To evaluate the integral $\int_{-4}^{5} f(x) \, dx$ using geometry, we need to find the area under the curve of $f(x)$ from $x = -4$ to $x = 5$. We'll do this by breaking down the graph into geometric shapes, such as triangles and semi-circles, and calculating the area of each.

Step-by-Step Solution:

1. Identify the Shapes:

   • From $x = -4$ to $x = -2$, there is a half-circle below the $x$-axis.
   • From $x = -2$ to $x = 2$, there is a curve below the $x$-axis that resembles a semi-circle (but appears to be a different radius).
   • From $x = 2$ to $x = 5$, there is a triangle above the $x$-axis.

2. Calculate the Areas:

   • Area of the Half-Circle (from $x = -4$ to $x = -2$): The radius of the half-circle is 2. The formula for the area of a circle is $A = \pi r^2$, so a half-circle would be: $A = \frac{1}{2} \pi (2^2) = 2\pi.$ Since the half-circle is below the $x$-axis, its contribution to the integral is $-2\pi$.

   • Area of the Curve (from $x = -2$ to $x = 2$): This curve looks like a larger half-circle with a radius of 3. Thus: $A = \frac{1}{2} \pi (3^2) = \frac{9\pi}{2}.$ Since it is also below the $x$-axis, the contribution is $-\frac{9\pi}{2}$.

   • Area of the Triangle (from $x = 2$ to $x = 5$): The triangle has a base of 3 (from $x = 2$ to $x = 5$) and a height of 3 (up to $y = 3$). The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$: $A = \frac{1}{2} \times 3 \times 3 = 4.5.$

3. Combine the Areas: $\text{Total Area} = -2\pi - \frac{9\pi}{2} + 4.5.$

   We need to evaluate this numerically, with $\pi \approx 3.14159$: $-2\pi = -6.28318, \quad -\frac{9\pi}{2} = -14.13717, \quad 4.5.$ $\text{Total} = -6.28318 - 14.13717 + 4.5 \approx -15.92035.$

4. Rounding: The final result, rounded to the nearest thousandth, is: $\boxed{-15.920}$

Would you like more details or explanations on any of the steps?

Related Questions:

1. How do you determine the sign of the area when integrating using geometry?
2. What is the general method for calculating the area under a curve if it's not a simple geometric shape?
3. Can integrals represent the area above the curve if the function is negative?
4. How would the solution change if the curve was not symmetric about the $x$-axis?
5. What other methods can you use to approximate the area under a complex curve?

Tip:

When calculating areas using geometry, always pay attention to whether the sections lie above or below the $x$-axis to assign the correct sign to each part.