$|2x - 1| \le y \le |x^2 - x|$

$2\left|x - \frac{1}{2}\right| \le y \le \left|\left(x - \frac{1}{2}\right)^2 - \frac{1}{4}\right| \qquad x \in [0,1] \qquad x - \frac{1}{2} = t$

$2|x| \le y \le \left|t^2 - \frac{1}{4}\right| \qquad t \in \left[-\frac{1}{2}, \frac{1}{2}\right]$

$-x^2 + \frac{1}{4} = 2x$

$x^2 + 2x - \frac{1}{4} = 0$

$x = \frac{-2 \pm \sqrt{5}}{2} = \frac{-2 + \sqrt{5}}{2}$

$\int_{0}^{\frac{\sqrt{5}-2}{2}} \left(\frac{1}{4} - x^2 - 2x\right) dx$

$\left(\frac{x}{4} - \frac{x^3}{3} - x^2\right)_{0}^{\frac{\sqrt{5}-2}{2}} \qquad \alpha = \frac{\sqrt{5}-2}{2}$

$\frac{\alpha}{4} - \frac{\alpha^3}{3} - \alpha^2 \qquad \alpha^2 = \frac{9-4\sqrt{5}}{4}$

$\frac{\alpha}{4} - \frac{\alpha^3}{3} - \alpha^2$

$\frac{\alpha}{12}(3 - 4\alpha^2) - \alpha^2$

$\frac{1}{12}\left(\frac{\sqrt{5}-2}{2}\right)[3 - 9 + 4\sqrt{5}] - \left(\frac{9-4\sqrt{5}}{4}\right)$

$\frac{1}{2}(\sqrt{5}-2)(-3 + 2\sqrt{5}) - \frac{3-4\sqrt{5}}{4}$

JEE MainsJEE AdvanceBITSATJEE Mains | PYQ

JEE Mains JEE Advance BITSAT
JEE Mains | PYQ