The area of the region between x – y + 2 = 0 and the curve$\mathrm{x} = \sqrt{\mathrm{y}}$ , Y-axis is

$\begin{array}{r}x+2=y,y=x^2\\ \Rightarrow x^2-x-2=0\end{array}$

$$\begin{gathered} \mathrm{The} \mathrm{two} \mathrm{curves} \mathrm{intersect} \mathrm{at} \mathrm{x}=2,-1 \\ \left(\mathrm{i}.\mathrm{e}\right) \mathrm{at} (2,4) \mathrm{and}(-1,1) \end{gathered}$$

$Area={\int }_0^2\left(x+2-x^2\right)dx={\left(\frac{x^2}{2}+2x-\frac{x^3}{3}\right)}_0^2=2+4-\frac{8}{3}=\frac{10}{3}$