The problem asks to evaluate the integral $\int_{0}^{9} f(x) \, dx$, and provides a graph of the function $f(x)$ over the interval from $x = 0$ to $x = 9$.

Step 1: Analyze the Graph

The graph shows the function $f(x)$ over the interval from $x = 0$ to $x = 9$, divided into regions with distinct geometric shapes, specifically:

1. A triangle from $x = 0$ to $x = 3$.
2. A semicircle from $x = 3$ to $x = 6$.
3. Another triangle from $x = 6$ to $x = 9$.

Step 2: Calculate the Area of Each Region

Since the integral represents the area under the curve:

1. Triangle from $x = 0$ to $x = 3$:

• Base = 3 units
• Height = 2 units
• Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 2 = 3$

2. Semicircle from $x = 3$ to $x = 6$:

• Radius $r = 1$ (as the height of the semicircle is 2 units)
• Area = $\frac{1}{2} \times \pi r^2 = \frac{1}{2} \times \pi \times 1^2 = \frac{\pi}{2}$

3. Triangle from $x = 6$ to $x = 9$ (Below the x-axis, so area is negative):

• Base = 3 units
• Height = 1 unit
• Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 1 = 1.5$
• Since it is below the x-axis, this area is $-1.5$.

Step 3: Sum the Areas

The definite integral is the sum of these areas:

$\int_{0}^{9} f(x) \, dx = \text{Area 1} + \text{Area 2} + \text{Area 3} = 3 + \frac{\pi}{2} - 1.5 = 1.5 + \frac{\pi}{2}$

Final Answer

$\int_{0}^{9} f(x) \, dx = 1.5 + \frac{\pi}{2}$

Would you like more details or have any questions?

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Here are some related questions you might find helpful:

1. How do you calculate the area under a curve for non-geometric shapes?
2. Why is the area below the x-axis considered negative in integrals?
3. What are the properties of definite integrals?
4. How do integrals relate to the concept of accumulated area?
5. Can you explain the difference between definite and indefinite integrals?
6. How do different shapes affect the integration process?
7. What are some common mistakes to avoid when calculating areas under a curve?
8. How does the Fundamental Theorem of Calculus connect differentiation and integration?

Tip: When dealing with definite integrals, it's useful to break down the area into known geometric shapes, calculate their areas, and sum them according to their sign (positive above the x-axis, negative below).