CBSE Class 12 Mathematics sample paper 2026-27: question 34(B)

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Step-by-step solution

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1. Figure (described): $y^2=2x$ is a right-opening parabola with vertex at the origin. The line $x=2y$ passes through the origin. The enclosed region lies in the first quadrant, between the upper branch $y=\sqrt{2x}$ (above) and the line $y=\frac{x}{2}$ (below), from $x=0$ to $x=8$.

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2. Points of intersection: put $x=2y$ in $y^2=2x$ to get $y^2=4y\Rightarrow y=0$ or $y=4$. This gives the points $(0,0)$ and $(8,4)$.

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3. Required area $={\int }_0^8\left(\sqrt{2x}-\frac{x}{2}\right)dx$.

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4. Integrating, the area is ${\left[\sqrt{2}\cdot \frac{2}{3}{x}^{3/2}-\frac{x^2}{4}\right]}_0^8$.

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5. Now $8^{3/2}=16\sqrt{2}$, so the area is $\sqrt{2}\cdot \frac{2}{3}\cdot 16\sqrt{2}-\frac{64}{4}=\frac{64}{3}-16=\frac{16}{3}$ sq. units.

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Answer: $\frac{16}{3}$ sq. units

Revise the chapter

• Chapter: Application of Integrals (notes, formulas, where marks are lost and every past question)