- #1

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(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?

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- #1

- 171

- 0

(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?

- #2

Office_Shredder

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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?

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

and if so how?

Last edited:

- #4

CompuChip

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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".

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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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