Nasty areas - 2

Calculus Level 5

Let $\displaystyle{f(x) = \max\{x^2 , (1-x)^2 , 2x(1-x)}\}$

where $\displaystyle{0 \le x \le 1}$ .

The area of the region bounded by the curves $y=f(x),x-axis,x=0,x=1$ can be expressed as $\dfrac{a}{b}$ . Give $\displaystyle{a+b}$

1 solution

Tom Engelsman
Jul 20, 2019

If one plots these three parabolas in the $xy$ -plane, one finds that:

$f(x) = (1-x)^2$ for $0 \le x \le \frac{1}{3} \cup 2x(1-x)$ for $\frac{1}{3} \le x \le \frac{2}{3} \cup x^2$ for $\frac{2}{3} \le x \le 1$

and the required area computes to:

$\int_{0}^{\frac{1}{3}} (1-x)^2 dx + \int_{\frac{1}{3}}^{\frac{2}{3}} 2x(1-x) dx + \int_{\frac{2}{3}}^{1} x^2 dx = \boxed{\frac{17}{27}}.$