Definite integral of a periodic funtion, offset by the period, equals the original definite integral

Question

Suppose $f: \mathbb R \to \mathbb R$ is Riemann integrable on every finite interval and periodic with period $T>0$. Then for every interval $[a,b]$: $$ \int_a^b f = \int_c^d f,$$ where $c = a+T$ and $d = b+T$.

I don't understand why, if so, this is true.

How van I explain it? I could do it if I knew that the antiderivative is periodic, but do we even know that there is an antiderivative?

Answer 1

By the change variable $x=u-T$ so $du=dx$ we have $$\int_a^b f(x)dx=\int_{a+T}^{b+T}f(u-T)du=\int_c^df(u)du$$

Answer 2

Okay, let's think the graphical interpretation of integration .It's very logical and I don't know a formal proof for this . But I can explain it to you with the help of an example.
for eg: Consider - $$f(x)= sin x$$
If I were to integrate sin x from 0 to 2pi . Without even working out we can say that it is 0 as the area above x-axis is the same as the area below x axis. Now lets take this integral one step further.
Integrate this from 0+100pi to 2pi+100pi.
Quite obviously again area above x axis will be equal to the area below x axis.(Number of cycles does not make a difference) . And so,we can conclude:

$$\int_{0}^{2pi}sin(x)=\int_{0+nT}^{2pi+nT}sin(x)dx=0 $$where n is any arbit natural number.