Definite integral with change of limits

Question

If a function is defined such as $f(x+k)=f(x)$ for $k∈\mathbb{Z}^+ $(positive integer) and $\int\limits_{0}^{k} f(x) dx=I$. Then find $\int\limits_{0}^{k^2-k} f(x) dx$ in terms of $k$ and $I$.

Answer 1

$f(x+k)=f(x)$ means $f$ is periodic with period $k$

so given

$I=\int_0^kf(x)dx$

$2I=\int_0^{2k}f(x)dx=\int_0^kf(x)dx+\int_k^{2k}f(x)dx$

Then

$$\int_0^{k(k-1)}f(x)dx=\int_0^{k}f(x)dx+\int_k^{2k}f(x)dx+...+\int_{k(k-2)}^{k(k-1)}f(x)dx=(k-1)I$$