\(\int_{-\pi}^\pi x^2(\sin x) d x=\)

\(\int_{-\pi}^\pi x^2(\sin x) d x=\)
  1. \(\pi^2\)
  2. \(\frac{\pi^2}{2}\)
  3. 0
  4. \(2 \pi^2\)

Solution

\(\int_{-\pi}^\pi x^2(\sin x) d x\) \(x^2(\sin x)\) is an odd function So, \(\int_{-\pi}^\pi x^2(\sin x) d x=0\)

Asked in: AP EAMCET 2020 (17 Sep Shift 1)

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