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Math Help - Integral of an Even Function

  1. #1
    Senior Member AllanCuz's Avatar
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    Integral of an Even Function

    This is probably going to be one of those questions where I can't believe I was so dumb but...What I'm trying to do is calculate the fourier series of some function. This is okay, but I'm having a problem with my integral.

    In the fourier series Ao is given by

    Ao = \frac {1}{L} \int_{-L}^{L} f(x)dx

    The fourier series I'm attempting to calculate is for

    f(x) = 1 - |x| for  -2 \le x \le 2

    Let us note that this function is even

    f(-1)=0=f(1)

    So Ao should simplify to

    Ao = \frac {2}{L} \int_{0}^{L} f(x)dx

    We then plug in the values and obtain

    Ao = \int_{0}^{2} (1-|x|)dx

    Ao = 2 - \frac {x^2}{2} evaluated from 0-->2

    This of course yields

    Ao = 2 - 2 = 0

    Now, I was fiddling with this. And had I not changed my integral I get a different answer.

    Ao = \frac {1}{L} \int_{-L}^{L} f(x)dx

    Ao = \frac {1}{2} \int_{-2}^{2} (1-|x|)dx

    Ao = \frac {1}{2} (4-\frac {x^2}{2} ) evaluted from -2-->2

    Ao = \frac {1}{2} (4-[\frac {4}{2} - \frac {4}{2}]

    Ao = 2

    I'm sure that I have violated some law in the second evaluation, i don't remember what is is though. Is this because the equation isn't valid on the left side of the XY domain and that the derivative at 0 doesn't exist? I think I vaguely remember something about this from first year.
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  2. #2
    Super Member Deadstar's Avatar
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    Yes it's not continuous at x=0 and you can't just integrate it.

    When dealing with |x| type things usually you can split them up into...

    \int_{-L}^0 -x dx + \int_0^L x dx

    However since it's gonna be even in your case you can just do...

    2\int_0^L x dx

    (obviously with 1-x though)
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  3. #3
    Super Member Deadstar's Avatar
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    Edit
    Last edited by Deadstar; April 13th 2010 at 04:58 PM. Reason: nevermind
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by Deadstar View Post
    Actually hold on a sec.

    The answer should be 2 I think but your method is wrong...
    My book has it as equal to 0. Which is what prompted me to think about it some more. I figured it had something to do with not being contious at 0, my simple brain just couldn't rationalize why :P

    Let's work it out in full

    \int_{-2}^0 (1+x)dx + \int_{0}^{-2} (1-x)dx

    2+ \frac {x^2}{2} evaluated from -2-->0 and

     -2 - \frac {x^2}{2} evaluated rom 0-->-2

    Yeah that equals 0. Thanks mate for helping me think back. I should probably brush up on some definitions :P
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  5. #5
    Super Member Deadstar's Avatar
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    Quote Originally Posted by AllanCuz View Post
    My book has it as equal to 0. Which is what prompted me to think about it some more. I figured it had something to do with not being contious at 0, my simple brain just couldn't rationalize why :P

    Let's work it out in full

    \int_{-2}^0 (1+x)dx + \int_{0}^{-2} (1-x)dx

    2+ \frac {x^2}{2} evaluated from -2-->0 and

     -2 - \frac {x^2}{2} evaluated rom 0-->-2

    Yeah that equals 0. Thanks mate for helping me think back. I should probably brush up on some definitions :P
    Ya I edited that post cos I realized I was being stupid about 3 seconds after I posted it. Plot the graph and you'll see its 0.
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