For the integral, let $s = r/m$. Then $m\,ds = dr$ and
$$\int_0^m f\left(\frac{r}{m}\right)\,dr = m\int_0^1 f(s)\,ds.$$
For the sum it is not so easy, because it equals $f(\frac{1}{m}) + f(\frac{2}{m}) + \dotsb + f(\frac{m-1}{m})$; you need the $m$ inside there, and the number of terms is dictated by the bounds on $\sum$.
For the integral, let $s = r/m$. Then $m\,ds = dr$ and $$\int_0^m f\left(\frac{r}{m}\right)\,dr = m\int_0^1 f(s)\,ds.$$
For the sum it is not so easy, because it equals $f(\frac{1}{m}) + f(\frac{2}{m}) + \dotsb + f(\frac{m-1}{m})$; you need the $m$ inside there, and the number of terms is dictated by the bounds on $\sum$.