Find $f'(x) = 2\cos(2x) = 0 \Rightarrow 2x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$. Hence find the required area.
Step-by-Step Solution
Key Concept: Use critical points from derivatives to determine bounds, then integrate using trigonometric identities.
Given $f'(x) = 2\cos(2x) = 0$ at $x = \frac{\pi}{4}$. The required area is $\int_{\frac{\pi}{4}}^{\pi} \sin^2(2x)dx = \int_{\frac{\pi}{4}}^{\pi} \frac{1-\cos(4x)}{2}dx = \left[\frac{x}{2} - \frac{\sin(4x)}{8}\right]_{\frac{\pi}{4}}^{\pi} = \frac{3\pi}{8}$.
Correct Answer: 2.3