To solve this problem, we need to evaluate the definite integral $\int_{-3}^{3} f(x) \, dx$ by interpreting the graph geometrically, as it represents the area under the curve $y = f(x)$ from $x = -3$ to $x = 3$.

Step-by-Step Solution:

1. Identify Areas Under the Curve:

   • From $x = -3$ to $x = 0$, the curve is below the $x$-axis, which represents a negative area.
   • From $x = 0$ to $x = 3$, the curve is above the $x$-axis, representing a positive area.

2. Calculate Each Region Separately:

   • Region 1 (from $x = -3$ to $x = 0$):

     • This region forms a triangle with a base of 3 units (from $-3$ to $0$) and a height of 3 units.
     • The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$.
     • So, the area for this triangle is $-\frac{1}{2} \times 3 \times 3 = -\frac{9}{2}$.

   • Region 2 (from $x = 0$ to $x = 3$):

     • This region also forms a triangle with a base of 3 units (from $0$ to $3$) and a height of 3 units.
     • The area of this triangle is $\frac{1}{2} \times 3 \times 3 = \frac{9}{2}$.

3. Add the Areas Together:

   • The total area under $f(x)$ from $-3$ to $3$ is $-\frac{9}{2} + \frac{9}{2} = 0$.

Final Answer:

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

Would you like a detailed breakdown of each step or have any questions on this?

Related Questions:

1. What does it mean for the integral of a function to be zero over a symmetric interval?
2. How can we interpret the integral geometrically when a function has regions both above and below the $x$-axis?
3. Why does the area below the $x$-axis count as negative in an integral?
4. How would the result change if the curve was only below or only above the $x$-axis?
5. What are some applications of calculating areas using integrals in real-world scenarios?

Tip:

Remember, when evaluating integrals with symmetry, it can often simplify calculations, especially when areas cancel out due to equal regions above and below the $x$-axis.