The area of the region bounded by the line y = mx ( m >0), the curve x^2+ y^2=4 and the x -axis in the first quadrant is (π)/2 units. Using…

$$\begin{aligned} & \mathrm{x}^{2}+\mathrm{y}^{2}=4 \text { and } \mathrm{y}=\mathrm{m} \mathrm{x} \\ & \Rightarrow \mathrm{x}^{2}+\mathrm{m}^{2} \mathrm{x}^{2}=4 \Rightarrow \mathrm{x}=\frac{2}{\sqrt{1+\mathrm{m}^{2}}} \end{aligned} $$ x - coordinate of the required point of intersection is $\displaystyle \frac{2}{\sqrt{1+\mathrm{m}^{2}}}$. According to question, $$\int_{0}^{\frac{2}{\sqrt{1+\mathrm{m}^{2}}}} \mathrm{m} \mathrm{x} d \mathrm{x}+\int_{\frac{2}{\sqrt{1+\mathrm{m}^{2}}}}^{2} \sqrt{4-\mathrm{x}^{2}} \mathrm{dx}=\frac{\pi}{2} $$ $\displaystyle \left.\Rightarrow \mathrm{m} \frac{\mathrm{x}^{2}}{2}\right|_{0} ^{\frac{2}{\sqrt{1+\mathrm{m}^{2}}}}+\frac{\mathrm{x}}{2} \sqrt{4-\mathrm{x}^{2}}+\left.2 \sin ^{-1} \frac{\mathrm{x}}{2}\right|_{\frac{2}{\sqrt{1+\mathrm{m}^{2}}}} ^{2}=\frac{\pi}{2}$ $\displaystyle \Rightarrow \frac{2 \mathrm{m}}{1+\mathrm{m}^{2}}+\pi-\frac{2 \mathrm{m}}{1+\mathrm{m}^{2}}-2 \sin ^{-1} \frac{1}{\sqrt{1+\mathrm{m}^{2}}}=\frac{\pi}{2}$