The least integer which is greater than or equal to the area of region in x - y plane satisfying
${\mathrm{x}}^6-{\mathrm{x}}^2+{\mathrm{y}}^2\le 0$ is

$$\begin{gathered} Required area \\ =4{\int }_0^1 \sqrt{{\mathrm{x}}^2-{\mathrm{x}}^6}\mathrm{dx}={\int }_0^1 4\mathrm{x}\sqrt{1-{\mathrm{x}}^4}\mathrm{dx} \\ put x^2=\sin \theta \Rightarrow 2x dx=\cos \theta d\theta \\ =2{\int }_0^{\frac{\pi }{2}} \sqrt{1-{\sin }^2\theta } .\cos \theta d\theta \\ =2{\int }_0^{\frac{\pi }{2}} {\cos }^2\theta d\theta \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$