int_{0}^{1000} e^{x - [x]} , dx is equal to: | vedclass

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(C) The function $f(x) = e^{x - [x]}$ is a periodic function with period $T = 1$,where $[x]$ denotes the greatest integer function.Using the property

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$1000(e - 1)$

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(C) The function $f(x) = e^{x - [x]}$ is a periodic function with period $T = 1$,where $[x]$ denotes the greatest integer function.Using the property of definite integrals for periodic functions,$\int_{0}^{nT} f(x) \, dx = n \int_{0}^{T} f(x) \, dx$,we have:$\int_{0}^{1000} e^{x - [x]} \, dx = 1000 \int_{0}^{1} e^{x - [x]} \, dx$.For $0 Substituting this into the integral:$1000 \int_{0}^{1} e^x \, dx = 1000 [e^x]_{0}^{1}$.Evaluating the definite integral:$1000 (e^1 - e^0) = 1000 (e - 1)$.

$t$ (in second)	$0$	$2$	$4$	$6$	$8$	$10$
$v$ (in m/s)	$0$	$12$	$16$	$20$	$35$	$60$

$\int_{0}^{1000} e^{x - [x]} \, dx$ is equal to:

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