The function f(x) is defined as f(x) = max $\{4-x^2,|x-2|,{(x-2)}^{1/3}\}$  for $x\in [-2,4]$ . If the area bounded by the curve and x–axis is A, then the value of 4A=___________

$y=4-x^2$  is a parabola EMA.
$y=|x-2|$  is a pair of straight lines ABF and ACD.
$y={(x-2)}^{1/3}$  is the curve IHABG.
Thus from the graph
$f\left(x\right)=\max \{4-x^2,|x-2|,{(x-2)}^{1/3}\}$  is
 $f\left(x\right)=\{\begin{array}{ccc}2-x& if& -2\le x<-1\\ 4-x^2& if& -1\le x<2\\ {(x-2)}^{1/3}& if& 2\le x<3\\ x-2& if& 3\le x\le 4\end{array}$
Now area bounded by the curve and x–axis
= $\underset{-2}{\overset{-1}{\int }}(2-x)dx+\underset{-1}{\overset{2}{\int }}(4-x^2)dx+\underset{2}{\overset{3}{\int }}{(x-2)}^{1/3}dx+\underset{3}{\overset{4}{\int }}(x-2)dx$
= ${[2x-\frac{x^2}{2}]}_{-2}^{-1}+{[4x-\frac{x^3}{3}]}_{-1}^2+{\left[\frac{3}{4}{(x-2)}^{4/3}\right]}_2^3+{[\frac{x^2}{2}-2x]}_3^4$
=$\frac{7}{2}+9+\frac{3}{4}+\frac{3}{2}=\frac{59}{4}$ sq. units