Let the area of the region \ (x, y): |2x - 1| ≤ y ≤ | x ^ 2 - x|, 0...

Given,

The area of the region $\left\{\left(x, y\right): \left|2x - 1\right| \le y \le \left|{x}^{2} - x\right|, 0 \le x \le 1\right\}$ be $A$,

Now solving,

$x - {x}^{2} = 2x - 1$

$⇒ {x}^{2} + x - 1 = 0$

$⇒ x = \dfrac{- 1 \pm \sqrt{5}}{2}$

And equating $x - {x}^{2} = 1 - 2x$, we get $x = \dfrac{3 \pm \sqrt{5}}{2}$

Now plotting the diagram of $y \ge \left|2x - 1\right|$ and $y \le \left|{x}^{2} - x\right|$,

Now from diagram, both curve are symmetric about $x = \dfrac{1}{2}$

Hence, area of the bounded region is given by,

$A = \left|2{∫}_{\dfrac{3 - \sqrt{5}}{2}}^{\dfrac{1}{2}}\left(\left(x - {x}^{2}\right) - \left(1 - 2x\right)\right)dx\right|$

$⇒ A = \left|2{∫}_{\dfrac{3 - \sqrt{5}}{2}}^{\dfrac{1}{2}}\left(- {x}^{2} + 3x - 1\right)dx\right|$

$⇒ A = \left|2{\left(\dfrac{- {x}^{3}}{3} + \dfrac{3}{2}{x}^{2} - x\right)}_{\dfrac{3 - \sqrt{5}}{2}}^{\dfrac{1}{2}}\right|$

$⇒ A = \dfrac{5\sqrt{5} - 11}{6}$

$⇒ 6A + 11 = 5\sqrt{5}$

$⇒ {\left(6A + 11\right)}^{2} = 125$.