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. $\frac{9}{17}$
• C. $\frac{18}{17}$
• D. None of these

Answer 1

Answer: (D)

None of these

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 $A={\int }_0^{1}f\left(x\right)dx={\int }_0^{\frac{1}{3}}{(1-x)}^{2}dx+{\int }_{\frac{1}{3}}^{\frac{2}{3}}2x(1-x)dx+{\int }_{\frac{2}{3}}^1{x}^{2}dx$
$={(-\frac{1}{3}{(1-x)}^3)}_0^{\frac{1}{3}}+{(x^2-\frac{2x^3}{3})}_{\frac{2}{3}}^{\frac{2}{3}}+{\left(\frac{x^3}{3}\right)}_{\frac{2}{3}}^1$
$=\frac{19}{81}+\frac{13}{81}+\frac{19}{81}=\frac{17}{27}$