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$\int_0^{1000} e^{x-[x]}$ is equal to
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Verified Answer
The correct answer is:
$1000(\mathrm{e}-1)$
Hins : $I=1000 \int_0^1 \mathrm{e}^{x-[x]}$
$=1000 \int_0^1 e^x d x=1000\left(e^x\right)_0^1=100(e-1)$
Period of function is 1
$=1000 \int_0^1 e^x d x=1000\left(e^x\right)_0^1=100(e-1)$
Period of function is 1
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