Discussion
The area under one quarter-period of the sine, \( \int_{0}^{\pi/2}\sin x\,dx \), comes to
Answer: D
Integrate and evaluate at the two limits:
\[ \int_{0}^{\pi/2} \sin x\,dx \;=\; \left[-\cos x\right]_{0}^{\pi/2} \;=\; -\cos\frac{\pi}{2} + \cos 0 \;=\; 0 + 1 \;=\; 1. \]
The value two belongs to the integral over the full half-period from zero to \( \pi \), which covers twice the area of this quarter-period
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