\( \int_{0}^{2}\left[x^{2}\right] d x \)

Consider the given integral.
\(I=\int_{0}^{2}\left[x^{2}\right] d x\)
Since,
\(\begin{aligned}
&{\left[\mathrm{x}^{2}\right]=0 \text { for } \mathrm{x} \in[0,1]} \\
&{\left[\mathrm{x}^{2}\right]=1 \text { for } \mathrm{x} \in[1, \sqrt{2}]} \\
&{\left[\mathrm{x}^{2}\right]=0 \text { for } \mathrm{x} \in[\sqrt{2}, \sqrt{3}]} \\
&{\left[\mathrm{x}^{2}\right]=\text { for } \mathrm{x} \in[\sqrt{3}, 2]}
\end{aligned}\)
Therefore,
\(\begin{aligned}
&I=\int_{0}^{1}\left[x^{2}\right] d x+\int_{1}^{\sqrt{2}}\left[x^{2}\right] d x+\int_{\sqrt{2}}^{\sqrt{3}}\left[x^{2}\right] d x+\int_{\sqrt{3}}^{2}\left[x^{2}\right] d x \\
&I=\int_{0}^{1} 0 d x+\int_{1}^{\sqrt{2}} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x+\int_{\sqrt{3}}^{2} 3 d x \\
&I=0+\left.x\right|_{1} ^{\sqrt{2}}+\left.2 x\right|_{\sqrt{2}} ^{\sqrt{3}}+\left.3 x\right|_{\sqrt{3}} ^{2} \\
&I=(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3}) \\
&I=\sqrt{2}-1+2 \sqrt{3}-2 \sqrt{2}+6-3 \sqrt{3} \\
&I=5-(\sqrt{2}+\sqrt{3})
\end{aligned}\)
Hence, the value of integral is \(5-(\sqrt{2}+\sqrt{3})\).