Question

Consider the functions $f, g: R \rightarrow R$ defined by
$f(x)=x^2+\frac{5}{12} \text { and } g(x)= \begin{cases} 2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4}, \\0, & |x|>\frac{3}{4} .\end{cases}$
If $\alpha$ is the area of the region
$\left\{( x , y ) \in R \times R :| x | \leq \frac{3}{4}, 0 \leq y \leq \min \{f( x ), g( x )\}\right\},$
then the value of $9 \alpha$ is _____

Answer 1

$ x^2+\frac{5}{12}=\frac{2-8 x}{3}$
$ x^2+\frac{8 x}{3}+\frac{5}{12}-2=0$

$ 12 x^2+32 x-19=0$
$ 12 x^2+38 x-6 x-19=0$
$ 2 x(6 x+19)-1(6 x+19)=0 $
$ (6 x+19)(2 x-1)=0$
$ x=\frac{1}{2}$
$\alpha=2 A _1+ A _2 $
$ \alpha=2\left(\int\limits_0^{1 / 2} x ^2+\frac{5}{12} dx +\frac{1}{2} \times \frac{1}{4} \times \frac{2}{3}\right)$
$ \Rightarrow \alpha=2\left[\left(\frac{ x ^3}{3}+\frac{5 x }{12}\right)_0^{1 / 2}+\frac{1}{12}\right] $
$ \Rightarrow \alpha=2\left[\frac{1}{24}+\frac{5}{24}+\frac{1}{12}\right]$
$ \Rightarrow \alpha=2\left[\frac{1+5+2}{24}\right] \Rightarrow \alpha=2 \times \frac{8}{24} \Rightarrow 9 \alpha=9 \times \frac{8}{12} $
$\Rightarrow 9 \alpha=6$