Correct Answer: Option A

Explanation

\(\int_0 ^{\frac{\pi}{2}}(-2cos x)dx = [-2sin x + c]_0 ^{\frac{\pi}{2}}\)
=(-2sin\(\frac{\pi}{2}+c+2sin0-c)\)
=-2sin90+c+2sin0-c
=-2(1)+2(0)
=-2

There is an explanation video available below.

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