Q66 / 120 · VITEEE 2012 · Mathematics

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

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

Solution

Let $\mathrm{I}=\int_{0}^{1} \mathrm{x}\left|\mathrm{x}-\frac{1}{2}\right| \mathrm{dx}$ $=-\int_{0}^{1 / 2} x\left(x-\frac{1}{2}\right) d x+\int_{1 / 2}^{1} x\left(x-\frac{1}{2}\right) d x$ $=\int_{0}^{1 / 2}\left(\frac{x}{2}-x^{2}\right) d x+\int_{1 / 2}^{1} x\left(x^{2}-\frac{x}{2}\right) d x$ $=\left[\frac{x^{2}}{4}-\frac{x^{3}}{3}\right]_{0}^{1 / 2}+\left[\frac{x^{3}}{3}-\frac{x^{2}}{4}\right]_{1 / 2}^{1}$ $=\left(\frac{1}{16}-\frac{1}{24}\right)+\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{24}+\frac{1}{16}\right)$ $=\left(\frac{6-4}{96}\right)+\left(\frac{32-24-4+6}{96}\right)$ $=\frac{12}{96}=\frac{1}{8}$

Hence, the correct answer is option: 'C'.

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