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The value of $I=\int_{0}^{1} x\left|x-\frac{1}{2}\right| d x$ is
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$\frac{1}{8}$
Let $\mathrm{I}=\int_{0}^{1} \mathrm{x}\left|\mathrm{x}-\frac{1}{2}\right| \mathrm{dx}$
$=-\int_{0}^{1 / 2} x\left(x-\frac{1}{2}\right) d x+\int_{1 / 2}^{1} x\left(x-\frac{1}{2}\right) d x$
$=\int_{0}^{1 / 2}\left(\frac{x}{2}-x^{2}\right) d x+\int_{1 / 2}^{1} x\left(x^{2}-\frac{x}{2}\right) d x$
$=\left[\frac{x^{2}}{4}-\frac{x^{3}}{3}\right]_{0}^{1 / 2}+\left[\frac{x^{3}}{3}-\frac{x^{2}}{4}\right]_{1 / 2}^{1}$
$=\left(\frac{1}{16}-\frac{1}{24}\right)+\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{24}+\frac{1}{16}\right)$
$=\left(\frac{6-4}{96}\right)+\left(\frac{32-24-4+6}{96}\right)$
$=\frac{12}{96}=\frac{1}{8}$
$=-\int_{0}^{1 / 2} x\left(x-\frac{1}{2}\right) d x+\int_{1 / 2}^{1} x\left(x-\frac{1}{2}\right) d x$
$=\int_{0}^{1 / 2}\left(\frac{x}{2}-x^{2}\right) d x+\int_{1 / 2}^{1} x\left(x^{2}-\frac{x}{2}\right) d x$
$=\left[\frac{x^{2}}{4}-\frac{x^{3}}{3}\right]_{0}^{1 / 2}+\left[\frac{x^{3}}{3}-\frac{x^{2}}{4}\right]_{1 / 2}^{1}$
$=\left(\frac{1}{16}-\frac{1}{24}\right)+\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{24}+\frac{1}{16}\right)$
$=\left(\frac{6-4}{96}\right)+\left(\frac{32-24-4+6}{96}\right)$
$=\frac{12}{96}=\frac{1}{8}$
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