Let $$f(x) = |x - 2|$$ and $$g(x) = f(f(x))$$, $$x \in [0, 4]$$. Then $$\int_0^3 (g(x) - f(x)) dx$$ is equal to

We have $$f(x)=|x-2|$$. The definition of absolute value is $$|t|=\begin{cases}t,& t\ge 0\\-t,& t<0\end{cases}$$, so we first write $$f(x)$$ in simple algebraic form for each part of the domain $$[0,4]$$.

When $$0\le x\le 2$$, the expression $$x-2\le 0$$. Hence $$f(x)=-(x-2)=2-x$$.

When $$2\le x\le 4$$, the expression $$x-2\ge 0$$. Hence $$f(x)=x-2$$.

Next we define $$g(x)=f(f(x))$$, that is $$g(x)=\left|\,f(x)-2\,\right|$$. We substitute the two pieces of $$f(x)$$ found above.

For $$0\le x\le 2$$ we already have $$f(x)=2-x$$, so

$$f(x)-2=(2-x)-2=-x\le 0,\qquad g(x)=|-x|=x.$$

For $$2\le x\le 4$$ we have $$f(x)=x-2$$, so

$$f(x)-2=(x-2)-2=x-4\le 0,\qquad g(x)=|x-4|=4-x.$$

Thus, on the interval we need, $$[0,3]$$, the two functions are

$$ \begin{aligned} 0\le x\le 2:&\qquad f(x)=2-x,\; g(x)=x,\\[4pt] 2\le x\le 3:&\qquad f(x)=x-2,\; g(x)=4-x. \end{aligned}$$

We have to evaluate the definite integral

$$\int_{0}^{3}\bigl(g(x)-f(x)\bigr)\,dx.$$

Because the formulas change at $$x=2$$, we split the integral:

$$\int_{0}^{3}(g-f)\,dx=\int_{0}^{2}(g-f)\,dx+\int_{2}^{3}(g-f)\,dx.$$

First sub-integral: $$0\le x\le 2$$.

Here $$g(x)=x$$ and $$f(x)=2-x$$, so

$$g(x)-f(x)=x-(2-x)=2x-2=2(x-1).$$

Integrate term-by-term:

$$ \int_{0}^{2}2(x-1)\,dx =2\int_{0}^{2}(x-1)\,dx =2\left[\frac{x^{2}}{2}-x\right]_{0}^{2} =2\left(\left(\frac{4}{2}-2\right)-0\right) =2(2-2)=0. $$

Second sub-integral: $$2\le x\le 3$$.

Here $$g(x)=4-x$$ and $$f(x)=x-2$$, so

$$g(x)-f(x)=(4-x)-(x-2)=4-x-x+2=6-2x.$$

Integrate:

$$ \int_{2}^{3}(6-2x)\,dx =\left[6x- x^{2}\right]_{2}^{3} =\bigl(6\cdot3-9\bigr)-\bigl(6\cdot2-4\bigr) =(18-9)-(12-4)=9-8=1. $$

Adding the two results,

$$\int_{0}^{3}(g-f)\,dx=0+1=1.$$

Hence, the correct answer is Option A.