<s>[INST] Proving $\int_0^{n^2} \left\lfloor\sqrt t\right\rfloor \,dt=\frac{n(n-1)(4n+1)}{6}$ I want to prove 

$$\int_{0}^{n^{2}} \left\lfloor \sqrt{t} \right\rfloor \,dt=\frac{n(n-1)(4n+1)}{6}$$

Is it correct to say that $\left\lfloor \sqrt{t} \right\rfloor=\sqrt{(k-1)}$ and $(k-1)^{2} < t < k^{2}$, where
$$\int_{0}^{n^{2}} \left\lfloor \sqrt{t} \right\rfloor \,dt =\sum_{k=1}^{n} \sqrt{(k-1)} \cdot (k^2 - (k-1)^2)$$
I would know how to do the rest, but how did we know that $\left\lfloor \sqrt{t} \right\rfloor=\sqrt{(k-1)}$, is it something you assume depending on the function? Because for a previous function, there was
$$\int_{0}^{n} \left\lfloor t^{2} \right\rfloor \,dt$$
and it was set $\left\lfloor t^{2} \right\rfloor=(k-1)^2$, I would really appreciate if someone would kindly explain this to me.
 [/INST] You have 
$$\int_0^{n^2} \lfloor \sqrt{t} \rfloor \mathrm{dt} = \sum_{k=0}^{n-1} \int_{k^2}^{(k+1)^2} \lfloor \sqrt{t} \rfloor \mathrm{dt}$$
But if $k^2 < t < (k+1)^2$, then $\lfloor \sqrt{t} \rfloor = k$. So 
$$\int_0^{n^2} \lfloor \sqrt{t} \rfloor \mathrm{dt} = \sum_{k=0}^{n-1} k((k+1)^2-k^2) = \sum_{k=0}^{n-1} k(2k+1)  = 2 \frac{n(n-1)(2n-1)}{6}+\frac{(n-1)n}{2} $$
i.e.
$$\int_0^{n^2} \lfloor \sqrt{t} \rfloor \mathrm{dt} = \frac{n(n-1)(4n+1)}{6}$$
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