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

Question

Let the area of the region

${(x,y):|2x-1|\le y\le |{x}^{2}-x|,0\le x\le 1}$ be $A$.

Then $(6A+11{)}^{2}$ is equal to

Answer: 125

Accepted Answer

For $B$,

$\begin{gathered} x-{x}^{2}=2x-1 \ {x}^{2}+x-1=0 \ x=\frac{-1+\sqrt{5}}{2} \end{gathered}$

Area $=2($ area of $BCE)$

$A=2{\int }_{\frac{\sqrt{5}-1}{2}}^{\frac{1}{2}}(x-{x}^{2})-(2x-1)dx$

$=2{\int }_{\frac{\sqrt{5}-1}{2}}^{\frac{1}{2}}(1-x-{x}^{2})dx={2(x-\frac{{x}^{2}}{2}-\frac{{x}^{3}}{3})|}_{{]}_{\frac{\sqrt{5}-1}{2}}^{5}}^{\frac{1}{2}}$

$\begin{gathered} =2(\frac{\sqrt{5}-1}{2}-\frac{1}{2})-{(\frac{\sqrt{5}-1}{2})}^{2} -\frac{1}{4}\frac{1}{2} \ \ -[[{(\frac{\sqrt{5}-1}{2})}^{3}-\frac{1}{8}]]\frac{1}{3} \end{gathered}$

$A=\frac{-11+5\sqrt{5}}{6}$

$\Rightarrow (6A+11{)}^{2}=125$

Solutions · 1