Consider $A = \int_0^{\pi/2} \frac{\sin(2x)}{1}dx$, then
Step-by-Step Solution
Key Concept: Using the inequality $\sin\theta < \theta$ to establish upper bounds for integrals involving trigonometric functions
Since $\sin\theta < \theta$ for all $\theta \in (0, \frac{\pi}{2})$, we have $\sin(2x) < 2x$. Therefore $A = 2\int_0^{\pi/4} \frac{\sin(2x)}{x} dx < 2\int_0^{\pi/4} \frac{2x}{x} dx = 2\int_0^{\pi/4} 2\,dx = 2 \cdot 2 \cdot \frac{\pi}{4} = \pi$. Thus $A < 2(\frac{\pi}{2}) = \pi$.
Correct Answer: C