An integrable and periodic function f(x) For a periodic function we have: $$\int_{b}^{b+a}f(t)dt = \int_{b}^{na}f(t)dt+\int_{na}^{b+a}f(t)dt = \int_{b+a}^{(n+1)a}f(t)dt+\int_{an}^{b+a}f(t)dt = \int_{na}^{(n+1)a}f(t)dt = \int_{0}^{a}f(t)dt.$$ , but I don't understand how we obtain $\int _{b+a}^{\left(n+1\right)a}\:f\left(t\right)\:dt=\int _b^{na}\:f\left(t\right)dt$ in our equality?

What does it mean for a function to have a period $a$ ? Informally, it means that the function values gets repeated after an increment of $a$ in the $x-$value. (domain value). Notationally,

$$a\textrm{ is period of }f\iff f(x)=f(x+a)~\forall~x,x+a\in\textrm{Dom(f)}$$

That gives us $na=na+a=n(a+1)$ and $b=b+a$ since $na$ and $b$ are domain values ($x-$values) for $f$ and $a$ is the period.

So, the definite integral remains the same.

Here's a simple diagram:

![Image](