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$\int_{0}^{100} e^{x-[x]} d x$ is equal to
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Verified Answer
The correct answer is:
$100(e-1)$
Let $I=\int_{0}^{100} e^{x-[x]} d x$
$=100 \int_{0}^{1} e^{x-[x]} d x$
$[\because x-[x]$ is a periodic function of period 1 and $\int_{0}^{mT} f(x) d x=m \int_{0}^{T} f(x) d x,$ where $T$ is period of $f(x)]$ $=100 \int_{0}^{1} e^{x} d x[\because x-[x]=x$ for $0 < x < 1]$
$=100\left[e^{x}\right]_{0}^{1}$
$=100\left[e^{1}-e^{0}\right]$
$=100(e-1)$
$=100 \int_{0}^{1} e^{x-[x]} d x$
$[\because x-[x]$ is a periodic function of period 1 and $\int_{0}^{mT} f(x) d x=m \int_{0}^{T} f(x) d x,$ where $T$ is period of $f(x)]$ $=100 \int_{0}^{1} e^{x} d x[\because x-[x]=x$ for $0 < x < 1]$
$=100\left[e^{x}\right]_{0}^{1}$
$=100\left[e^{1}-e^{0}\right]$
$=100(e-1)$
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