If ∫ _0^2(√2x-√2x-x^2)dx= ∫ _0^1(1-√1-y^2- y^2 2)dy+∫ _1^2(2- y^2 2)dy+I, then I equal to

Given,

$${\int }_{0}^{2}(\sqrt{2x}-\sqrt{2x-{x}^{2}})dx=$$∫02​(2x​−2x−x2​)dx=

$${\int }_{0}^{1}(1-\sqrt{1-{y}^{2}}-\frac{{y}^{2}}{2})dy+{\int }_{1}^{2}(2-\frac{{y}^{2}}{2})dy+I$$∫01​(1−1−y2​−2y2​)dy+∫12​(2−2y2​)dy+I

Solving LHS $\displaystyle ={\int }_{0}^{2}(\sqrt{2x}-\sqrt{2x-{x}^{2}})dx=\frac{8}{3}-\frac{\pi }{2}$=∫02​(2x​−2x−x2​)dx=38​−2π​

Now, solving RHS $\displaystyle ={\int }_{0}^{1}(1-\sqrt{1-{y}^{2}}-\frac{{y}^{2}}{2})dy+{\int }_{1}^{2}(2-\frac{{y}^{2}}{2})dy+I$=∫01​(1−1−y2​−2y2​)dy+∫12​(2−2y2​)dy+I

$$=I+\frac{5}{3}-\frac{\pi }{4}$$=I+35​−4π​

Now equating LHS and RHS we get, $\displaystyle I=1-\frac{\pi }{4}={\int }_{0}^{1}(1-\sqrt{1-{y}^{2}})dy$I=1−4π​=∫01​(1−1−y2​)dy as $\displaystyle {\int }_{0}^{1}(\sqrt{1-{y}^{2}})dy=\frac{\pi }{4}$∫01​(1−y2​)dy=4π​