To calculate the integral $\int_0^2 f(x) \, dx$ from the given graph, we need to evaluate the area between the curve and the x-axis from $x = 0$ to $x = 2$.

Step-by-Step:

1. Analyze the graph:

   • From $x = 0$ to $x = 2$, the function appears to be linear and sloping downward.

2. Equation of the line: We can find the slope ($m$) and y-intercept ($b$) for the segment between $x = 0$ and $x = 2$ using the endpoints:

   • Endpoint 1: $(0, 2)$
   • Endpoint 2: $(2, -2)$

   The slope $m$ is: $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 2}{2 - 0} = -2$

   So, the equation of the line is: $f(x) = -2x + 2$

3. Set up the integral: The definite integral is: $\int_0^2 (-2x + 2) \, dx$

4. Compute the integral: The antiderivative of $-2x + 2$ is: $\int (-2x + 2) \, dx = -x^2 + 2x + C$

   Evaluate at the bounds: [ \left[-x^2 + 2x \right]_0^2 = \left[ -(2)^2 + 2(2) \right] - \left[ -(0)^2 + 2(0) \right] ]

   Simplify: [ \left[ -4 + 4 \right] - [0] = 0 ]

Final Answer:

$\int_0^2 f(x) \, dx = 0$

This means the positive and negative areas cancel each other out over this interval.

Let me know if you'd like further details or additional examples!

Here are some related questions for practice:

1. How do we compute integrals when the curve crosses the x-axis?
2. What is the total area (ignoring signs) from $x = 0$ to $x = 2$?
3. How would the integral change if we computed $\int_{-2}^4 f(x) \, dx$?
4. What happens to the integral when a function is completely above or below the x-axis?
5. How can we verify the calculation using numerical methods or approximation?

Tip: Always visualize the graph of a function before integrating to identify regions of positive and negative area!