Question

Let $f\left(x\right)=max[x^2,(x-1{)}^2,2x(1-x)]$. If area of the region bounded by the curve $y=f\left(x\right),x-\text{axis},x=0$ and $x=1$ is $A$, then the value of $54A$ is

Answer 1

Let us draw the graph in the interval $[0,1]$
$f\left(x\right)=\{\begin{array}{cc}(x-1{)}^2,& 0\le x\le \frac{1}{3}\\ 2x(1-x),& \frac{1}{3}\le x\le \frac{2}{3}\\ x^2,& \frac{2}{3}\le x\le 1\end{array}$

Required area is,
$A=\underset{0}{\overset{1/3}{\int }}(x-1{)}^{2}dx+\underset{1/3}{\overset{2/3}{\int }}2x(1-x)dx+\underset{2/3}{\overset{1}{\int }}{x}^{2}dx$
$={\left[\frac{(x-1{)}^3}{3}\right]}_0^{1/3}+{[x^2-\frac{2x^3}{3}]}_{1/3}^{2/3}+{\left[\frac{x^3}{3}\right]}_{2/3}^1$
$=\frac{19}{81}+\frac{13}{81}+\frac{19}{81}$
$=\frac{51}{81}A=\frac{17}{27}$

$\therefore 54A=34$