Integral Calculus – Area Under the Curve – JEE Main 03 April 2025 Shift 2

Solution:

$$|x-y| \le y \le 4\sqrt{x}$$

$$y = |x-y|$$

$$y^2 = (x-y)^2$$

$$y^2 = x^2 – 2xy + y^2$$

$$x^2 – 2xy = 0$$

$$x = 0 \text{ or } y = \frac{x}{2}$$

$$4\sqrt{x} = \frac{x}{2}$$

$$16x = \frac{x^2}{4}$$

$$x^2 – 64x = 0 \Rightarrow x = 0, 64$$

$$A = \int_{0}^{64} \left(4\sqrt{x} – \frac{x}{2}\right) dx$$

$$A = \left[ \frac{4x^{3/2}}{3/2} – \frac{x^2}{4} \right]_{0}^{64}$$

$$A = \left[ \frac{8}{3}x^{3/2} – \frac{x^2}{4} \right]_{0}^{64}$$

$$A = \frac{8}{3}(8^3) – \frac{64^2}{4}$$

$$A = \frac{8}{3}(512) – 1024$$

$$A = 64^2 \left(\frac{1}{12}\right)$$

$$A = \frac{1024}{3}$$

Ans. (2)