Using integration, find the area of the ellipse (x^2)/16+(y^2)/4=1, included between the lines x =-2 and x =2.

$$\begin{aligned} \text { Area }= & 4 \int_{0}^{2} y d \mathrm{x} \\ = & 4\left[\frac{1}{2} \int_{0}^{2} \sqrt{4^{2}-\mathrm{x}^{2}} d \mathrm{x}\right] \\ = & 2\left[\frac{\mathrm{x}}{2} \sqrt{4^{2}-\mathrm{x}^{2}}+8 \sin ^{-1}\left(\frac{\mathrm{x}}{4}\right)\right]_{0}^{2} \\ = & 2\left[\sqrt{12}+\frac{8 \pi}{6}\right]=4 \sqrt{3}+\frac{8 \pi}{3} \end{aligned} $$