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Question: Answered & Verified by Expert
$\int_{-2}^{1}[x+1] d x=$
(Where $[x]$ is greatest integer function not greater than $x$ )
MathematicsDefinite IntegrationMHT CETMHT CET 2020 (15 Oct Shift 1)
Options:
  • A $1$
  • B $0$
  • C $-1$
  • D $2$
Solution:
1121 Upvotes Verified Answer
The correct answer is: $0$
$\begin{aligned} \int_{-2}^{1}[x+1] d x &=\int_{-2}^{-1}([x]+1) d x+\int_{-1}^{0}([x]+1) d y+\int_{0}^{1}([x]+1) d x \\ &=\int_{-2}^{-1}(-2+1) d x+\int_{-1}^{0}(-1+1) d x+\int_{0}^{1}(0+1) d x \\ &=-[x]_{-2}^{-1}+0+[x]_{0}^{1}=-(-1+2)+0+(1-0) \\ &=0 \end{aligned}$

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