The given problem involves evaluating the limit of a Riemann sum, which corresponds to a definite integral. The integral provided is:

$\int_3^9 \left(4 - \sqrt{9 - (x - 6)^2}\right) \, dx$

Step 1: Recognizing the Integral

The function inside the integral is $4 - \sqrt{9 - (x - 6)^2}$. Notice that $\sqrt{9 - (x - 6)^2}$ represents the upper semicircle of a circle with a radius of $3$, centered at $x = 6$. Therefore:

• The term $9 - (x - 6)^2$ is the equation of a circle of radius $3$.
• The integral evaluates the area between the curve $y = 4$ and this semicircle from $x = 3$ to $x = 9$.

Step 2: Breaking the Integral into Two Parts

We can split the integral into: $\int_3^9 \left(4 - \sqrt{9 - (x - 6)^2}\right) \, dx = \int_3^9 4 \, dx - \int_3^9 \sqrt{9 - (x - 6)^2} \, dx$

First Term: $\int_3^9 4 \, dx$

The integral of a constant is straightforward: $\int_3^9 4 \, dx = 4(x) \Big|_3^9 = 4(9 - 3) = 24$

Second Term: $\int_3^9 \sqrt{9 - (x - 6)^2} \, dx$

The term $\sqrt{9 - (x - 6)^2}$ describes the upper semicircle with radius $3$. The total area of this semicircle is: $\text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi (3)^2 = \frac{9\pi}{2}$

Since the limits of integration ($x = 3$ to $x = 9$) cover the entire semicircle, the integral becomes: $\int_3^9 \sqrt{9 - (x - 6)^2} \, dx = \frac{9\pi}{2}$

Step 3: Combining the Results

Now, subtract the two parts: $\int_3^9 \left(4 - \sqrt{9 - (x - 6)^2}\right) \, dx = 24 - \frac{9\pi}{2}$

Final Answer:

$\boxed{24 - \frac{9\pi}{2}}$

Would you like a detailed breakdown of any part? Here are 5 related questions for further exploration:

1. How do you evaluate the area of a semicircle or circle using integration?
2. Can we generalize this method for integrals involving semicircles with different radii and centers?
3. What happens if the function inside the integral changes to $\sqrt{(x - h)^2 - r^2}$?
4. How would you numerically approximate the integral using Riemann sums?
5. How does symmetry simplify the evaluation of integrals involving circles?

Tip: When dealing with integrals involving circles, always visualize the geometry—it often reveals shortcuts or simplifications!