Using integration, find the area of region bounded by line y =√ 3 x, the curve y =√(4-x^2) and y -axis in first quadrant.

Correct Figure Shaded region Point of intersection at $\displaystyle \mathbf{x}=\mathbf{1} \operatorname{ar}(O A B)=\int_{0}^{1} \sqrt{4-x^{2}} d x-\int_{0}^{1} \sqrt{3} x d x \left.\left.=\left(\frac{x \sqrt{4-x^{2}}}{2}+2 \sin ^{-1}\left(\frac{x}{2}\right)\right)\right]_{0}^{1}-\frac{\sqrt{3}}{2} x^{2}\right]_{0}^{1}$ $\displaystyle =\frac{\sqrt{3}}{2}+2 \times \frac{\pi}{6}-\frac{\sqrt{3}}{2}=\frac{\pi}{3}$