Mathematics - Area Under Curves Question with Solution | TestHub

$$\begin{array}{rl}& \text{ Area of circle }=\pi (\sqrt{2}{)}^2=2\pi \\ & \text{ Area of }OCADO=2\{\text{ Area of }OCAO\}\\ & =2\text{ area of }OCB+\text{ area of }BCA\}\\ & =2{\int }_0^1{y}_{p}dx+2{\int }_1^{\sqrt{2}}{y}_{c}dx\\ & \text{ where }{y}_p=\sqrt{x}\text{ and }{y}_c=\sqrt{2-x^2}\\ & \therefore \text{ Required Area }\\ & =2\int \frac{1}{x}dx+2{\int }_0^{\sqrt{2}}\sqrt{2-x^2}dx\\ & =2\left[\frac{2}{3}\cdot 1-0\right]+2{\left[\frac{x\sqrt{2-x^2}}{2}+{\sin }^{-1}\frac{x}{\sqrt{2}}\right]}_1^{\sqrt{2}}\end{array}$$ $$\begin{array}{rl}& =\frac{4}{3}+2\left\{\frac{\pi }{2}-\frac{\pi }{4}-\frac{1}{2}\right\}=\frac{4}{3}+2\left\{\frac{\pi }{4}-\frac{1}{2}\right\}=\frac{3\pi +2}{6}\\ & \text{ Bigger area }=2\pi -\frac{3\pi +2}{6}=\frac{9\pi -2}{6}\\ & \therefore \text{ Required Ratio }=\frac{9\pi -2}{3\pi +2}\text{ i.e., }9\pi -2:3\pi +2\end{array}$$