Chọn phương án D.

Phương trình hoành độ giao điểm: $$-x^2+4x-3=0\Leftrightarrow\left[\begin{array}{ll}x=1 &\in[0;3]\\ x=3\end{array}\right.$$Vậy \(S=\displaystyle\int\limits_{0}^{1}\left|-x^2+4x-3\right|\mathrm{\,d}x+\displaystyle\int\limits_{1}^{3}\left|-x^2+4x-3\right|\mathrm{\,d}x\).

Theo đó ta có $$\begin{aligned}
S&=\displaystyle\int\limits_{0}^{1}\left(x^2-4x+3\right)\mathrm{\,d}x+\displaystyle\int\limits_{1}^{3}\left(-x^2+4x-3\right)\mathrm{\,d}x\\
&=\left(\dfrac{x^3}{3}-2x^2+3x\right)\bigg|_0^1+\left(-\dfrac{x^3}{3}+2x^2-3x\right)\bigg|_1^3\\
&=\dfrac{4}{3}+\dfrac{4}{3}=\dfrac{8}{3}.
\end{aligned}$$