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Math Help - Integration

  1. #1
    Senior Member pankaj's Avatar
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    Integration

    Given f(x) is an odd function and defined everywhere,periodic with period 2 and integrable on every interval.

    Let g(x)=\int_{0}^xf(t)dt.

    Then prove that g(2n)=0 for every integer n.
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  2. #2
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    Quote Originally Posted by pankaj View Post
    Given f(x) is an odd function and defined everywhere,periodic with period 2 and integrable on every interval.

    Let g(x)=\int_{0}^xf(t)dt.

    Then prove that g(2n)=0 for every integer n.
    Given that f is defined everywhere and integrable on every interval we can consider

    I = \int _{-1}^{1} f(t) dt

    f being odd implies (easy to prove if you're not allowed to state) I = 0

    Now you can see that a simple substitution should give you that I = g(2).

    Using the periodicity of f you should can show that any integral of f over an interval of length 2 has the same value. Therefore

     g(2) = \int_0^2 f(t) dt = \int_{2}^4 f(t) dt = \int_4^6 f(t) = \ldots = \int_{2n-2}^{2n} f(t) dt = 0

    Adding together these integrals gives g(2n) = 0

    Hope this helps.
    Last edited by pomp; September 8th 2009 at 09:59 AM.
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