Given that $f(x) = \sqrt{1-x^2}$ and $f$ is integrable on $[0,1]$, prove that
$$\displaystyle \lim_{n\to \infty} \dfrac{1}{n^2} \sum_{k=1}^{n}\sqrt{n^2 -k^2} = \int_{0}^{1} f.$$
I am not exactly sure where to begin with this problem nor do I really understand what my plan of solving it should be. I appreciate your help and nudges in the right direction.
On
The left side can be converted into a Riemann sum: $$ \lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n \sqrt{n^2 - k^2} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sqrt{1 - (k/n)^2} $$ Since $f(x)$ is integrable, the limit exists and equals to $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sqrt{1 - (k/n)^2} = \int_0^1 dx\,\sqrt{1-x^2} = \int_0^1 f(x)\,dx $$
hint: $\sqrt{n^2-k^2} = n\sqrt{1-\left(\frac{k}{n}\right)^2}$