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

$\frac{4\pi}{3} - \sqrt{3}$

$\text{Area of minor segment} = \int_{1}^{2} \left(\sqrt{4 - x^2} - (-\sqrt{4 - x^2})\right)\,dx$

$= 2\int_{1}^{2} \sqrt{4 - x^2}\,dx$

$= 2 \left[ \frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\left(\frac{x}{2}\right) \right]_{1}^{2}$

$= 2 \left[ \left(0 + 2\cdot \frac{\pi}{2}\right) - \left(\frac{1}{2}\sqrt{3} + 2\cdot \frac{\pi}{6}\right) \right]$

$= 2 \left[ \pi - \left(\frac{\sqrt{3}}{2} + \frac{\pi}{3}\right) \right]$

$= 2 \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)$

$= \frac{4\pi}{3} - \sqrt{3}$

$\textbf{The area of the minor segment is } \frac{4\pi}{3} - \sqrt{3}.$