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

$\sqrt{3} + \frac{2\pi}{3}$

The correct answer is Option (1) → $\sqrt{3} + \frac{2\pi}{3}$

$x^2 + y^2 = 4$

$∴y = \sqrt{4 - x^2}$

Required area $= 2 \int\limits_{1}^{2} y \, dx$

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

$= 2 \left[ \frac{x}{2} \sqrt{4-x^2} + \frac{1}{2} \times 4 \sin^{-1} \frac{x}{2} \right]_0^1$

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

$= \left( \sqrt{3} + \frac{2\pi}{3} \right) \text{ unit}^2$