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

[coverband]

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?

 

[Office_Shredder]

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

\frac{ \int_{-a}^{a}|x|dx }{ |a| } = |a|
\frac{ \int_{-a}^{a} xdx } { a } = 0

Like that?

 

[coverband]

Basically is
(int|x|dx)/(|x|) = (intxdx)/( x )

and if so how?

 

[CompuChip]

First let's clear up the notation.
The numerator
\int f(x) \, dx
is a number, it does not depend on x. The denominator f(x) does.
Do you then claim that
\frac{ \int |f(y)| \, dy}{|x|} = \frac{\int f(y) \, dy}{x}
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".

 

[praharmitra]

First let's clear up the notation.
The numerator
\int f(x) \, dx
is a number, it does not depend on x. The denominator f(x) does.
Do you then claim that
\frac{ \int |f(y)| \, dy}{|x|} = \frac{\int f(y) \, dy}{x}
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