Question

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

• A. $\frac{12}{25}sq.units.$
• B. $\frac{1}{25}sq.units.$
• C. $\frac{12}{375}sq.units.$
• D. $\frac{36}{75}sq.units.$

Answer 1

The correct option is A $\frac{12}{25}sq.units.$

Graph for the curves
$y=2x-x^2=-(x-1{)}^2+1,4y=(x-2{)}^2\text{and}y=0$ is,


Point of intersection for the curves with equation
$y=2x-x^2\cdots \left(1\right)4y=(x-2{)}^2\cdots (2)$
Using equation (1) and (2),
$4(2x-x^2)=(x-2{)}^2\Rightarrow 8x-4x^2=x^2-4x+4\Rightarrow 5x^2-12x+4=0\Rightarrow 5x^2-10x-2x+4=0\Rightarrow (x-2\left)\right(5x-2)=0\Rightarrow x=2\text{and}x=\frac{2}{5}\Rightarrow y=0\text{and}y=\frac{16}{25}$

Required area bounded by the curves,
$Area=\underset{0}{\overset{2/5}{\int }}(2x-x^2)dx+\underset{2/5}{\overset{2}{\int }}\frac{(x-2{)}^2}{4}dx={(x^2-\frac{x^3}{3})}_0^{2/5}+{\left(\frac{(x-2{)}^3}{12}\right)}_{2/5}^2=\frac{4}{25}-\frac{8}{375}+\frac{128}{375}=\frac{180}{375}=\frac{12}{25}sq.units.$