Mathematics - Area Under Curves Question with Solution | TestHub

Required area $=4$ (Area of the shaded region in first quadrant) $$=4{\int }_0^{1\sqrt{2}}\left(y_1-y_2\right)dx=4{\int }_0^{1\sqrt{2}}\left(\sqrt{1-x^2}-x\right)dx$$ $=4{\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x-\frac{x^2}{2}\right]}_0^{1/\sqrt{2}}$ $=4\left[\frac{1}{2\sqrt{2}}\times \frac{1}{\sqrt{2}}+\frac{1}{2}\times \frac{\pi }{4}-\frac{1}{4}\right]$ $=4\left[\frac{1}{4}+\frac{\pi }{8}-\frac{1}{4}\right]=\frac{4\pi }{8}=\frac{\pi }{2}$ sq units