MCQ
$\int_0^\pi x \sin x d x=$
  • $\pi$
  • B
    $0$
  • C
    1
  • D
    $\pi^2$

Answer

Correct option: A.
$\pi$
(A)
Let $I =\int_0^\pi x \sin x d x$ … (i)
$\therefore \quad I =\int_0^\pi(\pi-x) \sin (\pi-x) d x$
$\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$
$\therefore \quad I=\int_0^\pi(\pi-x) \sin x d x$ ...(ii)
Adding (i) and (ii), we get
$2 I =\pi \int_0^\pi \sin x d x=\pi[-\cos x]_0^\pi=2 \pi$
$\Rightarrow I =\pi$

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