Thank you for reporting, we will resolve it shortly
Q.
Let the area of the region $\left\{(x, y):|2 x-1| \leq y \leq\left|x^2-x\right|, 0 \leq x \leq 1\right\}$ be $A$. Then $(6 A +11)^2$ is equal to ____
$y \geq|2 x -1|, \,\,\,\, y \leq\left| x ^2- x \right|$
Both curve are symmetric about $x=\frac{1}{2}$ Hence
$A=2 \int\limits_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}\left(\left(x-x^2\right)-(1-2 x)\right) d x$
$A=2 \int\limits_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}\left(-x^2+3 x-1\right) d x=2\left(\frac{-x^3}{3}+\frac{3}{2} x^2-x\right)_{\frac{3-\sqrt{5}}{2}}^{\frac{1}{2}}$
On solving $6 A +11=5 \sqrt{5}$
$(6 A +11)^2=125$
