Question Detail

To evaluate ∫x sin x dx, we use integration by parts, ∫u dv = uv - ∫v du.Let u = x, so du = dx.Let dv = sin x dx, so v = ∫sin x dx = -cos x.∫x sin x dx = x(-cos x) - ∫(-cos x) dx= -x cos x + ∫cos x dx= -x cos x + sin xNow, evaluate the definite integral from 0 to π:[-x cos x + sin x]0π = (-π cos π + sin π) - (-0 cos 0 + sin 0)= (-π(-1) + 0) - (0 + 0)= πThis matches option B.