Mcbiles, Emily – Homework 1 – Due: Sep 5 2006, 3:00 am – Inst: David Fonken
1
This
printout
should
have
15
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
be±ore answering.
The due time is Central
time.
001
(part 1 o± 1) 10 points
Stewart Section 4.10, Example 2, page 301
Find all ±unctions
g
such that
g
0
(
x
) =
x
2
+ 3
x
+ 1
√
x
.
1.
g
(
x
) =
√
x
µ
1
5
x
2
+
x
+ 1
¶
+
C
2.
g
(
x
)
=
2
√
x
µ
1
5
x
2
+
x
+ 1
¶
+
C
cor
rect
3.
g
(
x
) =
√
x
(
x
2
+ 3
x
+ 1
)
+
C
4.
g
(
x
) = 2
√
x
(
x
2
+ 3
x
+ 1
)
+
C
5.
g
(
x
) = 2
√
x
µ
1
5
x
2
+
x

1
¶
+
C
6.
g
(
x
) = 2
√
x
(
x
2
+ 3
x

1
)
+
C
Explanation:
A±ter division
g
0
(
x
) =
x
3
/
2
+ 3
x
1
/
2
+
x

1
/
2
,
so we can now fnd an antiderivative o± each
term separately. But
d
dx
µ
ax
r
r
¶
=
ax
r

1
±or all
a
and all
r
6
= 0. Thus
2
5
x
5
/
2
+ 2
x
3
/
2
+ 2
x
1
/
2
= 2
√
x
µ
1
5
x
2
+
x
+ 1
¶
is an antiderivative o±
g
0
. Consequently,
g
(
x
) = 2
√
x
µ
1
5
x
2
+
x
+ 1
¶
+
C
with
C
an arbitrary constant.
keywords: antiderivative, power ±unctions
002
(part 1 o± 1) 10 points
Find the most general antiderivative,
F
, o±
the ±unction
f
(
x
) = 9
x
2

10
x
+ 8
.
1.
F
(
x
) = 9
x
3

10
x
2
+ 8
x
+
C
2.
F
(
x
) = 3
x
3

5
x
2
+ 8
x
+
C
correct
3.
F
(
x
) = 3
x
3

5
x
2
+ 8
x
4.
F
(
x
) = 3
x
3
+ 5
x
2
+ 8
x
+
C
5.
F
(
x
) = 3
x
3
+ 5
x
2
+ 8
x
Explanation:
Since
d
dx
x
r
=
rx
r

1
,
the most general antiderivative o±
f
is the
±unction
F
(
x
) = 9
µ
x
3
3
¶

10
µ
x
2
2
¶
+ 8
x
+
C
with
C
an arbitrary constant. Consequently,
F
(
x
) = 3
x
3

5
x
2
+ 8
x
+
C
.
keywords: antiderivative, polynomial
003
(part 1 o± 1) 10 points
Consider the ±ollowing ±unctions:
(
A
)
F
1
(
x
) = sin
2
x,
(
B
)
F
2
(
x
) =

cos 2
x
4
,
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View Full DocumentMcbiles, Emily – Homework 1 – Due: Sep 5 2006, 3:00 am – Inst: David Fonken
2
(
C
)
F
3
(
x
) =
cos
2
x
2
.
Which are antiderivatives of
f
(
x
) = sin
x
cos
x
?
1.
F
1
and
F
3
only
2.
F
2
and
F
3
only
3.
F
2
only
correct
4.
F
1
only
5.
F
3
only
6.
none of them
7.
F
1
and
F
2
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
)
Antiderivative.
(
C
)
Not antiderivative.
keywords: antiderivative, trig function, dou
ble angle formula, T/F,
004
(part 1 of 1) 10 points
Find
f
(
t
) when
f
0
(
t
) = 2 cos
1
3
t

4 sin
2
3
t
and
f
(
π
2
) = 9.
1.
f
(
t
) = 6 sin
1
3
t
+ 2 cos
2
3
t
+ 5
2.
f
(
t
) = 6 cos
1
3
t
+ 2 sin
2
3
t
+ 5
3.
f
(
t
) = 6 sin
1
3
t
+ 6 cos
2
3
t
+ 3
correct
4.
f
(
t
) = 10 sin
1
3
t

6 cos
2
3
t
+ 7
5.
f
(
t
) = 10 cos
1
3
t

6 sin
2
3
t
+ 7
6.
f
(
t
) = 6 cos
1
3
t
+ 6 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
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 Fall '08
 RAdin
 Trigonometry, Derivative, Cos, David Fonken, Mcbiles

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