At every $x$ , the curve takes the larger value of

$y_1=|x|,y_2=x|x-2|.$

So we must determine which function lies above on different intervals.

(i) Interval $x<0$

$|x|=-x,|x-2|=2-x$

$x|x-2|=x(2-x)=2x-x^2<0$

Since $|x|>0$ and $x|x-2|<0$ ,

$\mathrm{Max}=|x|$ on $[-2,0]$ .

(ii) Interval $0\le x\le 2$

Here

$|x|=x,x|x-2|=x(2-x).$

Compare:

$x(2-x)\ge x\Rightarrow x(1-x)\ge 0$

This holds for $0\le x\le 1$ .

Hence

$0\le x\le 1$ : $max=2x-x^2$

$1\le x\le 2$ : $max=x$

(iii) Interval $x\ge 2$

|x|=x,  x|x-2|=x(x-2)

Compare:

$x(x-2)\ge x\Rightarrow x(x-3)\ge 0$

True for $x\ge 3$ .

So

$2\le x\le 3$ : $max=x$

$3\le x\le 4$ : $max=x^2-2x$

So, Area as a piecewise integral

$A={\int }_{-2}^0|x|dx+{\int }_0^1(2x-x^2)dx+{\int }_1^{3}xdx+{\int }_3^4(x^2-2x)dx$

$=2+\frac{2}{3}+4+\frac{16}{3}$

$=12$ Square Units