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Periodic function in integral

  1. Apr 15, 2012 #1
    ıf the function f :R->R is cont. and periodic with a period T>0 then

    Are integral from nT and zero f(x) dx and n(integral from T to zero f(x)dx are egual to each other ?

    I proved by giving examle that it is true. I thinl it is not right way How can ı prove this?

  2. jcsd
  3. Apr 15, 2012 #2
    You can prove it graphically.By observing overall sum of areas.Areas above x-axis are positive, while below x- axis are negative.So when from nT to 0 all the positive and negative areas get added and cancel each other except from T to 0.That's your RHS.
  4. Apr 15, 2012 #3
    Thank you for your answer.sorry but ı couldn't understand how to do this as you said.can you tell me more explicitly :)
  5. Apr 15, 2012 #4


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    I don't see that "positive" and "negative" has anything to do with this.
    [tex]\int_0^{nT}f(t)dt= \int_0^T f(t)dt+ \int_T^{2T} f(t)dt+ \int_{2T}^{3T} f(t)dt+ \cdot\cdot\cdot+ \int_{(n-1)T}^{nT} f(t)dt[/tex]

    There are n integrals and, because f is periodic with period T, they are all equal to
    [tex]\int_0^T f(t)dt[/tex]
  6. Apr 15, 2012 #5
    Thank you for your answer.I want to question about your answer. ıs answer integral from T to 0 f(t)dt or n[integral from T to 0 f(t)dt] ?
  7. Apr 15, 2012 #6


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    If you have n values, each equal to A, what is the sum?
  8. Apr 15, 2012 #7
    I understand now,thank you for your efforts.
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