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Math Help - Absolute integral

  1. #1
    Super Member dhiab's Avatar
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    Absolute integral

    Calculate this integral :
    They ask for the graphic expliquation .
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  2. #2
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    Hi!


    Quote Originally Posted by dhiab View Post
    Calculate this integral :
    use integration by parts

    it is \int u*v' dx = u*v - \int u' v dx

    Let v':= |ln(x)| => v = \frac{1}{x}

    u = \frac{1}{x} => u' = - \frac{1}{x^2}


    Thus

    \int \frac{|ln(x)|}{x} dx = \int |ln(x)|*\frac{1}{x} dx


     = \left[ \displaystyle \frac{1}{x}*\frac{1}{x}\right]^{\displaystyle a}_{\displaystyle 1/a} - \int (-\frac{1}{x^2}*\frac{1}{x}) dx

    You should be able to solve this...


    Quote Originally Posted by dhiab View Post
    They ask for the graphic expliquation .
    What is that supposed to mean?

    Rapha
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  3. #3
    Super Member dhiab's Avatar
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    Quote Originally Posted by Rapha View Post
    Hi!




    use integration by parts

    it is \int u*v' dx = u*v - \int u' v dx

    Let v':= |ln(x)| => v = \frac{1}{x}

    u = \frac{1}{x} => u' = - \frac{1}{x^2}


    Thus

    \int \frac{|ln(x)|}{x} dx = \int |ln(x)|*\frac{1}{x} dx


     = \left[ \displaystyle \frac{1}{x}*\frac{1}{x}\right]^{\displaystyle a}_{\displaystyle 1/a} - \int (-\frac{1}{x^2}*\frac{1}{x}) dx

    You should be able to solve this...






    Rapha
    Hello :
    the value of the integral varies according to the absoulue value
    What is that supposed to mean? : problem of area
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  4. #4
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    Quote Originally Posted by dhiab View Post
    Calculate this integral :
    They ask for the graphic expliquation .
    \int\limits_{1/e}^e {\frac{{\left| {\ln (x)} \right|}}<br />
{x}dx}  = \int\limits_{1/e}^1 {\frac{{ - \ln (x)}}<br />
{x}dx}  + \int\limits_1^e {\frac{{\ln (x)}}<br />
{x}dx = } \left. {\frac{{ - \ln ^2 (x)}}<br />
{2}} \right|_{1/e}^1  + \left. {\frac{{\ln ^2 (x)}}<br />
{2}} \right|_1^e <br />
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  5. #5
    Super Member dhiab's Avatar
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    Quote Originally Posted by Rapha View Post
    Hi!

    use integration by parts

    it is \int u*v' dx = u*v - \int u' v dx

    Let v':= |ln(x)| => v = \frac{1}{x}

    u = \frac{1}{x} => u' = - \frac{1}{x^2}

    Thus

    \int \frac{|ln(x)|}{x} dx = \int |ln(x)|*\frac{1}{x} dx

     = \left[ \displaystyle \frac{1}{x}*\frac{1}{x}\right]^{\displaystyle a}_{\displaystyle 1/a} - \int (-\frac{1}{x^2}*\frac{1}{x}) dx

    You should be able to solve this...
    HELLO :
    Last edited by mr fantastic; June 19th 2009 at 05:36 PM. Reason: Closed quote.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dhiab View Post
    HELLO :
    yes, this is correct. it is why Plato's answer has two integrals in it. as his post indicates, a good old fashioned u-substitution of u = \ln x gets the job done. make sure you understand what he did and finish up the problem
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