Key Concepts and Formulas

• Definition of Absolute Value: The absolute value function $|a|$ is defined as: $|a| = \begin{cases} a & \text{if } a \ge 0 \\ -a & \text{if } a < 0 \end{cases}$ This is crucial for breaking down the piecewise nature of the functions $f(x)$ and $g(x)$.
• Function Composition: For two functions $f$ and $g$, the composition $g(f(x))$ means substituting the entire expression of $f(x)$ into the variable $x$ of the function $g$. In this problem, we have $g(x) = f(f(x))$.
• Properties of Definite Integrals: We will use the linearity of integration, $\int (a \cdot h(x) + b \cdot k(x)) dx = a \int h(x) dx + b \int k(x) dx$, and the ability to split an integral over intervals, $\int_a^c h(x) dx = \int_a^b h(x) dx + \int_b^c h(x) dx$. We will also need to evaluate integrals of simple polynomial functions.

Step-by-Step Solution

Step 1: Define $f(x)$ as a piecewise function. The given function is $f(x) = |x - 2|$. We can rewrite this using the definition of absolute value: $f(x) = \begin{cases} x - 2 & \text{if } x - 2 \ge 0 \implies x \ge 2 \\ -(x - 2) & \text{if } x - 2 < 0 \implies x < 2 \end{cases}$ So, $f(x) = \begin{cases} x - 2 & \text{if } x \ge 2 \\ 2 - x & \text{if } x < 2 \end{cases}$

Step 2: Determine the piecewise definition of $g(x) = f(f(x))$. We need to evaluate $f(f(x))$. This means we substitute $f(x)$ into the definition of $f$. We need to consider the two cases for $f(x)$:

Case 1: $x \ge 2$. In this case, $f(x) = x - 2$. Now we need to evaluate $f(x-2)$. We need to know if $x-2 \ge 2$ or $x-2 < 2$. If $x-2 \ge 2$, which means $x \ge 4$, then $f(x-2) = (x-2) - 2 = x - 4$. If $x-2 < 2$, which means $x < 4$, then $f(x-2) = 2 - (x-2) = 2 - x + 2 = 4 - x$. Since we are considering $x \in [0, 4]$, for $x \ge 2$, we have two sub-cases: - If $2 \le x < 4$, then $f(x) = x-2$. Since $0 \le x-2 < 2$, $f(f(x)) = f(x-2) = 2 - (x-2) = 4-x$. - If $x = 4$, then $f(x) = 4-2 = 2$. $f(f(4)) = f(2) = |2-2| = 0$. Also, from $4-x$, we get $4-4=0$. So the formula $4-x$ holds for $x=4$ as well.

Case 2: $x < 2$. In this case, $f(x) = 2 - x$. Now we need to evaluate $f(2-x)$. We need to know if $2-x \ge 2$ or $2-x < 2$. If $2-x \ge 2$, which means $-x \ge 0$, or $x \le 0$. Since we are in the domain $x < 2$, this implies $x \le 0$. - If $x \le 0$, then $f(x) = 2-x$. Since $2-x \ge 2$, $f(f(x)) = f(2-x) = (2-x) - 2 = -x$. If $2-x < 2$, which means $-x < 0$, or $x > 0$. Since we are in the domain $x < 2$, this implies $0 < x < 2$. - If $0 < x < 2$, then $f(x) = 2-x$. Since $0 < 2-x < 2$, $f(f(x)) = f(2-x) = 2 - (2-x) = x$.

Combining these cases for $x \in [0, 4]$: $g(x) = \begin{cases} -x & \text{if } 0 \le x \le 0 \implies x = 0 \\ x & \text{if } 0 < x < 2 \\ 4 - x & \text{if } 2 \le x < 4 \\ 0 & \text{if } x = 4 \end{cases}$ Let's re-evaluate carefully at the boundaries. If $x=0$, $f(0)=|0-2|=2$. $g(0)=f(f(0))=f(2)=|2-2|=0$. Our formula $-x$ gives $-0=0$. If $x \to 0^+$, $f(x) \to 2^-$. $g(x) = f(f(x)) \to f(2^-) = |2^- - 2| = 2^-$. This contradicts $g(x)=x$ which tends to 0. Let's recheck.

Let's re-evaluate $g(x) = f(f(x))$ for $x \in [0, 4]$.

We know $f(x) = \begin{cases} x - 2 & \text{if } x \ge 2 \\ 2 - x & \text{if } x < 2 \end{cases}$.

Consider $x \in [0, 2)$: Here $f(x) = 2 - x$. The range of $f(x)$ for $x \in [0, 2)$ is $(0, 2]$. So, we need to evaluate $f(y)$ where $y = 2-x$ and $y \in (0, 2]$. If $y \in (0, 2)$, which means $2-x \in (0, 2)$, so $0 < 2-x < 2$, which implies $0 < x < 2$. In this case, $f(y) = 2 - y$. So $g(x) = f(2-x) = 2 - (2-x) = x$. If $y = 2$, which means $2-x = 2$, so $x = 0$. In this case, $f(y) = f(2) = |2-2| = 0$. So $g(0) = 0$. Thus, for $x \in [0, 2)$, $g(x) = x$.

Consider $x \in [2, 4]$: Here $f(x) = x - 2$. The range of $f(x)$ for $x \in [2, 4]$ is $[0, 2]$. So, we need to evaluate $f(y)$ where $y = x-2$ and $y \in [0, 2]$. If $y \in [0, 2)$, which means $0 \le x-2 < 2$, so $2 \le x < 4$. In this case, $f(y) = 2 - y$. So $g(x) = f(x-2) = 2 - (x-2) = 4 - x$. If $y = 2$, which means $x-2 = 2$, so $x = 4$. In this case, $f(y) = f(2) = |2-2| = 0$. So $g(4) = 0$. From the formula $4-x$, for $x=4$, we get $4-4=0$. Thus, for $x \in [2, 4]$, $g(x) = 4 - x$.

Combining the piecewise definitions for $g(x)$ on $[0, 4]$: $g(x) = \begin{cases} x & \text{if } 0 \le x < 2 \\ 4 - x & \text{if } 2 \le x \le 4 \end{cases}$

Step 3: Set up the integral for $\int_0^3 (g(x) - f(x)) dx$. We need to evaluate the integral from $0$ to $3$. We will split the integral based on the piecewise definitions of $f(x)$ and $g(x)$. The interval of integration is $[0, 3]$. In $[0, 2)$, $f(x) = 2-x$ and $g(x) = x$. In $[2, 3]$, $f(x) = x-2$ and $g(x) = 4-x$.

So, the integral can be split as: $\int_0^3 (g(x) - f(x)) dx = \int_0^2 (g(x) - f(x)) dx + \int_2^3 (g(x) - f(x)) dx$

Step 4: Evaluate the first part of the integral: $\int_0^2 (g(x) - f(x)) dx$. For $x \in [0, 2)$, $g(x) = x$ and $f(x) = 2 - x$. $g(x) - f(x) = x - (2 - x) = x - 2 + x = 2x - 2$ $\int_0^2 (2x - 2) dx = \left[ \frac{2x^2}{2} - 2x \right]_0^2 = \left[ x^2 - 2x \right]_0^2$ $= (2^2 - 2(2)) - (0^2 - 2(0)) = (4 - 4) - (0) = 0$

Step 5: Evaluate the second part of the integral: $\int_2^3 (g(x) - f(x)) dx$. For $x \in [2, 3]$, $g(x) = 4 - x$ and $f(x) = x - 2$. $g(x) - f(x) = (4 - x) - (x - 2) = 4 - x - x + 2 = 6 - 2x$ $\int_2^3 (6 - 2x) dx = \left[ 6x - \frac{2x^2}{2} \right]_2^3 = \left[ 6x - x^2 \right]_2^3$ $= (6(3) - 3^2) - (6(2) - 2^2) = (18 - 9) - (12 - 4)$ $= 9 - 8 = 1$

Step 6: Combine the results from Step 4 and Step 5. The total integral is the sum of the two parts: $\int_0^3 (g(x) - f(x)) dx = \int_0^2 (g(x) - f(x)) dx + \int_2^3 (g(x) - f(x)) dx$ $= 0 + 1 = 1$

Common Mistakes & Tips

• Incorrectly determining the piecewise definition of $g(x)$: The composition $f(f(x))$ requires careful consideration of the nested absolute values and the different intervals. Always check the range of the inner function to determine which part of the outer function's definition to apply.
• Errors in integration of piecewise functions: Ensure that the integral is split correctly at the points where the definitions of $f(x)$ and $g(x)$ change.
• Algebraic errors in simplifying $g(x) - f(x)$: Double-check the subtraction of the piecewise functions, especially when dealing with negative signs.

Summary

The problem involves evaluating a definite integral of the difference between two composite absolute value functions, $g(x) = f(f(x))$ and $f(x)$, over the interval $[0, 3]$. The solution begins by defining $f(x)$ and $g(x)$ as piecewise functions. This is done by carefully applying the definition of the absolute value function and considering the nested nature of $g(x)$. The interval of integration $[0, 3]$ is then split into sub-intervals $[0, 2)$ and $[2, 3]$ where the piecewise definitions of $f(x)$ and $g(x)$ are consistent. The integral is then evaluated over each sub-interval by substituting the appropriate piecewise expressions and performing standard integration techniques. Finally, the results from each sub-interval are summed to obtain the total value of the definite integral.

The final answer is $\boxed{1}$.