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Mathematics · 2024 · 5 marks
CBSE 2024 · Region 5 · Set 1 · Q33
Find the area of the region bounded by the curve 4 x^2+y^2=36 using integration.
Find the area of the region bounded by the curve $\displaystyle 4 x^{2}+y^{2}=36$ using integration.
Marking-scheme solution
The given equation can be written as: $\displaystyle \frac{x^{2}}{9}+\frac{y^{2}}{36}=1$, which is an ellipse.
Correct Graph Area of the region bounded by the curve
$$\begin{aligned}
& =4 \times \frac{6}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x \\
& =8\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_{0}^{3} \\
& =18 \pi
\end{aligned}
$$
Correct Graph Area of the region bounded by the curve
$$\begin{aligned}
& =4 \times \frac{6}{3} \int_{0}^{3} \sqrt{9-x^{2}} d x \\
& =8\left[\frac{x}{2} \sqrt{9-x^{2}}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_{0}^{3} \\
& =18 \pi
\end{aligned}
$$All Application of Integrals questions2024 question paperNCERT solutions: Application of IntegralsPractice Application of Integrals →
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CBSE Class 12 Mathematics past-paper question from the 2024 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.