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Math Help - Characteristic of a periodic function

  1. #1
    Member Pranas's Avatar
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    Characteristic of a periodic function

    How to prove that

    \displaystyle \[I(a) = \int\limits_a^{a + T} {f(u)du} \]

    Assuming for every \displaystyle \[u\] we have \displaystyle \[f(u) = f(u + T)\]

    is nothing more than

    \displaystyle \[I(a) = C\] (const.)

    I mean it's quite intuitive, but how to write it? Thanks
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  2. #2
    Member Pranas's Avatar
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    Europe. Lithuania.
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    I believe I figured it out

    \displaystyle \[\int\limits_a^{a + T} {f(u)du}  = \int\limits_a^{a + T} {f(u)du}  - \int\limits_0^a {f(u)du}  + \int\limits_0^a {f(u)du}  = \int\limits_a^{a + T} {f(u)du}  - \int\limits_T^{a + T} {f(u + T)du}  + \int\limits_0^a {f(u)du}  = \]

    \displaystyle \[ = \int\limits_a^{a + T} {f(u)du}  + \int\limits_{a + T}^T {f(u)du}  + \int\limits_0^a {f(u)du}  = \int\limits_a^T {f(u)du}  + \int\limits_0^a {f(u)du}  = \int\limits_0^T {f(u)du}  = C\]
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