M408D SPRING 2015 HW #1 - cheatham (sc36975) HW01 um...

cheatham (sc36975) – HW01 – um – (53890) 1
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before answering.
001 10.0 points
Evaluate the integral
I =
9
1
3
√
x (
√
x + 2) 2
dx .
1. I =
8
15
2. I =
3
5
3. I =
2
3
4. I =
4
5
correct
5. I =
11
15
Explanation:
Set u
2
= x . Then 2 u du = dx , while
x = 1 = ⇒ u = 1
x = 9 = ⇒ u = 3 .
In this case,
I = 6
3
1
1
( u + 2) 2
du = 6 −
1
u + 2
3
1
.
Thus
I = − 6
1
5
−
1
3
=
4
5
.
002 10.0 points
The graph of f has slope
df
dx
= x 2 x
2
+ 1
and passes through the point (2 , 3). Find the
y -intercept of this graph.
1. y -intercept = − 2
2. y -intercept = −
5
3
3. y -intercept = −
8
3
4. y -intercept = −
7
3
5. y -intercept = −
4
3
correct
Explanation:
The function f satisfies the equations
f ( x ) = x 2 x
2
+ 1 dx, f (2) = 3 .
To evaluate the integral set u = 2 x
2
+ 1. For
then du = 4 x dx , in which case
f ( x ) =
1
4
u
1 / 2
du =
1
6
u
3 / 2
+ C
=
1
6
(2 x
2
+ 1)
3 / 2
+ C ,
where C has to be chosen so that
f (2) = 3 , i . e ., C +
9
2
= 3 .
Thus
f ( x ) =
1
6
(2 x
2
+ 1)
3 / 2
− 27 + 3 .
Consequently, the graph has
y -intercept = −
4
3
.
003 10.0 points
Evaluate the integral
I =
ln 2
0
e
x
√
4 e x + 1
dx .
1. I =
1
2
3 −
√
5 correct
2. I =
3
4
3 +
√
5

cheatham (sc36975) – HW01 – um – (53890) 2
3. I =
1
4
3 +
√
5
4. I =
1
2
3 +
√
5
5. I =
3
4
3 −
√
5
6. I =
1
4
3 −
√
5
Explanation:
Set u = 4 e
x
+ 1. Then du = 4 e
x
dx , while
x = 0 = ⇒ u = 5
x = ln 2 = ⇒ u = 9 .
In this case,
I =
1
4
9
5
1
u
1 / 2
du =
1
2
u
1 / 2
9
5
.
Consequently,
I =
1
2
3 −
√
5 .
004 10.0 points
Evaluate the integral
I =
12
4
x − 6
√
x − 3
dx.
1. I = 6
2. I =
16
3
correct
3. I =
20
3
4. I = 7
5. I =
46
3
Explanation:
Set u = x − 3. Then du = dx, so
12
4
x − 6
√
x − 3
dx =
9
1
u − 3
√
u
du
=
9
1
u
1 / 2
− 3 u
− 1 / 2
du
Consequently,
I =
9
1
u
1 / 2
− 3 u
− 1 / 2
du
=
2
3
u
3 / 2
− 6 u
1 / 2
9
1
=
16
3
.
005 10.0 points
Determine the integral
I =
1
1 + 9( x − 5) 2
dx .
1. I = 3 tan
− 1
x − 5
3
+ C
2. I = sin
− 1
3( x − 5) + C
3. I = 3 sin
− 1
x − 5
3
+ C
4. I =
1
3
tan
− 1
3( x − 5) + C correct
5. I = tan
− 1
3( x − 5) + C
6. I =
1
3
sin
− 1
3( x − 5) + C
Explanation:
Since
d
dx
tan
− 1
x =
1
1 + x 2
,
the substitution u = 3( x − 5) is suggested.
For then du = 3 dx , in which case
I =
1
3
1
1 + u 2
du =
1
3
tan
− 1
u + C ,
with C an arbitrary constant. Consequently,
I =
1
3
tan
− 1
3( x − 5) + C .

cheatham (sc36975) – HW01 – um – (53890) 3
keywords:
006 10.0 points