Question

The function
$f\left(x\right)=max\{x^2,(1-x{)}^2,2x(1-x)\}$ where $0\le x\le 1$
then area of the region bounded by the curve $y=f\left(x\right),x-$axis and $x=0,x=1$ is equals

• A. $\frac{27}{17}\text{sq. units}$
• B. $\frac{17}{27}\text{sq. units}$
• C. $\frac{18}{27}\text{sq. units}$
• D. $\frac{19}{17}\text{sq. units}$

Answer 1

The correct option is B $\frac{17}{27}\text{sq. units}$


Solving the curve equations for the points of intersection:
$y=x^2=2x(1-x)$
$\Rightarrow x=0,\frac{2}{3}$
and $y=(1-x{)}^2=2x(1-x)$
$\Rightarrow x=1,\frac{1}{3}$
From the figure it is clear that
$f\left(x\right)=\{\begin{array}{cc}(1-x{)}^2,& 0\le x\le \frac{1}{3}\\ 2x(1-x),& \frac{1}{3}<x\le \frac{2}{3}\\ x^2,& \frac{2}{3}<x\le 1\end{array}$
The required area $A$ is
$A=\underset{0}{\overset{\frac{1}{3}}{\int }}(1-x{)}^{2}dx+\underset{\frac{1}{3}}{\overset{\frac{2}{3}}{\int }}2x(1-x)dx+\underset{\frac{2}{3}}{\overset{1}{\int }}{x}^{2}dx={\left[\frac{(1-x{)}^3}{-3}\right]}_0^{1/3}+{\left[x^2\right]}_{1/3}^{2/3}-2{\left[\frac{x^3}{3}\right]}_{1/3}^{2/3}+{\left[\frac{x^3}{3}\right]}_{2/3}^1=[-\frac{8}{81}+\frac{1}{3}]+[\frac{4}{9}-\frac{1}{9}]-[\frac{16}{81}-\frac{2}{81}]+[\frac{1}{3}-\frac{8}{81}]=\frac{19}{81}+\frac{1}{3}-\frac{14}{81}+\frac{19}{81}=\frac{17}{27}\text{sq. units}$