JEE Main 2025 Apr 3 Shift 2, Mathematics Q11: Area as Definite Integral

JEE Main2025Apr 3, Shift 2Mathematics

The area of the region ${(x, y) : ∣ x - y ∣ \leq y \leq 4 x}$ is

A
$512$
B
$\frac{1024}{3}$
C
$\frac{512}{3}$
D
$\frac{2048}{3}$

Solution

The condition $∣ x - y ∣ \leq y$ is equivalent to $- y \leq x - y \leq y$ , i.e. $0 \leq x \leq 2 y$ , or $y \geq x /2$ together with $y \geq 0$ and $x \geq 0$ .

So the region is bounded below by $y = x /2$ and above by $y = 4 x$ , for $x \geq 0$ .

Intersection of the two upper-lower curves: $4 x = x /2 \Rightarrow 8 x = x \Rightarrow x = 64$ (taking $x > 0$ ). At $x = 64$ , $y = 32$ .

Area $= \int_{0}^{64} (4 x - \frac{x}{2}) d x = [\frac{8 x ^{3/2}}{3} - \frac{x ^{2}}{4}]_{0}^{64}$ .

$= \frac{8 \cdot 512}{3} - \frac{4096}{4} = \frac{4096}{3} - 1024 = \frac{4096 - 3072}{3} = \frac{1024}{3}$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q11: Area as Definite Integral

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