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Mathematics · 2023 · 5 marks
CBSE 2023 · Region 2 · Set 3 · Q35
Find the area of the smaller region bounded by the curves (x^2)/25+(y^2)/16=1 and x/5+y/4=1, using integration.
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.
Marking-scheme solution
Correct Figure
$$\begin{aligned}
& \text { Required Area }=\frac{4}{5} \int_{0}^{5} \sqrt{25-x^{2}} d x-\frac{4}{5} \int_{0}^{5}(5-x) d x \\
& \left.\left.=\frac{4}{5}\left(\frac{x}{2} \sqrt{25-x^{2}}+\frac{25}{2} \sin ^{-1} \frac{x}{5}\right)\right]_{0}^{5}+\frac{2}{5}(5-x)^{2}\right]_{0}^{5} \\
& =\frac{4}{5}\left(\frac{25 \pi}{4}\right)-10=5 \pi-10
\end{aligned}
$$
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