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SOLUTIONS FOR HOMEWORK PROBLEMS 16.8
MATH 132  WI00
4
When you are given some ﬁxed
x
s then they will deﬁnitely be limit values for
your integral; hence for our problem we’ll have, being given
x
= 2 and
x
= 4:
R
4
2
x
+ 5
dx
=
x
2
2
+ 5
x

4
2
= (
4
2
2
+ 5
*
4)

(
2
2
2
+ 5
*
2) = 28

12 = 16
•
8
Same trick here,
x
=

2 and
x
=

1, so:
R

1

2
2
x
2

x dx
= 2
x
3
3

x
2
2


1

2
=
(2
(

1)
3
3

(

1)
2
2
)

(2
(

2)
3
3

(

2)
2
2
) = (2

1
3

1
2
)

(2

8
3

4
2
) =

7
6

(

44
6
) =
37
6
The reason I went into that much depth into computing the result is that I want
to make sure you notice and remember that: ”
Negative
numbers raised to
odd
powers remain
negative
, but the same
negative
numbers raised to
even
powers
become
positive
” !! So a negative number will give up its negative sign when it
has a even power (2, 4, 6, etc).
•
20
Notice that
e
2
is a constant .
.. I know you know, but just wanted to make sure
:).
R
e
2
1
1
x
dx
= ln
x

e
2
1
= ln (
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This note was uploaded on 07/26/2011 for the course MATH 132 taught by Professor Staff during the Spring '08 term at Ohio State.
 Spring '08
 Staff
 Math, Calculus

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