int_0^{1000} e^{x-[x]} , dx is equal to | vedclass

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(A) The function $f(x) = e^{x-[x]}$ is a periodic function with period $T = 1$, because $[x+1] = [x]+1$. Thus, $f(x+1) = e^{(x+1)-[x+1]} = e^{x+1-[x]-

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$1000(e-1)$

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(A) The function $f(x) = e^{x-[x]}$ is a periodic function with period $T = 1$, because $[x+1] = [x]+1$. Thus, $f(x+1) = e^{(x+1)-[x+1]} = e^{x+1-[x]-1} = e^{x-[x]} = f(x)$. Using the property of definite integrals $\int_0^{nT} f(x) \, dx = n \int_0^T f(x) \, dx$, where $n = 1000$ and $T = 1$: $I = \int_0^{1000} e^{x-[x]} \, dx = 1000 \int_0^1 e^{x-[x]} \, dx$. For $x \in [0, 1)$, $[x] = 0$, so $e^{x-[x]} = e^x$. $I = 1000 \int_0^1 e^x \, dx$. $I = 1000 [e^x]_0^1 = 1000(e^1 - e^0) = 1000(e-1)$.

$\int_0^{1000} e^{x-[x]} \, dx$ is equal to

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