Hard· Area between two curves· +4 / −1If ∫02(2x−2x−x2)dx=∫01(1−1−y2−y22)dy+∫12(2−y22)dy+I\displaystyle\int_0^2\left(\sqrt{2x}-\sqrt{2x-x^2}\right)dx=\int_0^1\left(1-\sqrt{1-y^2}-\frac{y^2}{2}\right)dy+\int_1^2\left(2-\frac{y^2}{2}\right)dy+I∫02(2x−2x−x2)dx=∫01(1−1−y2−2y2)dy+∫12(2−2y2)dy+I, then III equals :A∫01(1+1−y2)dy\displaystyle\int_0^1\left(1+\sqrt{1-y^2}\right)dy∫01(1+1−y2)dyB∫01(y22−1−y2+1)dy\displaystyle\int_0^1\left(\frac{y^2}{2}-\sqrt{1-y^2}+1\right)dy∫01(2y2−1−y2+1)dyC∫01(1−1−y2)dy\displaystyle\int_0^1\left(1-\sqrt{1-y^2}\right)dy∫01(1−1−y2)dyD∫01(y22+1−y2+1)dy\displaystyle\int_0^1\left(\frac{y^2}{2}+\sqrt{1-y^2}+1\right)dy∫01(2y2+1−y2+1)dyCheck answerJust show me the answer
