To evaluate the integral $\int_{-3}^{3} f(x) \, dx$ based on the given graph, we need to find the area under the curve from $x = -3$ to $x = 3$.

1. Split the integral into parts based on the curve's sections:

   • From $x = -3$ to $x = 0$, the graph is above the $x$-axis.
   • From $x = 0$ to $x = 3$, the graph is below the $x$-axis.

2. Calculate the area for each section separately:

   • The area from $x = -3$ to $x = 0$ appears as a right triangle with a base of 3 units and a height of 3 units. The area of this triangle is: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 3 = 4.5$
   • The area from $x = 0$ to $x = 3$ is also a triangular region but below the $x$-axis, giving it a negative sign. It has the same base and height as the first triangle, so its area is: $\text{Area} = -\frac{1}{2} \times 3 \times 3 = -4.5$

3. Combine the areas:

   • The total area from $x = -3$ to $x = 3$ is the sum of these two areas: $\int_{-3}^{3} f(x) \, dx = 4.5 + (-4.5) = 0$

Final Answer:

$\int_{-3}^{3} f(x) \, dx = 0$

Would you like further details or have any questions about this solution?

Here are some related questions for further exploration:

1. How would the result change if only the area from $x = -3$ to $x = 0$ was considered?
2. What is the significance of a zero result in an integral over a symmetric interval?
3. How would you compute the integral if the curve above and below the $x$-axis were not symmetric?
4. Can you use numerical integration methods to approximate the area if the shape was irregular?
5. How does the definite integral relate to the concept of net area versus total area?

Tip: Always check symmetry in graphs when computing definite integrals over intervals centered at zero, as it can simplify calculations.