Q229 / 270 · NDA 2016 (Phase 1) · Mathematics

What is $\int_{-2}^2 x d x-\int_{-2}^2[x] d x$ equal to, where $[\cdot]$ is the greatest integer function?

  • A
    0
  • B
    1
  • Correct answer
    C
    2
  • D
    4

Solution

$\begin{aligned} & \int_{-2}^2 x d x-\int_{-2}^2[x] d x \\ & =\left[\frac{x^2}{2}\right]_{-2}^2-\int_{-2}^{-1}[x] d x-\int_{-1}^0[x] d x-\int_0^1[x] d x-\int_1^2[x] d x \\ & =\frac{1}{2}[4-4]-(-2)-(-1)-0-(1) \\ & =2+1-1=2\end{aligned}$

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

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