Q88 / 150 · MHT CET 2022 (06 Aug Shift 2) · Mathematics

$\int_0^2[x] d x+\int_0^2|x-1| d x=$ (where $[x]$ denotes the greatest integer function.)

  • A
    3
  • B
    4
  • C
    1
  • Correct answer
    D
    2

Solution

$\begin{aligned} & \int_0^2[x] d x+\int_0^2[x-1] d x \\ & =\int_0^1 0 . d x+\int_1^2 1 . d x+\int_0^1(1-x) d x+\int_1^2(x-1) d x \\ & =[0]_0^1+[x]_1^2+\left[x-\frac{x^2}{2}\right]_0^1+\left[\frac{x^2}{2}-x\right]_1^2 \\ & =0+1+\frac{1}{2}+\frac{1}{2}=2\end{aligned}$

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

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