Prove that $\int_0^1\lfloor nx\rfloor^2 dx = \frac{1}{n}\sum_{k=1}^{n-1} k^2$

Question

First of all apologies for the typo I made in an earlier question, I decided to delete that post and reformulate it

I am asked to prove that

$$\int_{(0,1)} \lfloor nx\rfloor^2\,\mathrm{d}x =\frac{1}{n}\sum_{k=1}^{n-1} k^2$$

If $x \in (0,1)$, then the values of $\lfloor nx\rfloor^2$ will be integers on the interval $[0,n^2-1]$.

For any given $k$ in this interval, we will have $\lfloor nx\rfloor^2 = k$ for $\lfloor nx \rfloor \in [\sqrt{k}, \sqrt{k+1})$

After this I am not sure how to proceed

Answer 1

Instead of working with the squares work with $\lfloor nx \rfloor$ first. Notice that for $x \in [k/n, (k+1)/n)$ we have $\lfloor nx \rfloor = k$. So we can split the original integral as: $$\int_0^1 \lfloor nx \rfloor^2 \,dx = \sum_{k=0}^{n-1}\int_{\frac{k}{n}}^\frac{k+1}{n} \lfloor nx\rfloor^2\,dx = \sum_{k=0}^{n-1}\int_{\frac{k}{n}}^\frac{k+1}{n} k^2\,dx=\frac{1}{n}\sum_{k=1}^{n-1}k^2$$ which is exactly what we wanted to prove.