Find the area of the minor segment of the circle $\displaystyle x^{2}+y^{2}=4$ cut off by the line $\displaystyle x=1$, using integration.

$$\text { Required area }=2 \int_{1}^{2} \sqrt{4-x^{2}} d x =2\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1}\left(\frac{x}{2}\right)\right]_{1}^{2} =2\left[\left\{0+2\left(\frac{\pi}{2}\right)\right\}-\left\{\frac{1}{2} \sqrt{3}+2 \cdot \frac{\pi}{6}\right\}\right] =2\left(\pi-\frac{\sqrt{3}}{2}-\frac{\pi}{3}\right) =\left(\frac{4 \pi}{3}-\sqrt{3}\right) $$