CBSE 2024 · Region 1 · Set 1 · Q34 · 5 marks
Using integration, find the area of the ellipse $\displaystyle \frac{\mathrm{x}^{2}}{16}+\frac{y^{2}}{4}=1$, included between the lines $\displaystyle \mathrm{x}=-2$ and $\displaystyle \mathrm{x}=2$.
Marking-scheme solution
$$\begin{aligned}
\text { Area }= & 4 \int_{0}^{2} y d \mathrm{x} \\
= & 4\left[\frac{1}{2} \int_{0}^{2} \sqrt{4^{2}-\mathrm{x}^{2}} d \mathrm{x}\right] \\
= & 2\left[\frac{\mathrm{x}}{2} \sqrt{4^{2}-\mathrm{x}^{2}}+8 \sin ^{-1}\left(\frac{\mathrm{x}}{4}\right)\right]_{0}^{2} \\
= & 2\left[\sqrt{12}+\frac{8 \pi}{6}\right]=4 \sqrt{3}+\frac{8 \pi}{3}
\end{aligned}
$$
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