JEE Main 2026 Apr 6 Shift 2, Mathematics Q19: Area as Definite Integral

JEE Main2026Apr 6, Shift 2Mathematics

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

Solution

Find intersection points: $x^{2} - 8 x = - x$ $\Rightarrow$ $x^{2} - 7 x = 0$ $\Rightarrow$ $x = 0$ or $x = 7$ .

On $[0, 7]$ the upper curve is $y = - x$ and the lower is $y = x^{2} - 8 x$ . Area:

$A = \int_{0}^{7} [- x - (x^{2} - 8 x)] d x = \int_{0}^{7} (7 x - x^{2}) d x$

$= [\frac{7 x ^{2}}{2} - \frac{x ^{3}}{3}]_{0}^{7} = \frac{343}{2} - \frac{343}{3} = \frac{343}{6}$ .

Concept behind this question

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