\(\int_{-\pi}^\pi x^2(\sin x) d x=\)
\(\int_{-\pi}^\pi x^2(\sin x) d x=\)
- \(\pi^2\)
- \(\frac{\pi^2}{2}\)
- 0
- \(2 \pi^2\)
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)
Practice more Definite Integration questions on Aicharya