If [.] denotes the greatest integer function, | vedclass

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(B) The function $f(x) = e^{x-[x]}$ is a periodic function with period $T = 1$. We can split the integral over the interval $[0, 1000]$ into $1000$ in

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$1000(e-1)$

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(B) The function $f(x) = e^{x-[x]}$ is a periodic function with period $T = 1$. We can split the integral over the interval $[0, 1000]$ into $1000$ integrals of length $1$: $\int_0^{1000} e^{x-[x]} dx = \sum_{k=0}^{999} \int_k^{k+1} e^{x-[x]} dx$. For $x \in [k, k+1)$,$[x] = k$,so the integrand becomes $e^{x-k}$. Thus,$\int_k^{k+1} e^{x-k} dx = [e^{x-k}]_k^{k+1} = e^{(k+1)-k} - e^{k-k} = e^1 - e^0 = e-1$. Since there are $1000$ such intervals,the total sum is $1000 	imes (e-1)$.

If $[.]$ denotes the greatest integer function,then $\int_0^{1000} e^{x-[x]} dx=$

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