y^{2} \ge 2x(Areaoutsidetheparabola)<br/><br/>x^{2} + y^{2} \le 4x(Areainsidethecircle)<br/><br/><imgalt=""src="/api/proxy−image?url=httpsx^{2} + y^{2} = 4xandy^{2} = 2x<br/><br/>⇒ x^{2} + 2x = 4x<br/><br/>⇒ x^{2} = 2x<br/><br/>⇒ x = 0, x = 2<br/><br/>Ifx=0,theny=0andifx=2,theny=±2.
Coordinates of A(0,0)andB(2,2)
As x≥0y≥0 only area above x-axis would be considered.
Hence, area
= \left[\mathop{∫}_{0}^{2}\sqrt{4x - x^{2}}dx - \sqrt{2}\mathop{∫}_{0}^{2}\sqrt{x} dx\right]<br/><br/>= \left[\mathop{∫}_{0}^{2}\sqrt{4 - { \left(x - 2\right)}^{2}}dx - \sqrt{2}\mathop{∫}_{0}^{2}\sqrt{x}dx\right]<br/><br/>= {\left[\left(\dfrac{x - 2}{2}\right)\sqrt{4x - x^{2}} + \dfrac{4}{2} . \sin^{- 1} \left(\dfrac{x - 2}{2}\right)\right]}_{0}^{2} - \sqrt{2}{\left(\dfrac{x^{\dfrac{3}{2}}}{\dfrac{3}{2}}\right)}_{0}^{2}<br/><br/>= \left[0 - 2\sin^{- 1} \left(-1\right)\right] - \sqrt{2} . \d ddfrac{2}{3}2\sqrt{2} ] =\left[\pi - \dfrac{8}{3}\right]& sq. units
