Third Problem for Floors

First, let's generalize:

$\begin{aligned} \int_0^{n^2}\lfloor\sqrt{t}\rfloor \ dt \hspace{0.5 cm} \text{let t = } \ u^2 \implies dt = 2u \ du \\ & = 2\int_0^{n}\lfloor u \rfloor u \ du \\ & = 2\sum_{k=0}^{n-1}k\int_{k}^{k+1} u \, du \\ & = \sum_{k=0}^{n-1} \Big( k \big( (k+1)^2 - k^2 \big) \Big ) \\ & = \cfrac{1}{6}n(n-1)(4n+1). \end{aligned}$

Substituting n = 6 $\begin{aligned} \implies \int_0^{6^2}\lfloor\sqrt{t}\rfloor \ dt \ & = \cfrac{1}{6}(6)(6-1)\big( 4(6)+1 \big ) \\ & = \boxed{125} \end{aligned}$