CBSE 2024 · Region 3 · Set 3 · Q35 · 5 marks
Using integration, find the area of the region enclosed between the curve $\displaystyle y=\sqrt{4-x^{2}}$ and the lines $\displaystyle x=-1, x=1$ and the $\displaystyle x$-axis.
Marking-scheme solution
Required area $\displaystyle = 2\int_{0}^{1}\sqrt{4-x^{2}}\,dx$$\displaystyle = 2\left[\dfrac{x}{2}\sqrt{4-x^{2}} + 2\sin^{-1}\dfrac{x}{2}\right]_{0}^{1}$$\displaystyle = \sqrt{3} + \dfrac{2\pi}{3}$Application of IntegralsArea under Simple CurvesApplylong_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.