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.