allery$\int_{\pi/6}^{\pi/3} \cos x\,dx$ equals:
A$(\sqrt 3 - 1)/2$
B$(\sqrt 3 + 1)/2$
C$1$
D$0$
Answer & Solution
Correct answer: A. $(\sqrt 3 - 1)/2$
$\int \cos x\,dx = \sin x$, so $[\sin x]_{\pi/6}^{\pi/3} = \sin(\pi/3) - \sin(\pi/6) = \sqrt 3/2 - 1/2 = (\sqrt 3 - 1)/2$.
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