The area of the region bounded by the curve $$y = \max\{|x|, x|x-2|\}$$, then x-axis and the lines $$x = -2$$ and $$x = 4$$ is equal to __________.

The curve is given by $$y=\max\{\lvert x\rvert,\;x\,\lvert x-2\rvert\}$$ between the vertical lines $$x=-2$$ and $$x=4$$. To find the bounded area we must, on every sub-interval, decide which of the two expressions is larger and then integrate that expression above the $$x$$-axis.

Write the two functions explicitly on each interval:
For $$\lvert x\rvert$$ we already have the standard definition.
For $$x\,\lvert x-2\rvert$$ we split at $$x=2$$.

Case 1: $$x\le 0$$ $$\lvert x\rvert=-x,\qquad\lvert x-2\rvert=2-x,\qquad x\,\lvert x-2\rvert=x(2-x)=2x-x^{2}\lt0$$ Here $$x\,\lvert x-2\rvert$$ is negative whereas $$\lvert x\rvert$$ is positive, so $$y=\lvert x\rvert=-x\quad\text{on}\;[-2,0].$$

Case 2: $$0\le x\le2$$ $$\lvert x\rvert=x,\qquad\lvert x-2\rvert=2-x,\qquad x\,\lvert x-2\rvert=2x-x^{2}.$$ Compare the two: $$x\; \text{vs}\; 2x-x^{2}\;\Longrightarrow\; (2x-x^{2})-x=x(1-x).$$ Therefore   • When $$0\le x\lt1$$, $$x(1-x)\gt0$$, so $$y=2x-x^{2}.$$   • When $$1\le x\le2$$, $$x(1-x)\le0$$, so $$y=x.$$

Case 3: $$2\le x\le4$$ $$\lvert x\rvert=x,\qquad\lvert x-2\rvert=x-2,\qquad x\,\lvert x-2\rvert=x^{2}-2x.$$ Compare: $$(x^{2}-2x)-x=x(x-3).$$ Hence   • When $$2\le x\lt3$$, $$x(x-3)\lt0$$, so $$y=x.$$   • When $$3\le x\le4$$, $$x(x-3)\ge0$$, so $$y=x^{2}-2x.$$

Collecting the pieces we integrate:

$$A_1=\int_{-2}^{0}(-x)\,dx=\Bigl[-\tfrac{x^{2}}{2}\Bigr]_{-2}^{0}=2$$

$$A_2=\int_{0}^{1}(2x-x^{2})\,dx=\Bigl[x^{2}-\tfrac{x^{3}}{3}\Bigr]_{0}^{1}=1-\tfrac13=\tfrac23$$

$$A_3=\int_{1}^{2}x\,dx=\Bigl[\tfrac{x^{2}}{2}\Bigr]_{1}^{2}=2-\tfrac12=\tfrac32$$

$$A_4=\int_{2}^{3}x\,dx=\Bigl[\tfrac{x^{2}}{2}\Bigr]_{2}^{3}=\tfrac92-\tfrac42=\tfrac52$$

$$A_5=\int_{3}^{4}(x^{2}-2x)\,dx=\Bigl[\tfrac{x^{3}}{3}-x^{2}\Bigr]_{3}^{4}=\tfrac{64}{3}-16=\tfrac{16}{3}$$

Add all parts:
$$A=A_1+A_2+A_3+A_4+A_5$$ $$\;=2+\tfrac23+\tfrac32+\tfrac52+\tfrac{16}{3}$$ Convert to a common denominator $$6$$: $$\;=\frac{12+4+9+15+32}{6}=\frac{72}{6}=12.$$

Hence the required area is $$\mathbf{12}$$.