Q74 / 120 · VITEEE 2018 · Mathematics

The value of $\int_{-1}^{1}(\mathrm{x}-[\mathrm{x}]) \mathrm{d} \mathrm{x}$ (where [.] denotes greatest integer function) is

  • A
    0
  • Correct answer
    B
    1
  • C
    2
  • D
    None of these

Solution

$\begin{aligned} \mathrm{I} &=\int_{-1}^{1}(\mathrm{x}-[\mathrm{x}]) \mathrm{d} \mathrm{x}=\int_{-1}^{1} \mathrm{x} \mathrm{dx}-\int_{-1}^{1}[\mathrm{x}] \mathrm{dx} \\ &=\left[\frac{\mathrm{x}^{2}}{2}\right]_{-1}^{1}-\left[\int_{-1}^{0}[\mathrm{x}] \mathrm{d} \mathrm{x}+\int_{0}^{1}[\mathrm{x}] \mathrm{d} \mathrm{x}\right] \\ &=\frac{1}{2}[1-1]-\left[\int_{-1}^{0}(-1) \mathrm{d} \mathrm{x}+\int_{0}^{1} 0 . \mathrm{dx}\right] \\ &=\left[\begin{array}{c}\text { If }-1 \leq \mathrm{x} < 0,[\mathrm{x}]=-1] \\ \text { If } 0 \leq \mathrm{x} < 1,[\mathrm{x}]=0]\end{array}\right.\\ &=[-\mathrm{x}]_{-1}^{0}-0=0-[-0-(-1)]=1 \end{aligned}$

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

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