For the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$, what is the area of the region in the first quadrant between $x = 0$ and $x = 2$?

Explanation

1. 01

   $y = \frac{1}{2}\sqrt{16 - x^{2}}$: $\frac{1}{2}\int_0^{2} \sqrt{16 - x^{2}}\,dx = \frac{1}{2}\left[\frac{x}{2}\sqrt{16 - x^{2}} + 8\sin^{-1}\frac{x}{4}\right]_0^{2}$.

2. 02

   $\frac{1}{2}\left(2\sqrt{3} + \frac{4\pi}{3}\right) = \sqrt{3} + \frac{2\pi}{3}$ square units.