Chọn phương án B.

Theo đề ta có $$\begin{aligned}
t=\sqrt{4-x^2}\Rightarrow&\,t^2=4-x^2\\
\Rightarrow&\,2t\mathrm{\,d}t=-2x\mathrm{\,d}x\\
\Rightarrow&\,x\mathrm{\,d}x=-t\mathrm{\,d}t.
\end{aligned}$$

• \(x=1\Rightarrow t=\sqrt{3}\)
• \(x=2\Rightarrow t=0\)

Khi đó $$\begin{aligned}
I&=\displaystyle\int\limits_{1}^{2}x\sqrt{4-x^2}\mathrm{\,d}x=\displaystyle\int\limits_{\sqrt{3}}^{0}\left(-t^2\right)\mathrm{\,d}t\\
&=\displaystyle\int\limits_{0}^{\sqrt{3}}t^2\mathrm{\,d}t=\dfrac{t^3}{3}\bigg|_0^{\sqrt{3}}=\sqrt{3}.
\end{aligned}$$