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$\int_{-2}^2|[x]| d x$ is equal to
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Verified Answer
The correct answer is:
4
We have
$\int_{-2}^2|[x]| d x=\int_{-2}^{-1}|[x]| d x +\int_{-1}^0|[x]| d x +\int_0^1|[x]| d x+\int_1^2[[x] \mid d x$
$=\int_{-2}^{-1}|-2| d x+\int_{-1}^0|-1| d x$ $+\int_0^1|0| d x+\int_1^2|1| d x$
$=2[x]_{-2}^{-1}+[x]_{-1}^0+0+[x]_1^2$
$=2[-1+2]+[0+1]+[2-1]$
$=2+1+1=4$
$\int_{-2}^2|[x]| d x=\int_{-2}^{-1}|[x]| d x +\int_{-1}^0|[x]| d x +\int_0^1|[x]| d x+\int_1^2[[x] \mid d x$
$=\int_{-2}^{-1}|-2| d x+\int_{-1}^0|-1| d x$ $+\int_0^1|0| d x+\int_1^2|1| d x$
$=2[x]_{-2}^{-1}+[x]_{-1}^0+0+[x]_1^2$
$=2[-1+2]+[0+1]+[2-1]$
$=2+1+1=4$
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