If $$\int_0^2 \left(\sqrt{2x} - \sqrt{2x - x^2}\right)dx = \int_0^1 \left(1 - \sqrt{1-y^2}-\frac{y^2}{2}\right)dy + \int_1^2 \left(2 - \frac{y^2}{2}\right)dy + I$$, then $$I$$ equal to

Given,

$$\int_0^2\left(\sqrt{2x}-\sqrt{2x-x^2}\right)dx=\int_0^1\left(1-\sqrt{1-y^2}-\frac{y^2}{2}\right)dy+\int_1^2\left(2-\frac{y^2}{2}\right)dy+I$$

Now,

$$\text{L.H.S.}=\int_0^2\sqrt{2x}\,dx-\int_0^2\sqrt{2x-x^2}\,dx$$

$$=\frac83-2\int_0^1\sqrt{1-y^2}\,dy$$

Also,

$$\text{R.H.S.}=\int_0^1\left(1-\frac{y^2}{2}\right)dy-\int_0^1\sqrt{1-y^2}\,dy+\int_1^2\left(2-\frac{y^2}{2}\right)dy+I$$

$$=\left[y-\frac{y^3}{6}\right]_0^1-\int_0^1\sqrt{1-y^2}\,dy+\left[2y-\frac{y^3}{6}\right]_1^2+I$$

$$=\frac56-\int_0^1\sqrt{1-y^2}\,dy+\frac56+I$$

$$=\frac53-\int_0^1\sqrt{1-y^2}\,dy+I$$

Comparing L.H.S. and R.H.S.,

$$\frac83-2\int_0^1\sqrt{1-y^2}\,dy=\frac53-\int_0^1\sqrt{1-y^2}\,dy+I$$

Hence,

$$I=1-\int_0^1\sqrt{1-y^2}\,dy$$

Therefore,

$$I=\int_0^1\left(1-\sqrt{1-y^2}\right)dy$$

Hence, the correct answer is

$$\boxed{\text{Option C}}$$