$\int_{0}^{1} \sqrt{4-x^{2}} d x=$
设 $y=\sqrt{4-x^{2}}$, 则 $x^{2}+y^{2}=4$, 其中 $0 \leq x \leq 1, y \geq 0$. $\int_{0}^{1} \sqrt{4-x^{2}} d x=$ 的几何意义为图中阴影面积,<ImageHere>设 $\angle B O C=\alpha$ ，易知 $\alpha=rac{\pi}{6}$ ，则 $S=S_{1}+S_{2}=rac{1}{2} \alpha r^{2}+rac{1}{2} O A \cdot A C=rac{1}{2} 	imes rac{\pi}{6} 	imes 4+rac{1}{2} 	imes 1 	imes \sqrt{3}=rac{\pi}{3}+rac{\sqrt{3}}{2}$.