A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{8}\]
D) none of these
Correct Answer: C
Solution :
We have, \[I=\int_{0}^{1}{x}\left| x-\frac{1}{2} \right|\,\,dx\] \[=\int_{0}^{1/2}{x}\left( \frac{1}{2}-x \right)dx+\int_{1/2}^{1}{x}\left( x-\frac{1}{2} \right)\,dx\] \[=\int_{0}^{1/2}{\left( \frac{x}{2}-{{x}^{2}} \right)dx}+\int_{1/2}^{1}{\left( {{x}^{2}}-\frac{x}{2} \right)dx}\] \[=\left[ \frac{{{x}^{2}}}{4}-\frac{{{x}^{3}}}{3} \right]_{0}^{1/2}+\left[ \frac{1}{3}-\frac{1}{4}-\left( \frac{1}{24}-\frac{1}{16} \right) \right]\] \[=\frac{2}{16}-\frac{2}{24}+\frac{1}{12}\] \[=\frac{6-4+4}{48}=\frac{6}{48}\] \[=\frac{1}{8}\]
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