∫limits_0^2≤ft[x^2] d x is equal to

We have,

$\int\limits_0^2\left[x^2\right] d x=\int\limits_0^1\left[x^2\right] d x+\int\limits_1^{\sqrt{2}}\left[x^2\right] d x+\int\limits_{\sqrt{2}}^{\sqrt{3}}\left[x^2\right] d x+\int\limits_{\sqrt{3}}^2\left[x^2\right] d x$

$\Rightarrow \int\limits_0^2\left[x^2\right] d x=\int\limits_0^1 0 d x+\int\limits_1^{\sqrt{2}} 1 . d x+\int\limits_{\sqrt{2}}^{\sqrt{3}} 2 d x+\int\limits_{\sqrt{3}}^2 3 d x$

$= 0+1(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})=5-\sqrt{3}-\sqrt{2}$