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

Correct Figure $$\begin{aligned} & \text { Required Area }=\frac{4}{5} \int_{0}^{5} \sqrt{25-x^{2}} d x-\frac{4}{5} \int_{0}^{5}(5-x) d x \\ & \left.\left.=\frac{4}{5}\left(\frac{x}{2} \sqrt{25-x^{2}}+\frac{25}{2} \sin ^{-1} \frac{x}{5}\right)\right]_{0}^{5}+\frac{2}{5}(5-x)^{2}\right]_{0}^{5} \\ & =\frac{4}{5}\left(\frac{25 \pi}{4}\right)-10=5 \pi-10 \end{aligned} $$