Find the area of the smaller region bounded by the curves $\frac{x^2}{25} + \frac{y^2}{16} = 1$ and $\frac{x}{5} + \frac{y}{4} = 1$, using integration.

$5\pi - 10$ sq units

The correct answer is Option (2) → $5\pi - 10$ sq units

Eq. of line: $\frac{x}{5} + \frac{y}{4} = 1$

$y = 4 \left( 1 - \frac{x}{5} \right) = 4 - \frac{4x}{5}$

Eq. of ellipse: $\frac{x^2}{25} + \frac{y^2}{16} = 1$

$y = 4 \sqrt{1 - \frac{x^2}{25}}$

$= \frac{4}{5} \sqrt{25 - x^2}$

$\text{Required area} = \int\limits_{0}^{5} (y_{\text{ellipse}} - y_{\text{line}}) \, dx$

$= \int\limits_{0}^{5} \left\{ \frac{4}{5} \sqrt{25 - x^2} - \left( 4 - \frac{4x}{5} \right) \right\} dx$

$= \int\limits_{0}^{5} \frac{4}{5} \sqrt{25 - x^2} \, dx - \int\limits_{0}^{5} \left( 4 - \frac{4x}{5} \right) dx$

$= \frac{4}{5} \left[ \frac{x}{2} \sqrt{25 - x^2} + \frac{1}{2} \cdot 25 \sin^{-1} \frac{x}{5} \right]_{0}^{5} - \left[ 4x - \frac{4x^2}{10} \right]_{0}^{5}$

$= \frac{4}{5} \left[ \frac{25}{2} \sin^{-1} 1 \right] - [20 - 10]$

Substituting $\sin^{-1}(1) = \frac{\pi}{2}$:

$= \frac{4}{5} \times \frac{25\pi}{4} - 10$

$= (5\pi - 10)$

Required Area is $(5\pi - 10) \text{ units}^2$