For any natural number n, the value of ∫limits_0^n^2[√x], d x is

We have,

$I =\int\limits_0^{n^2}[\sqrt{x}] d x$

$\Rightarrow I =\sum\limits_{r=1}^n \int\limits_{(r-1)^2}^{r^2}[\sqrt{x}] d x$

$\Rightarrow I =\sum\limits_{r=1}^n \int\limits_{(r-1)^2}^{r^2}(r-1) d x$

$\Rightarrow I=\sum\limits_{r=1}^n(r-1)(2 r-1)$

$\Rightarrow I=\sum\limits_{r=1}^n\left(2 r^2-3 r+1\right)$

$\Rightarrow I=2 \times \frac{n(n+1)(2 n+1)}{6}-3 \times \frac{n(n+1)}{2}+n$

$\Rightarrow I=\frac{n}{6}\left\{2\left(2 n^2+3 n+1\right)-9 n-9+6\right\}$

$\Rightarrow I=\frac{n}{6}\left(4 n^2-3 n-1\right)$

$\Rightarrow I=\frac{n}{6}(n-1)(4 n+1)=\frac{n(n-1)(4 n+1)}{6}$