allery$\displaystyle\int_0^\pi x \sin x\, dx = ?$
A$0$
B$\pi$
C$\pi/2$
D$2\pi$
Answer & Solution
Correct answer: B. $\pi$
**Integration by parts:** $u = x$, $dv = \sin x\, dx$.
$[-x\cos x]_0^\pi + \int_0^\pi \cos x\, dx = \pi + [\sin x]_0^\pi = \pi + 0 = \pi$.
(Or use $\int_0^\pi x f(\sin x)\, dx = (\pi/2)\int_0^\pi f(\sin x)\, dx$ rule.)
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