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$\int_{-2}^{2.24}[x] d x=$
where $[x]$ is the greatest integer function
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where $[x]$ is the greatest integer function
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Verified Answer
The correct answer is:
$-2$
(D)
$\begin{aligned} \int_{-2}^{2}[\mathrm{x}] \mathrm{dx} &=\int_{-2}^{-1}-2 \mathrm{dx}+\int_{-1}^{0}-1 \mathrm{dx}+\int_{0}^{1} 0 \mathrm{dx}+\int_{1}^{2} 1 \mathrm{dx} \\ &=-2[\mathrm{x}]_{-2}^{-1}+(-1)[\mathrm{x}]_{-1}^{0}+1[\mathrm{x}]_{1}^{2} \\ &=-2[-1-(-2)]-1[-(-1)]+1[2-1] \\ &=-2[1]-1[1]+1[1]=-2-1+1=-2 \end{aligned}$
$\begin{aligned} \int_{-2}^{2}[\mathrm{x}] \mathrm{dx} &=\int_{-2}^{-1}-2 \mathrm{dx}+\int_{-1}^{0}-1 \mathrm{dx}+\int_{0}^{1} 0 \mathrm{dx}+\int_{1}^{2} 1 \mathrm{dx} \\ &=-2[\mathrm{x}]_{-2}^{-1}+(-1)[\mathrm{x}]_{-1}^{0}+1[\mathrm{x}]_{1}^{2} \\ &=-2[-1-(-2)]-1[-(-1)]+1[2-1] \\ &=-2[1]-1[1]+1[1]=-2-1+1=-2 \end{aligned}$
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