Step 1: Rewrite the equation of the ellipse
The equation of the ellipse is:\[9x^2 + 25y^2 = 225 \implies y = \pm \frac{3}{5} \sqrt{25 - x^2}\]Step 2: Set up the integral for the area
The area of the region bounded by the ellipse, the X-axis, and the lines \( x = -2 \) and \( x = 2 \) is given by:\[{Area} = 2 \int_{0}^{2} \frac{3}{5} \sqrt{25 - x^2} \, dx\]Step 3: Simplify the integral
Let \( I = \int \sqrt{a^2 - x^2} \, dx \), where \( a = 5 \). Using the standard formula:\[\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C\]Step 4: Evaluate the integral
Substitute \( a = 5 \) and evaluate \( \int_{0}^{2} \sqrt{25 - x^2} \, dx \):\[\int_{0}^{2} \sqrt{25 - x^2} \, dx = \left[ \frac{x}{2} \sqrt{25 - x^2} + \frac{25}{2} \sin^{-1}\left(\frac{x}{5}\right) \right]_{0}^{2}\]At \( x = 2 \):\[\frac{2}{2} \sqrt{25 - 2^2} + \frac{25}{2} \sin^{-1}\left(\frac{2}{5}\right) = \sqrt{21} + \frac{25}{2} \sin^{-1}\left(\frac{2}{5}\right)\]At \( x = 0 \):\[\frac{0}{2} \sqrt{25 - 0^2} + \frac{25}{2} \sin^{-1}(0) = 0\]Thus:\[\int_{0}^{2} \sqrt{25 - x^2} \, dx = \sqrt{21} + \frac{25}{2} \sin^{-1}\left(\frac{2}{5}\right)\]Step 5: Final area calculation
Multiply by \( \frac{6}{5} \) to account for \( \frac{3}{5} \) and the factor of 2:\[{Area} = \frac{6}{5} \left(\sqrt{21} + \frac{25}{2} \sin^{-1}\left(\frac{2}{5}\right)\right)\]Step 6: Final result
The area bounded by the ellipse, the X-axis, and the lines \( x = -2 \) and \( x = 2 \) is:\[\frac{6\sqrt{21}}{5} + 15 \sin^{-1}\left(\frac{2}{5}\right)\]