[(|f(x)|)^(-1)]*integral(|f(x)|dx) = [(f(x))^(-1)]*integral(f(x)dx)?

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  • #1
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I heard [(|f(x)|)^(-1)]*integral(|f(x)|dx) = [(f(x))^(-1)]*integral(f(x)dx)

(i.e. when dividing the integral of the absolute value of a function by the absolute value of the function, the absolute value signs cancel out)

Is this true

If so, how so?
 

Answers and Replies

  • #2
Office_Shredder
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Are these antiderivatives? If so, the presence of unknown constants are going to blow this whole thing out of the water. Regardless, let's pick an example

f(x) = x

[tex] \frac{ \int_{-a}^{a}|x|dx }{ |a| } = |a| [/tex]
[tex]\frac{ \int_{-a}^{a} xdx } { a } = 0[/tex]

Like that?
 
  • #3
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Basically is
(int|x|dx)/(|x|) = (intxdx)/( x )

and if so how?
 
Last edited:
  • #4
CompuChip
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First let's clear up the notation.
The numerator
[tex]\int f(x) \, dx[/tex]
is a number, it does not depend on x. The denominator f(x) does.
Do you then claim that
[tex]\frac{ \int |f(y)| \, dy}{|x|} = \frac{\int f(y) \, dy}{x}[/tex]
for all x?

Secondly, is the integration definite (e.g. implied over the whole domain) or indefinite (i.e. you need an arbitrary integration constant to be added)?

Finally, Office_Shredder has given a counter-example to what was supposedly your statement, so to answer your question: "No, basically it is not".
 
  • #5
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First let's clear up the notation.
The numerator
[tex]\int f(x) \, dx[/tex]
is a number, it does not depend on x. The denominator f(x) does.
Do you then claim that
[tex]\frac{ \int |f(y)| \, dy}{|x|} = \frac{\int f(y) \, dy}{x}[/tex]
for all x?

Secondly, is the integration definite (e.g. implied over the whole domain) or indefinite (i.e. you need an arbitrary integration constant to be added)?

Finally, Office_Shredder has given a counter-example to what was supposedly your statement, so to answer your question: "No, basically it is not".

first of all, why should the numerator be a "number" if the integral is indefinite, then it gives a function!! only if the num was definite, would it be a number
 

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