Accumulation
Authors: Bill Davis, Horacio Porta and Jerry Uhl
©19962007
Publisher: Math Everywhere, Inc.
Version 6.0
2.04 Transforming Integrals
BASICS
The really interesting stuff in this lesson is
the material on bellshaped curves in B.3).
The material in B.2) will be beneficial to you if you are going
on to vector calculus.
Both B.2) and B.3) require some experience with B.1).
B.1) Breaking more of the code of the integral:
Transforming integrals
·
B.1.a) Combining the fundamental formula and the chain rule
How do you know that the integrals
Ÿ
a
b
f '
@
u
@
x
DD
u'
@
x
D
„
x
and
Ÿ
u
@
a
D
u
@
b
D
f '
@
u
D
„
u
are equal?
·
Answer:
The best way to see why they are equal is to calculate both of them.
The second is the easier because the fundamental formula tells you that
Ÿ
u
@
a
D
u
@
b
D
f '
@
u
D
„
u
=
f
@
u
D
»
u
@
a
D
u
@
b
D
=
f
@
u
@
b
DD

f
@
u
@
a
DD
.
For the first integral, the chain rule tells you that the derivative of
f
@
u
@
x
DD
is
f '
@
u
@
x
DD
u'
@
x
D
.
The fundamental formula steps in to say
Ÿ
a
b
f '
@
u
@
x
DD
u'
@
x
D
„
x
=
f
@
u
@
x
DD
»
a
b
=
f
@
u
@
b
DD

f
@
u
@
a
DD
.
Now do you know why
Ÿ
a
b
f '
@
u
@
x
DD
u'
@
x
D
„
x
=
Ÿ
u
@
a
D
u
@
b
D
f '
@
u
D
„
u ?
Reason: They are both equal to
f
@
u
@
b
DD

f
@
u
@
a
DD
.
A point of anxiety might come up:
In the integral
Ÿ
u
@
a
D
u
@
b
D
f '
@
u
D
„
u,
the lone symbol
u is treated as a variable and
u
@
a
D
and
u
@
b
D
are
numbers. In the integral,
Ÿ
a
b
f '
@
u
@
x
DD
u'
@
x
D
„
x,
u
@
x
D
is a function,
x is a variable, and
a and
b are numbers.
This should not cause you any trouble.
·
B.1.b)
Now you know
Ÿ
a
b
f '
@
u
@
x
DD
u'
@
x
D
„
x
=
Ÿ
u
@
a
D
u
@
b
D
f '
@
u
D
„
u.
What practical use is this?
·
Answer:
Notational magic allows you to take the more complicated integral and
replace it with the less complicated but equal integral.
Pair them up as follows:
f '
@
u
@
x
DD
<>
f '
@
u
D
u'
@
x
D
„
x <>
„
u
Ÿ
a
b
<>
Ÿ
u
@
a
D
u
@
b
D
.
Lots of folks like to call this a "transformation" of a hard integral into
an easy integral.
·
B.1.c)
Here is
Mathematica
's calculation of
Ÿ
0
p
Cos
A
x
2
E
2
x
„
x:
Clear
@
x
D
;
MathematicaCalculation
=
‡
0
p
Cos
A
x
2
E
2x
„
x
Sin
A
p
2
E
Use a transformation to explain where this result comes from.
·
Answer:
The integral at the center stage is
Ÿ
0
p
Cos
A
x
2
E
2
x
„
x.
The key is that
2
x is the derivative of
x
2
, so go with
u
@
x
D
=
x
2
.
This gives the pairings
Cos
A
x
2
E
<>
Cos
@
u
D
2
x
„
x
=
u'
@
x
D
„
x <>
„
u
Ÿ
0
p
=
Ÿ
a
b
<>
Ÿ
u
@
a
D
u
@
b
D
=
Ÿ
0
2
p
2
So
Ÿ
0
p
Cos
A
x
2
E
2
x
„
x
=
Ÿ
0
2
p
2
Cos
@
u
D
„
u.
In other words, the substitution
u
@
x
D
=
x
2
transforms
Ÿ
0
p
Cos
A
x
2
E
2
x
„
x into
Ÿ
0
2
p
2
Cos
@
u
D
„
u.
The second integral is cake.
You look for a function
f
@
u
D
with
f '
@
u
D
=
Cos
@
u
D
.
Here's one:
Clear
@
f, u
D
;
f
@
u
_
D
=
Sin
@
u
D
Sin
@
u
D
Check:
f'
@
u
D
Cos
@
u
D
And now you can say with confidence and authority that
Ÿ
0
p
Cos
A
x
2
E
2
x
„
x
=
Ÿ
0
2
p
2
Cos
@
u
D
„
u
is given by:
f
A
p
2
E

f
A
0
2
E
Sin
A
p
2
E
Check with
Mathematica
's calculation.
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 Spring '08
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