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Math Help - absolute value in primitive log

  1. #1
    Junior Member
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    absolute value in primitive log

    So when you integrate to get a log you need to have absolute value signs.
    Why is that?
    Obviously you can't have a negative log, but why is this correct as opposed to just defining that x>0 in log(x)?
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  2. #2
    MHF Contributor

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    Well, for one thing, you can't always require that x be postive!

    For example, find the area bounded by the graphs of y= 1/x, y= 0, x= -2, and x= -1.

    That is [tex]\int_{-2}^{-1} (0-\frac{1}{x})dx= -\int_{-2}^{-1}\frac{dx}{x}[/itex].

    Now, we have -ln(|x|)|_{-2}^{-1}= -(ln(|-1|)- ln(|2|)= ln(2)

    That is the same as we would get for the area bounded by y= 1/x, y= 0, x=1, and x= 2, which, because of the symmetry is what we should get.
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