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

$\frac{1}{8}$

The explanation for the correct answer.

Evaluate the integral.

$$\begin{gathered} I={\int }_0^{1}x|x-\frac{1}{2}|dx \\ ={\int }_0^{\frac{1}{2}}x\left|x-\frac{1}{2}\right|dx+{\int }_{\frac{1}{2}}^{1}x\left|x-\frac{1}{2}\right|dx \\ =-{\int }_0^{\frac{1}{2}}x\left(x-\frac{1}{2}\right)dx+{\int }_{\frac{1}{2}}^{1}x\left(x-\frac{1}{2}\right)dx \\ =-{\int }_0^{\frac{1}{2}}\left(x^2-\frac{x}{2}\right)dx+{\int }_{\frac{1}{2}}^1\left(x^2-\frac{x}{2}\right)dx \\ =-{\left[\frac{x^3}{3}-\frac{x^2}{4}\right]}_0^{\frac{1}{2}}+{\left[\frac{x^3}{3}-\frac{x^2}{4}\right]}_{\frac{1}{2}}^1 \\ =-\left[\frac{1}{24}-\frac{1}{16}\right]+\left[\frac{1}{12}-\frac{1}{24}+\frac{1}{16}\right] \\ =\frac{1}{12}-\frac{1}{12}+\frac{1}{8} \\ =\frac{1}{8} \end{gathered}$$

Hence, $\mathrm{Option}\left(\mathrm{C}\right)$ is correct .