Plotting the diagram of the given function $\mathrm{f}\left(\mathrm{x}\right)=\max \left\{{\mathrm{x}}^2,1+\left[x\right]\right\}$ we get,

Now when $x\in [0,1)\Rightarrow f\left(x\right)=1$

When $x\in [1,\sqrt{2})\Rightarrow f\left(x\right)=2$

And when $x\in \left[\sqrt{2},2\right]\Rightarrow f\left(x\right)=x^2$

So, from above diagram area can be calculated as,

$A={\int }_0^{1}1.dx+{\int }_1^{\sqrt{2}}2dx+{\int }_{\sqrt{2}}^2{x}^{2}dx$

$\Rightarrow A=1+2\sqrt{2}-2+\frac{8}{3}-\frac{2\sqrt{2}}{3}$

$\Rightarrow A=\frac{5}{3}+\frac{4\sqrt{2}}{3}$