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

Question

Let $A$ be the area of the region

${(x,y):y\ge {x}^{2},y\ge (1-x{)}^{2},y\le 2x(1-x)}$.

Then $540A$ is equal to :

Answer: 25

Accepted Answer

$y={x}^{2}$ and $y=2x(1-x)$

$\Rightarrow {x}^{2}=2x-2{x}^{2}$

$\Rightarrow x=0,x=\frac{2}{3}$

Now,

$y=(1-x{)}^{2}$ and $y=2x(1-x)$

$\Rightarrow 1+{x}^{2}-2x=2x-2{x}^{2}$

$\Rightarrow 3{x}^{2}-4x+1=0$

$x=1,x=\frac{1}{3}$

$	herefore $ $A=\int _{\frac{1}{3}}^{\frac{2}{3}}(2x-2{x}^{2})dx-\int _{\frac{1}{3}}^{\frac{1}{2}}{(1-x)}^{2}dx+\int _{\frac{1}{2}}^{\frac{2}{3}}{x}^{2}dx$

= ${({x}^{2}-\frac{2{x}^{3}}{3})}_{\frac{2}{3}}^{\frac{1}{3}}-{{(\frac{{(x-1)}^{3}}{3})}_{\frac{1}{2}}^{\frac{1}{3}}+{(\frac{{x}^{3}}{3})}_{\frac{2}{3}}^{\frac{1}{2}}}$

$=\frac{5}{108}$

$	herefore $ $540A=25$

Solutions · 1