Using integration, find the area bounded by the ellipse 9x^2 + 25y^2…

The correct answer is Option (1) → $\frac{6}{5} \sqrt{21} + 15 \sin^{-1} \frac{2}{5} \text{ sq. units}$

$9x^2 + 25y^2 = 225$

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

$A = \int\limits_{-2}^{2} \frac{3}{5} \sqrt{25 - x^2} dx$

$A = \frac{6}{5} \int\limits_{0}^{2} \sqrt{25 - x^2} dx$

$A = \frac{6}{5} \left[ \frac{x}{2} \sqrt{25 - x^2} + \frac{25}{2} \sin^{-1} \frac{x}{5} \right]_{0}^{2}$

$A = \frac{6}{5} \left[ \frac{2}{2} \sqrt{25 - 2^2} + \frac{25}{2} \sin^{-1} \frac{2}{5}\right] - \left[ \frac{0}{2} \sqrt{25 - (0)^2} + \frac{25}{2} \sin^{-1} \frac{0}{5} \right]$

$A = \frac{6}{5} \left[ 1 \times \sqrt{21} + \frac{25}{2} \sin^{-1} \frac{2}{5} \right]$

$A = \frac{6}{5} \left[ \sqrt{21} + \frac{25}{2} \sin^{-1} \frac{2}{5} \right]$

$A = \frac{6}{5} \sqrt{21} + 15 \sin^{-1} \frac{2}{5} \text{ sq. units}$