Question

Find the area bouded by the curves $y=2x-x^2,4y=(x-2{)}^2$ and $y=0$.

Answer 1

Graph for the curves $y=2x-x^2,4y=(x-2{)}^2$ and $y=0$ is


Point of intersection of the curves with equation
$y=2x-x^2$ and $4y=(x-2{)}^2$ is given by
$4(2x-x^2)=(x-2{)}^2$
$\Rightarrow 5x^2-12x+4=0$
$\Rightarrow (x-2)(2-5x)=0$
$\Rightarrow x=2,\frac{2}{5}\Rightarrow y=0,\frac{16}{25}$
$\Rightarrow (2,0),(\frac{2}{5},\frac{16}{25})$

Required area is given by,
$=\underset{0}{\overset{\frac{2}{5}}{\int }}(2x-x^2)dx+\underset{\frac{2}{5}}{\overset{2}{\int }}\frac{(x-2{)}^2}{4}dx$
$={[x^2-\frac{x^3}{3}]}_0^{\frac{2}{5}}+{\left[\frac{(x-2{)}^3}{12}\right]}_{\frac{2}{5}}^2$
$=\frac{4}{25}-\frac{8}{375}+\frac{128}{375}=\frac{12}{25}\text{sq. units.}$