Proving Equality of Integrals with Periodic Functions

[e179285]

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?

Regards

 

[1994Bhaskar]

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.

 

[e179285]

Thank you for your answer.sorry but ı couldn't understand how to do this as you said.can you tell me more explicitly :)

 

[HallsofIvy]

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

There are n integrals and, because f is periodic with period T, they are all equal to
$$\int_0^T f(t)dt$$

 

[e179285]

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] ?

 

[HallsofIvy]

If you have n values, each equal to A, what is the sum?

 

[e179285]

I understand now,thank you for your efforts.