Question

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

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

Answer 1

Answer: (B)

$\frac{17}{27}$

Solving $y=x^2$ and $y=2x(1-x)$ we get $x=0,\frac{2}{3}$
And solving $y=(1-x{)}^2$ and $y=2x(1-x)$, we get $x=1,\frac{1}{3}$
Therefore
The required area $A={\int }_0^{1}f\left(x\right)dx={\int }_0^{\frac{1}{3}}{(1-x)}^{2}dx+{\int }_{\frac{1}{3}}^{\frac{2}{3}}2x(1-x)dx+{\int }_{\frac{2}{3}}^1{x}^{2}dx$
$={[-\frac{1}{3}{(1-x)}^3]}_0^{\frac{1}{3}}+{[x^2-\frac{2x^3}{3}]}_{\frac{1}{3}}^{\frac{2}{3}}+{\left[\frac{x^3}{3}\right]}_{\frac{2}{3}}^1$
$=\frac{19}{81}+\frac{13}{81}+\frac{19}{81}=\frac{17}{27}$