Chọn phương án A.

Phương trình hoành độ giao điểm:

• \(\sqrt{2x}=0\Leftrightarrow x=0\).
• \(\sqrt{2x}=2x-2\Leftrightarrow\begin{cases}2x-2\geq0\\ 2x=4x^2-8x+4\end{cases}\)\(\Leftrightarrow\begin{cases}x\geq1\\ \left[\begin{array}{ll}x=2 &\text{(nhận)}\\ x=\dfrac{1}{2} &\text{(loại)}\end{array}\right.\end{cases}\Leftrightarrow x=2\).
• \(2x-2=0\Leftrightarrow x=1\).

Khi đó: $$\begin{eqnarray*}
S&=&\displaystyle\int\limits_0^1\left|\sqrt{2x}-0\right|\mathrm{\,d}x+\displaystyle\int\limits_1^2\left|2x-2-\sqrt{2x}\right|\mathrm{\,d}x\\
&=&\displaystyle\int\limits_0^1\sqrt{2}x^{\tfrac{1}{2}}\mathrm{\,d}x+\displaystyle\int\limits_1^2\left(\sqrt{2}x^{\tfrac{1}{2}}-2x+2\right)\mathrm{\,d}x\\
&=&\dfrac{2\sqrt{2}}{3}\sqrt{x^3}\bigg|_0^1+\left(\dfrac{2\sqrt{2}}{3}\sqrt{x^3}-x^2+2x\right)\bigg|_1^2\\
&=&\dfrac{5}{3}.
\end{eqnarray*}$$