
+
C
5.
g
(
x
) = 2
√
x
(
2
x
2
+ 5
x
+ 1
)
+
C
6.
g
(
x
) =
√
x
p
2
5
x
2
+
5
3
x
+ 1
P
+
C
Explanation:
AFter division
g
′
(
x
) = 2
x
3
/
2
+ 5
x
1
/
2
+
x
−
1
/
2
,
so we can now fnd an antiderivative oF each
term separately. But
d
dx
p
ax
r
r
P
=
ax
r
−
1
For all
a
and all
r
n
= 0. Thus
4
5
x
5
/
2
+
10
3
x
3
/
2
+ 2
x
1
/
2
= 2
√
x
p
2
5
x
2
+
5
3
x
+ 1
P
is an antiderivative oF
g
′
. Consequently,
g
(
x
) = 2
√
x
p
2
5
x
2
+
5
3
x
+ 1
P
+
C
with
C
an arbitrary constant.
002
10.0 points
Consider the Following Functions:
(
A
)
F
1
(
x
) =
−
cos 2
x
4
,
(
B
)
F
2
(
x
) =
sin
2
x
2
,
(
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
and
F
3
only
3.
F
1
and
F
3
only
4.
F
2
only
5.
F
1
only
6.
all oF them
7.
none oF them
8.
F
1
and
F
2
only
correct
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
) Antiderivative.
(
C
) Not antiderivative.
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2
003
10.0 points
Find
f
(
t
) when
f
′
(
t
) = cos
1
3
t
−
2 sin
2
3
t
and
f
(
π
2
) = 2.
1.
f
(
t
) = 3 sin
1
3
t
+ cos
2
3
t
2.
f
(
t
) = 5 cos
1
3
t
−
3 sin
2
3
t
+ 1
3.
f
(
t
) = 3 sin
1
3
t
+ 3 cos
2
3
t
−
1
correct
4.
f
(
t
) = 3 cos
1
3
t
+ sin
2
3
t
5.
f
(
t
) = 5 sin
1
3
t
−
3 cos
2
3
t
+ 1
6.
f
(
t
) = 3 cos
1
3
t
+ 3 sin
2
3
t
−
1
Explanation:
The function
f
must have the form
f
(
t
) = 3 sin
1
3
t
+ 3 cos
2
3
t
+
C
where the constant
C
is determined by the
condition
f
p
π
2
P
= 3 sin
π
6
+ 3 cos
π
3
+
C
= 2
.
But by known trig values
sin
π
6
= cos
π
3
=
1
2
,
so 3 +
C
= 2. Consequently,
f
(
t
) = 3 sin
1
3
t
+ 3 cos
2
3
t
−
1
.
004
10.0 points
Find
f
(
x
) on
(
−
π
2
,
π
2
)
when
f
′
(
x
) = 7 + 5 tan
2
x
and
f
(0) = 4.
1.
f
(
x
) = 9
−
2
x
−
5 sec
x
2.
f
(
x
) = 4 + 2
x
+ 5 tan
x
correct
3.
f
(
x
) =
−
1 + 7
x
+ 5 sec
x
4.
f
(
x
) = 4 + 2
x
+ 5 tan
2
x
5.
f
(
x
) = 4
−
2
x
−
5 tan
x
6.
f
(
x
) =
−
1 + 7
x
+ 5 sec
2
x
Explanation:
The properties
d
dx
(tan
x
) = sec
2
x,
tan
2
x
= sec
2
x
−
1
,
suggest that we rewrite
f
′
(
x
) as
f
′
(
x
) = 2 + 5 sec
2
x,
for then the most general antiderivative of
f
′
is
f
(
x
) = 2
x
+ 5 tan
x
+
C,
with C an arbitrary constant. But if
f
(0) = 4,
then
f
(0) =
C
= 4
.
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 Spring '09
 GOGOLEV
 Derivative, Fundamental Theorem Of Calculus, Cos, Constant of integration, Núñez

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