Find the smaller area enclosed by the circle $x^2 + y^2 = 4$ and the line $x + y = 2$

The correct answer is $\pi-2$

Area = $\int_{0}^{2} \left[ \sqrt{4 - x^2} - (2 - x) \right]$ $dx$

= $\dfrac{x}{2} \sqrt{4 - x^2} $ + $\dfrac{4}{2} sin^{-1} \dfrac{x}{2}$ - $2x$ + $\dfrac{x^2}{2}$ $|_{0}^{2} $

= 0 + 2$\dfrac{\pi}{2} - 4 + 2 - 0$ = $\pi-2$