int_{-1}^1 x|x| , dx = | vedclass

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(C) The function $f(x) = x|x|$ is an odd function because $f(-x) = (-x)|-x| = -x|x| = -f(x)$.For an odd function,the definite integral over a symmetri

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(C) The function $f(x) = x|x|$ is an odd function because $f(-x) = (-x)|-x| = -x|x| = -f(x)$.For an odd function,the definite integral over a symmetric interval $[-a, a]$ is always zero.Alternatively,we can evaluate it as follows:$\int_{-1}^1 x|x| \, dx = \int_{-1}^0 x(-x) \, dx + \int_0^1 x(x) \, dx$$= -\int_{-1}^0 x^2 \, dx + \int_0^1 x^2 \, dx$$= -\left[\frac{x^3}{3}\right]_{-1}^0 + \left[\frac{x^3}{3}\right]_0^1$$= -\left(0 - \left(-\frac{1}{3}\right)\right) + \left(\frac{1}{3} - 0\right)$$= -\frac{1}{3} + \frac{1}{3} = 0$

$\int_{-1}^1 x|x| \, dx =$

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