To compute the integral $\int_{-1}^{9} f(x) \, dx$, we will analyze the area under the curve given in the graph. Here's the step-by-step solution:

Step 1: Break Down the Graph

The graph consists of:

1. A triangle from $x = -1$ to $x = 1$.
2. A semicircle from $x = 1$ to $x = 9$.

The integral can be computed as the sum of these two areas: $\int_{-1}^{9} f(x) \, dx = \text{Area of Triangle} + \text{Area of Semicircle}.$

Step 2: Compute the Area of the Triangle

The triangle's vertices are at:

• $(-1, 0)$,
• $(1, 1)$,
• $(1, 0)$.

The base of the triangle is from $-1$ to $1$, so its length is $2$, and the height is $1$. The area of a triangle is: $\text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 1 = 1.$

Step 3: Compute the Area of the Semicircle

The semicircle has:

• Radius $r = 4$ (since the diameter is from $x = 1$ to $x = 9$).
• The area of a semicircle is half the area of a circle: $\text{Area of Semicircle} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (4)^2 = 8\pi.$

Step 4: Combine the Results

Adding the two areas: $\int_{-1}^{9} f(x) \, dx = \text{Area of Triangle} + \text{Area of Semicircle} = 1 + 8\pi.$

Final Answer:

$\int_{-1}^{9} f(x) \, dx = 1 + 8\pi.$

Let me know if you need more explanation or related help! Here are some follow-up questions to expand understanding:

1. How do you compute the area of a quarter-circle?
2. What happens if the semicircle is below the $x$-axis?
3. How would you compute the integral if the triangle slanted downward?
4. Can you apply symmetry to simplify the calculation of the semicircle's area?
5. What are the units of this integral if the graph represents a physical scenario?

Tip: Always sketch a graph or label points to ensure accuracy when working with geometry-based integrals.