1. Start with the given integral

L=\int_0^2\left(\sqrt{2x}-\sqrt{2x-x^2}\right)dx

We simplify each term:

$$\sqrt{2x-x^2}=\sqrt{1-(x-1)^2}$$

because

$$2x-x^2=1-(x-1)^2.$$

So,

$$L=\int_0^2 \sqrt{2x}\,dx-\int_0^2 \sqrt{1-(x-1)^2}\,dx.$$

2. Substitute $y=\sqrt{2x}$ in the first integral

Then

$$y=\sqrt{2x}\implies x=\frac{y^2}{2},\qquad dx=y\,dy.$$

When $x=0$, $y=0$; when $x=2$, $y=2$.

Hence

$$\int_0^2 \sqrt{2x}\,dx=\int_0^2 y\cdot y\,dy=\int_0^2 y^2\,dy.$$

But this is not directly matching the given RHS form, so let us instead convert the whole integral carefully in terms of $y$ using the substitution

$$y=x-1.$$

For the second term,

$$\int_0^2\sqrt{1-(x-1)^2}\,dx=\int_{-1}^1\sqrt{1-y^2}\,dy=2\int_0^1\sqrt{1-y^2}\,dy.$$

Also,

$$\int_0^2\sqrt{2x}\,dx=\int_0^2 y^2\,dy=\frac{8}{3}.$$

So

$$L=\frac{8}{3}-2\int_0^1\sqrt{1-y^2}\,dy.$$

3. Now simplify the given right-hand side except for $I$

Given

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

Let

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

Compute the second integral:

$$\int_1^2\left(2-\frac{y^2}{2}\right)dy n=\left[2y-\frac{y^3}{6}\right]_1^2 =\left(4-\frac{8}{6}\right)-\left(2-\frac{1}{6}\right) =\frac{5}{6}.$$

Now,

$$\int_0^1\left(1-\sqrt{1-y^2}-\frac{y^2}{2}\right)dy =\int_0^1 1\,dy-\int_0^1\sqrt{1-y^2}\,dy-\frac12\int_0^1 y^2dy$$ $$=1-\int_0^1\sqrt{1-y^2}\,dy-\frac12\cdot\frac13 =\frac56-\int_0^1\sqrt{1-y^2}\,dy.$$

Therefore,

$$R=\left(\frac56-\int_0^1\sqrt{1-y^2}\,dy\right)+\frac56 =\frac53-\int_0^1\sqrt{1-y^2}\,dy.$$

So

$$I=L-R.$$

4. Compute $I$

We found

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

and

$$R=\frac53-\int_0^1\sqrt{1-y^2}\,dy.$$

Hence

$$I=\left(\frac83-2\int_0^1\sqrt{1-y^2}\,dy\right)-\left(\frac53-\int_0^1\sqrt{1-y^2}\,dy\right)$$ $$I=1-\int_0^1\sqrt{1-y^2}\,dy.$$

Now write $1$ as an integral:

$$1=\int_0^1 1\,dy.$$

Thus

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

5. Match with options

This is exactly Option C:

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

Therefore, the correct answer is C.