My Math Forum Periodic Integration Proof/Substitution

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 January 19th, 2010, 11:19 AM #1 Newbie   Joined: Dec 2009 Posts: 6 Thanks: 0 Periodic Integration Proof/Substitution Let f:R->R be a continuous function. Let T>0 be such that f(x+t) = f(x) for all x. Use an appropriate substitution to prove that for all real numbers a, $ \int _{a}^{a+T} f(x) dx = \int_ {0}^{T} f(x)dx$. ..I was thinking: $ \int _{a}^{a+T} f(x) dx = \int_ {a}^{T} f(x)dx = \int_ {0}^{T} f(x) dx$, but I can't think of a better way to explain it. Also, intuitively, I understand that f(x+T) = f(x), so for a periodic function F(x+T) = F(x), but again can't really translate that into a legible proof.
January 19th, 2010, 01:33 PM   #2
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Re: Periodic Integration Proof/Substitution

Quote:
 Originally Posted by semper Let f:R->R be a continuous function. Let T>0 be such that f(x+t) = f(x) for all x. Use an appropriate substitution to prove that for all real numbers a, $ \int _{a}^{a+T} f(x) dx = \int_ {0}^{T} f(x)dx$. ..I was thinking: $ \int _{a}^{a+T} f(x) dx = \int_ {a}^{T} f(x)dx = \int_ {0}^{T} f(x) dx$, but I can't think of a better way to explain it. Also, intuitively, I understand that f(x+T) = f(x), so for a periodic function F(x+T) = F(x), but again can't really translate that into a legible proof.
Two step process.
First: there is an integer n and some b between 0 and T so that b=a-nT.
Therefore by periodicity the integral from b to b+T is the same as the integral from a to a+T (the integrands are identical).
Second: The integrand from 0 to b is the same as the integrand from T to b+T, so the integral from 0 to T is the same as the integral from b to b+T.

 January 19th, 2010, 03:07 PM #3 Newbie   Joined: Dec 2009 Posts: 6 Thanks: 0 Re: Periodic Integration Proof/Substitution Very helpful. Thank you for your insight.
 January 19th, 2010, 03:50 PM #4 Global Moderator   Joined: Dec 2006 Posts: 21,034 Thanks: 2269 Each original T should be t. \begin{align*}\int_a\,^{a+t} f(x)dx\,&=\int_a\,^t f(x)dx\,+\,\int_t\,^{a+t} f(x)dx \\ &=\int_a\,^t f(x)dx\,+\,\int_0\,^a f(x\,+\,t)dx\text{ by the substitution }x\,\rightarrow\,x\,+\,t \\ &=\int_a\,^t f(x)dx\,+\,\int_0\,^a f(x)dx \\ &=\int_0\,^t f(x)dx.\end{align*}
 January 19th, 2010, 11:08 PM #5 Newbie   Joined: Nov 2009 Posts: 1 Thanks: 0 Re: Periodic Integration Proof/Substitution ...Hey! You're in Dr. J's class aren't you? Thank goodness someone posted this up, I was having major problems with this too.

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