Let $f(x)=$ Maximum $\left\{x^{2},(1-x)^{2}, 2 x(1-x)\right\}$, where $0 \leq x \leq 1$. If $\frac{m

Question

Let $f(x)=$ Maximum $\left\{x^{2},(1-x)^{2}, 2 x(1-x)\right\}$, where $0 \leq x \leq 1$. If $\frac{m}{n}$ (m,n are coprime postive integers) be the area of the region bounded by the curves $y=f(x), x$ - axis, $x=0$ and $x=1$, find $m+n$.

Professor PC Thomas Classes · Answer

We can rewrite $f(x)$ as\ $f(x)=\left\{\begin{array}{cc}(1-x)^{2}, & 0 \leq x \leq 1 / 3 \ 2 x(1-x), & 1 / 3<x \geq 2 / 3 \ x^{2}, & 2 / 3<x \leq 1\end{array}ight.$\ $	herefore \quad$ Required area\ $=\int_{0}^{1 / 3}(1-x)^{2} d x+\int_{1 / 3}^{2 / 3} 2 x(1-x) d x+\int_{2 / 3}^{1} x^{2} d x$\ $=\left[-\frac{(1-x)^{3}}{3}ight]_{0}^{1 / 3}+\left[x^{2}-\frac{2 x^{3}}{3}ight]_{1 / 3}^{2 / 3}+\left[\frac{x^{3}}{3}ight]_{2 / 3}^{1}$\ $=\frac{-8}{3 	imes 27}+\frac{1}{3}+\frac{4}{9}-\frac{16}{81}-\frac{1}{9}+\frac{2}{81}+\frac{1}{3}-\frac{8}{27 	imes 3}$\ $=-\frac{32}{81}+\frac{2}{81}+1$\ $=\frac{-30+81}{81}=\frac{51}{81}=\frac{17}{27}$ square units.\