2. $\int_{1}^{2} \sqrt{4-x^{2}} \mathrm{~d} x=$
2. $\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}$.

By the geometric meaning of definite integrals, we have
$$
\begin{array}{l}
\int_{1}^{2} \sqrt{4-x^{2}} \mathrm{~d} x=\frac{\pi}{6} \times 2^{2}-\frac{1}{2} \times 1 \times \sqrt{3} \\
=\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}
\end{array}
$$