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