To evaluate the integral
$\int_{-5}^4 f(x) \, dx$
using the graph, we calculate the areas under the curve and assign signs based on whether the region is above or below the $x$-axis.

The function $f(x)$ consists of piecewise linear segments and a semicircle:

Step 1: Break the regions into parts

1. Region 1: From $x = -5$ to $x = -6$ — a triangle.
2. Region 2: From $x = -6$ to $x = 0$ — a semicircle above the $x$-axis.
3. Region 3: From $x = 0$ to $x = 3$ — a triangle above the $x$-axis.
4. Region 4: From $x = 3$ to $x = 4$ — a triangle below the $x$-axis.

Step 2: Calculate areas

1. Region 1 (Triangle, $x = -5$ to $x = -6$):

   • Base = $1$, Height = $6$.
   • Area of a triangle = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 6 = 3$.
   • Since it's above the $x$-axis, area = $+3$.

2. Region 2 (Semicircle, $x = -6$ to $x = 0$):

   • Radius $r = 3$.
   • Area of a semicircle = $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (3)^2 = \frac{9\pi}{2}$.
   • Since it's above the $x$-axis, area = $+\frac{9\pi}{2}$.

3. Region 3 (Triangle, $x = 0$ to $x = 3$):

   • Base = $3$, Height = $6$.
   • Area = $\frac{1}{2} \times 3 \times 6 = 9$.
   • Since it's above the $x$-axis, area = $+9$.

4. Region 4 (Triangle, $x = 3$ to $x = 4$):

   • Base = $1$, Height = $6$.
   • Area = $\frac{1}{2} \times 1 \times 6 = 3$.
   • Since it's below the $x$-axis, area = $-3$.

Step 3: Add the areas

Now sum up all the areas:
$\text{Total area} = 3 + \frac{9\pi}{2} + 9 - 3.$
Simplify:
$\text{Total area} = \frac{9\pi}{2} + 9.$

Final Answer:

$\int_{-5}^4 f(x) \, dx = \frac{9\pi}{2} + 9.$

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Related Questions:

1. How do you find the area of a semicircle?
2. How do you determine the sign of an area in a definite integral?
3. What are the properties of a piecewise linear function?
4. How do you break an integral into parts geometrically?
5. What is the relationship between definite integrals and areas under curves?

Tip: For definite integrals, areas above the $x$-axis are positive, and areas below are negative!