Application of Integrals: The area of the region \(x,y):x^2-8xle yle-x\ is : | JEE Main 2026 | Yukti Vault

Yukti Vault

JEE Main 2026 (Online) 6th April Evening ShiftQuestion 1

The area of the region $\{(x,y):x^{2}-8x\le y\le-x\}$ is :

Application of Integrals

$\frac{343}{6}$✓ Correct answer$\frac{637}{6}$$\frac{437}{6}$$\frac{523}{6}$

Solution

$x^2-8x=-x\Rightarrow x=0,7$. Area $=\int_0^7[-x-(x^2-8x)]dx=\int_0^7(7x-x^2)dx=\tfrac{7\cdot49}{2}-\tfrac{343}{3}=\tfrac{343}{6}$. Answer: (A).

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