Step-by-Step Solution
Key Concept: Use the periodicity of $|\sin x|$ and properties of absolute value to evaluate the definite integral
The period of the function $|\sin x|$ is $\pi$, so $I = 6\int_0^\pi \sin^2 x \, dx$. We compute $I = 6\int_0^\pi \frac{1 - \cos(3x)}{2} \, dx = \frac{1}{2}\left[3\sin x - \sin(3x) + \frac{\sin x}{3}\right]_0^\pi = \frac{1}{2}\left(3 - (-3) + \left(\frac{1}{4} - \frac{1}{4}\right)\right) = \frac{1}{6}\left(6 - \frac{3}{2}\right) = \frac{3}{2} - \frac{3}{4} = 8$.
Correct Answer: 8