1. We need the area under

$y=\max\{|x|,\; x|x-2|\}$

from $x=-2$ to $x=4$, above the $x$-axis.

So the required area is

$A=\int_{-2}^{4} \max\{|x|,x|x-2|\}\,dx.$

Since the expression involves absolute values, we first simplify each part on suitable intervals.

2. Define

$f_1(x)=|x|, \qquad f_2(x)=x|x-2|.$

We compare $f_1$ and $f_2$ on intervals divided by the critical points $x=0$ and $x=2$.

3. For $x<0$:

$|x|=-x, \qquad |x-2|=2-x$ so

$f_2(x)=x(2-x)=2x-x^2.$

Since $x<0$, we have $2x-x^2<0$, while $|x|=-x>0$. Therefore,

$\max\{|x|,x|x-2|\}=|x|=-x \qquad (x<0).$

Hence on $[-2,0]$,

$y=-x.$

4. For $0\le x<2$:

$|x|=x, \qquad |x-2|=2-x$ so

$f_2(x)=x(2-x)=2x-x^2.$

Now compare with $f_1(x)=x$:

$2x-x^2 \ge x$ $\iff x-x^2\ge 0$ $\iff x(1-x)\ge 0.$

For $0\le x\le 1$, this is true; for $1\le x<2$, it is false.

So:

• on $[0,1]$, max is $2x-x^2$,
• on $[1,2]$, max is $x$.

5. For $x\ge 2$:

$|x|=x, \qquad |x-2|=x-2$ so

$f_2(x)=x(x-2)=x^2-2x.$

Compare with $f_1(x)=x$:

$x^2-2x\ge x$ $\iff x^2-3x\ge 0$ $\iff x(x-3)\ge 0.$

Since $x\ge 2$, this holds for $x\ge 3$ and fails for $2\le x\le 3$.

So:

• on $[2,3]$, max is $x$,
• on $[3,4]$, max is $x^2-2x$.

6. Therefore the function becomes