int_0^1 x left|x - frac{1}{2}right| dx = | vedclass

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(C) To evaluate the integral $I = \int_0^1 x \left|x - \frac{1}{2}\right| dx$,we split the integral at the point where the expression inside the absol

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$\frac{1}{8}$

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(C) To evaluate the integral $I = \int_0^1 x \left|x - \frac{1}{2}\right| dx$,we split the integral at the point where the expression inside the absolute value changes sign,which is $x = \frac{1}{2}$.For $0 \le x For $\frac{1}{2} \le x \le 1$,$\left|x - \frac{1}{2}\right| = x - \frac{1}{2}$.Thus,$I = \int_0^{1/2} x \left(\frac{1}{2} - x\right) dx + \int_{1/2}^1 x \left(x - \frac{1}{2}\right) dx$.Evaluating the first part:$\int_0^{1/2} \left(\frac{1}{2}x - x^2\right) dx = \left[\frac{x^2}{4} - \frac{x^3}{3}\right]_0^{1/2} = \left(\frac{1/4}{4} - \frac{1/8}{3}\right) = \frac{1}{16} - \frac{1}{24} = \frac{3 - 2}{48} = \frac{1}{48}$.Evaluating the second part:$\int_{1/2}^1 \left(x^2 - \frac{1}{2}x\right) dx = \left[\frac{x^3}{3} - \frac{x^2}{4}\right]_{1/2}^1 = \left(\frac{1}{3} - \frac{1}{4}\right) - \left(\frac{1/8}{3} - \frac{1/4}{4}\right) = \frac{1}{12} - \left(\frac{1}{24} - \frac{1}{16}\right) = \frac{1}{12} - \left(\frac{2 - 3}{48}\right) = \frac{1}{12} + \frac{1}{48} = \frac{4 + 1}{48} = \frac{5}{48}$.Adding both parts:$I = \frac{1}{48} + \frac{5}{48} = \frac{6}{48} = \frac{1}{8}$.

$\int_0^1 x \left|x - \frac{1}{2}\right| dx = $

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