Let the area of the region {(x, y) : |2x - 1| | vedclass

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(D) The region is defined by $|2x - 1| \leq y \leq |x^2 - x|$ for $0 \leq x \leq 1$.Since $|x^2 - x| = x - x^2$ for $0 \leq x \leq 1$,the inequality i

Vedclass · Accepted Answer

$125$

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(D) The region is defined by $|2x - 1| \leq y \leq |x^2 - x|$ for $0 \leq x \leq 1$.Since $|x^2 - x| = x - x^2$ for $0 \leq x \leq 1$,the inequality is $2|x - 1/2| \leq y \leq x - x^2$.The curves $y = |2x - 1|$ and $y = x - x^2$ intersect where $x - x^2 = |2x - 1|$.For $x \in [0, 1/2]$,$x - x^2 = 1 - 2x \implies x^2 - 3x + 1 = 0$,giving $x = \frac{3 - \sqrt{5}}{2}$.Due to symmetry about $x = 1/2$,the area $A$ is $2 \int_{\frac{3 - \sqrt{5}}{2}}^{1/2} ((x - x^2) - (1 - 2x)) dx$.$A = 2 \int_{\frac{3 - \sqrt{5}}{2}}^{1/2} (-x^2 + 3x - 1) dx = 2 \left[ -\frac{x^3}{3} + \frac{3x^2}{2} - x \right]_{\frac{3 - \sqrt{5}}{2}}^{1/2}$.Evaluating the integral,we get $A = \frac{5\sqrt{5} - 11}{6}$.Thus,$6A + 11 = 5\sqrt{5}$.Therefore,$(6A + 11)^2 = (5\sqrt{5})^2 = 125$.

Let the area of the region $\{(x, y) : |2x - 1| \leq y \leq |x^2 - x|, 0 \leq x \leq 1\}$ be $A$. Then $(6A + 11)^2$ is equal to $.......$.

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