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Math Help - improper integrals

  1. #1
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    improper integrals

    Hey everyone, i had a question but whenever i do the integral i keep on getting 0 but the answer is supposed to be -2

    the question is the integral ln(abs(x)) from -1 to 1

    i divided the integral into ln(-x) form -1 to 0 and ln(x) from 0 to 1
    i evaluated it but i keep on getting 1 + -1 = 0

    can anyone help me with this
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  2. #2
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    Re: improper integrals

    Quote Originally Posted by ubhutto View Post
    Hey everyone, i had a question but whenever i do the integral i keep on getting 0 but the answer is supposed to be -2
    the question is the integral ln(abs(x)) from -1 to 1
    i divided the integral into ln(-x) form -1 to 0 and ln(x) from 0 to 1
    i evaluated it but i keep on getting 1 + -1 = 0
    Note that \log(|x|) is an even function so \int_{ - 1}^1 {\log (|x|)dx}  = 2\int_0^1 {\log (x)dx} .

    Look at this webpage calculation.
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  3. #3
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    Re: improper integrals

    ʃ-11ln|x|dx=2 ʃ01lnxdx

    ʃ01lnxdx =limx→0 ʃx1lnxdx

    ʃx1lnxdx=xlnx-x|x1=-1-( xlnx-x)

    limx→0 ʃx1lnxdx = -1

    (limx→0xlnx=0)

    (I didn't think the web reference was that helpful)
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: improper integrals

    Quote Originally Posted by Hartlw View Post
    ʃ-11ln|x|dx=2 ʃ01lnxdx

    ʃ01lnxdx =limx→0 ʃx1lnxdx

    ʃx1lnxdx=xlnx-x|x1=-1-( xlnx-x)

    limx→0 ʃx1lnxdx = -1

    (limx→0xlnx=0)

    (I didn't think the web reference was that helpful)
    You are correct, but I would have chosen a different variable for the limit. x is confusing being both inside the integral and also as part of the limit.

    -Dan
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