Evaluate \(\int_0^{\frac{\pi}{2}}sin2xdx\)
The correct answer is A. 1
\(\int_0^{\frac{\pi}{2}}\)sin 2x dx = [-1/2cos 2x + C]\(_0^{\frac{\pi}{2}}\)
=[-1/2 cos 2 * π/2 + C] - [-1/2 cos 2 * 0]
= [-1/2 cos π] - [-1/2 cos 0]
= [-1/2x - 1] - [-1/2 * 1]
= 1/2 -(-1/2) = 1/2 + 1/2 = 1
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