The image shows a piecewise linear graph of the function $H(x)$ and asks to evaluate the definite integral:

$\int_{-8}^{8} H(x) \, dx$

This integral represents the net area between the graph of $H(x)$ and the x-axis from $x = -8$ to $x = 8$. Let's analyze the graph:

1. For $-10 \leq x \leq -5$, $H(x)$ is constant at $H(x) = 5$. The area of this part is the rectangle formed by height $5$ and width $3$ (since it extends from $x = -8$ to $x = -5$): $\text{Area}_1 = 5 \times 3 = 15$

2. For $-5 \leq x \leq 0$, $H(x)$ is a straight line that decreases linearly from 5 to -5, forming a triangle. The base of this triangle is 5 units, and the height is 5 units: $\text{Area}_2 = \frac{1}{2} \times 5 \times 5 = 12.5$ Since this triangle is above the x-axis, the area is positive.

3. For $0 \leq x \leq 5$, $H(x)$ increases linearly from -5 to 5, forming another triangle. The base and height are both 5 units, but this triangle is below the x-axis, so its area is negative: $\text{Area}_3 = -\frac{1}{2} \times 5 \times 5 = -12.5$

4. For $5 \leq x \leq 8$, $H(x)$ is constant at $H(x) = 5$. The area is another rectangle with a height of 5 and a width of 3: $\text{Area}_4 = 5 \times 3 = 15$

Now, summing these areas:

$\text{Total Area} = 15 + 12.5 - 12.5 + 15 = 30$

Thus, the value of the definite integral is:

$\int_{-8}^{8} H(x) \, dx = 30$

Would you like more details or have any questions?

Here are 5 follow-up questions:

1. How do you calculate the area of a triangle?
2. Why is the area below the x-axis considered negative?
3. What is the physical significance of a definite integral?
4. How would the result change if the function had a curve instead of straight lines?
5. What would the integral represent if it extended from $x = -10$ to $x = 10$?

Tip: Always keep track of whether a region is above or below the x-axis when calculating definite integrals, as it determines the sign of the area.