The value of I = int_{0}^{1} x left| x - frac | vedclass

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(C) Given $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$. Since the integrand involves an absolute value,we split the integral at $x = 1/2$: $

Vedclass · Accepted Answer

$1/8$

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(C) Given $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$. Since the integrand involves an absolute value,we split the integral at $x = 1/2$: $I = \int_{0}^{1/2} x \left( -\left( x - \frac{1}{2} \right) \right) dx + \int_{1/2}^{1} x \left( x - \frac{1}{2} \right) dx$. $I = \int_{0}^{1/2} \left( \frac{1}{2}x - x^2 \right) dx + \int_{1/2}^{1} \left( x^2 - \frac{1}{2}x \right) dx$. Evaluating the integrals: $I = \left[ \frac{x^2}{4} - \frac{x^3}{3} \right]_{0}^{1/2} + \left[ \frac{x^3}{3} - \frac{x^2}{4} \right]_{1/2}^{1}$. $I = \left( \frac{(1/2)^2}{4} - \frac{(1/2)^3}{3} \right) + \left( \left( \frac{1}{3} - \frac{1}{4} \right) - \left( \frac{(1/2)^3}{3} - \frac{(1/2)^2}{4} \right) \right)$. $I = \left( \frac{1}{16} - \frac{1}{24} \right) + \left( \frac{1}{12} - \left( \frac{1}{24} - \frac{1}{16} \right) \right)$. $I = \left( \frac{3-2}{48} \right) + \left( \frac{1}{12} - \left( \frac{2-3}{48} \right) \right) = \frac{1}{48} + \frac{1}{12} + \frac{1}{48} = \frac{2}{48} + \frac{4}{48} = \frac{6}{48} = \frac{1}{8}$.

The value of $I = \int_{0}^{1} x \left| x - \frac{1}{2} \right| dx$ is

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