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

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
Staff Emeritus
Gold Member
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?

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

and if so how?

Last edited:
CompuChip
Homework Helper
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 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