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Given the equation of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$, we need to express $y$ in terms of $x$. From the equation, we have $\frac{y^{2}}{4}=1-\frac{x^{2}}{16}$. Multiplying by 4, we get $y^{2}=4\left(1-\frac{x^{2}}{16}\right)=4-\frac{x^{2}}{4}$. Taking the square root, we have $y=\pm\sqrt{4-\frac{x^{2}}{4}}=\pm\frac{1}{2}\sqrt{16-x^{2}}$. Since we are finding the area, we consider the positive part, $y=\frac{1}{2}\sqrt{16-x^{2}}$.
The area of the ellipse between $x=-2$ and $x=2$ is given by the integral of $2y$ with respect to $x$ from $-2$ to $2$. $$Area = \int_{-2}^{2} 2y \, dx = \int_{-2}^{2} 2\left(\frac{1}{2}\sqrt{16-x^{2}}\right) \, dx = \int_{-2}^{2} \sqrt{16-x^{2}} \, dx$$
We use the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\left(\frac{x}{a}\right)+C$. In our case, $a=4$, so $$\int \sqrt{16-x^{2}} \, dx = \frac{x}{2}\sqrt{16-x^{2}}+\frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right)+C = \frac{x}{2}\sqrt{16-x^{2}}+8\sin^{-1}\left(\frac{x}{4}\right)+C$$ Now, we evaluate the definite integral: $$Area = \left[\frac{x}{2}\sqrt{16-x^{2}}+8\sin^{-1}\left(\frac{x}{4}\right)\right]_{-2}^{2}$$ $$Area = \left(\frac{2}{2}\sqrt{16-4}+8\sin^{-1}\left(\frac{2}{4}\right)\right) - \left(\frac{-2}{2}\sqrt{16-4}+8\sin^{-1}\left(\frac{-2}{4}\right)\right)$$ $$Area = \left(\sqrt{12}+8\sin^{-1}\left(\frac{1}{2}\right)\right) - \left(-\sqrt{12}+8\sin^{-1}\left(-\frac{1}{2}\right)\right)$$ Since $\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$ and $\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$, $$Area = \left(2\sqrt{3}+8\left(\frac{\pi}{6}\right)\right) - \left(-2\sqrt{3}+8\left(-\frac{\pi}{6}\right)\right)$$ $$Area = 2\sqrt{3}+\frac{4\pi}{3}+2\sqrt{3}+\frac{4\pi}{3} = 4\sqrt{3}+\frac{8\pi}{3}$$
The area of the ellipse between the lines $x=-2$ and $x=2$ is $4\sqrt{3}+\frac{8\pi}{3}$ square units.
Final Answer: $4\sqrt{3}+\frac{8\pi}{3}$
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