To solve the problem:

1. Understand the function $f(x)$:

   • For $0 \leq x < 4$, $f(x) = 2x - 8$, which is a linear function with a slope of 2 and y-intercept of -8.
   • For $4 \leq x \leq 8$, $f(x) = 8 - 2x$, which is also a linear function but decreasing (negative slope).

2. Sketch the graph:

   • $f(x) = 2x - 8$: This line passes through $(4, 0)$ and $(0, -8)$.
   • $f(x) = 8 - 2x$: This line passes through $(4, 0)$ and $(8, 0)$.
   • The graph forms two triangular regions: one below the x-axis ($0 \leq x \leq 4$) and one above the x-axis ($4 \leq x \leq 8$).

3. Calculate areas of these regions:

   • Region 1 ($0 \leq x \leq 4$):
     • Base = $4 - 0 = 4$
     • Height = $8$
     • Area = $\frac{1}{2} \times 4 \times 8 = 16$ (below x-axis, so it contributes $-16$ to the integral).
   • Region 2 ($4 \leq x \leq 8$):
     • Base = $8 - 4 = 4$
     • Height = $8$
     • Area = $\frac{1}{2} \times 4 \times 8 = 16$ (above x-axis, so it contributes $+16$ to the integral).

4. Evaluate the integral: $\int_0^8 f(x) \, dx = (-16) + (16) = 0$

Final Answer: $\int_0^8 f(x) \, dx = 0$

Do you want further clarification or detailed visualizations?

Related Questions:

1. How is the concept of signed area related to integrals?
2. Can this approach be generalized for any piecewise function?
3. Why is the integral zero if the positive and negative areas are equal?
4. How does symmetry in the graph simplify integral calculations?
5. What