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

  1. Feb 8, 2009 #1
    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?
     
  2. jcsd
  3. Feb 8, 2009 #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?
     
  4. Feb 8, 2009 #3
    Basically is
    (int|x|dx)/(|x|) = (intxdx)/( x )

    and if so how?
     
    Last edited: Feb 8, 2009
  5. Feb 12, 2009 #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".
     
  6. Feb 12, 2009 #5

    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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