int_{-pi}^{pi} x^2 sin x , dx = | vedclass

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(C) Let $f(x) = x^2 \sin x$. Check if the function is even or odd by evaluating $f(-x)$: $f(-x) = (-x)^2 \sin(-x) = x^2 (-\sin x) = -x^2 \sin x = -f(x

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(C) Let $f(x) = x^2 \sin x$. Check if the function is even or odd by evaluating $f(-x)$: $f(-x) = (-x)^2 \sin(-x) = x^2 (-\sin x) = -x^2 \sin x = -f(x)$. Since $f(-x) = -f(x)$,the function $f(x) = x^2 \sin x$ is an odd function. According to the property of definite integrals,if $f(x)$ is an odd function,then $\int_{-a}^{a} f(x) \, dx = 0$. Therefore,$\int_{-\pi}^{\pi} x^2 \sin x \, dx = 0$.

$\int_{-\pi}^{\pi} x^2 \sin x \, dx =$

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