Question

The value of I =${\int }_0^{1}x|x-\frac{1}{2}|dx$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{8}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{8}$

Let $I={\int }_0^{1}x|x-\frac{1}{2}|dx$
$=-{\int }_0^{1/2}x(\frac{1}{2}-x)dx+{\int }_{1/2}^{1}x(x-\frac{1}{2})dx$
$={\int }_0^{1/2}(\frac{x}{2}-x^2)dx+{\int }_{1/2}^1(x^2-\frac{x}{2})dx$
$={[\frac{x^2}{4}-\frac{x^3}{3}]}_0^{1/2}+{[\frac{x^3}{3}-\frac{x^2}{4}]}_{1/2}^1$
$=(\frac{1}{16}-\frac{1}{24})+(\frac{1}{3}-\frac{1}{4}-\frac{1}{24}+\frac{1}{16})$
$=\left(\frac{6-4}{96}\right)+\left(\frac{32-24-4+6}{96}\right)$
$=\frac{12}{96}=\frac{1}{8}$.

Hence, option C is correct.