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The given equation of the ellipse is $9x^{2}+25y^{2}=225$. We can rewrite this equation in the standard form by dividing both sides by 225:
$$\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1$$ This is an ellipse with semi-major axis $a = 5$ and semi-minor axis $b = 3$.
We need to find the area bounded by the ellipse, the lines $x = -2$, $x = 2$, and the x-axis. First, let's express $y$ in terms of $x$ from the ellipse equation:
$$\frac{y^{2}}{9} = 1 - \frac{x^{2}}{25}$$
$$y^{2} = 9\left(1 - \frac{x^{2}}{25}\right)$$ $$y = \pm 3\sqrt{1 - \frac{x^{2}}{25}}$$ Since we are considering the area bounded by the x-axis, we take the positive root: $$y = 3\sqrt{1 - \frac{x^{2}}{25}} = \frac{3}{5}\sqrt{25 - x^{2}}$$
The area $A$ is given by the integral of $y$ with respect to $x$ from $x = -2$ to $x = 2$: $$A = \int_{-2}^{2} \frac{3}{5}\sqrt{25 - x^{2}} \, dx$$
Since the function is even (i.e., $f(x) = f(-x)$), we can rewrite the integral as: $$A = 2 \int_{0}^{2} \frac{3}{5}\sqrt{25 - x^{2}} \, dx = \frac{6}{5} \int_{0}^{2} \sqrt{25 - x^{2}} \, dx$$
We use the formula $\int \sqrt{a^{2} - x^{2}} \, dx = \frac{x}{2}\sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$, with $a = 5$: $$A = \frac{6}{5} \left[ \frac{x}{2}\sqrt{25 - x^{2}} + \frac{25}{2}\sin^{-1}\left(\frac{x}{5}\right) \right]_{0}^{2}$$
$$A = \frac{6}{5} \left[ \left(\frac{2}{2}\sqrt{25 - 4} + \frac{25}{2}\sin^{-1}\left(\frac{2}{5}\right)\right) - \left(0 + \frac{25}{2}\sin^{-1}(0)\right) \right]$$ $$A = \frac{6}{5} \left[ \sqrt{21} + \frac{25}{2}\sin^{-1}\left(\frac{2}{5}\right) \right]$$
$$A = \frac{6\sqrt{21}}{5} + 15\sin^{-1}\left(\frac{2}{5}\right)$$
Final Answer: $\frac{6\sqrt{21}}{5} + 15\sin^{-1}\left(\frac{2}{5}\right)$
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