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Can you find the answer to the question with the empty box?

To find the solution, let's evaluate the given definite integral:

\[
A = \frac{1}{2} \left[ -6 \left( e^{-\theta/6} \right) \right]_{\pi/2}^{\pi}.
\]

We need to compute the expression by plugging in the limits of integration, \(\theta = \pi\) and \(\theta = \pi/2\), into \(e^{-\theta/6}\).

1. **Evaluate at \(\theta = \pi\):**
   \[
   e^{-\pi/6}.
   \]

2. **Evaluate at \(\theta = \pi/2\):**
   \[
   e^{-\pi/12}.
   \]

Now, the expression becomes:
\[
A = \frac{1}{2} \left[ -6 \left( e^{-\pi/6} - e^{-\pi/12} \right) \right].
\]

Simplifying:
\[
A = -3 \left( e^{-\pi/6} - e^{-\pi/12} \right).
\]

This is the final result for the area, which can be simplified further if needed. Would you like a more detailed explanation or numerical approximation? 

### Relative Questions:
1. How do you evaluate definite integrals using substitution methods?
2. What are the properties of exponential functions in definite integrals?
3. Why is the \(\frac{1}{2}\) factor used when integrating polar equations for area?
4. How does changing the bounds affect the result of a definite integral?
5. What are some common applications of polar integrals in physics and engineering?

### Tip:
When working with definite integrals, carefully substitute the bounds and simplify step by step to avoid errors.

Learn how to solve a problem involving definite integration with exponential functions. This problem demonstrates the use of the fundamental theorem of calculus in the context of polar coordinates. Suitable for high school and introductory college students.

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