$\int_0^{\pi/2} \sin x\,dx$ equals:
A$0$
B$1$
C$2$
D$\pi/2$
Answer & Solution
Correct answer: B. $1$
$\int_0^{\pi/2} \sin x\,dx = [-\cos x]_0^{\pi/2} = -\cos(\pi/2) + \cos 0 = 0 + 1 = 1$.
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