The value of sum_{n=1}^{100} int_{n-1}^{n} e^ | vedclass

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(A) Let $f(x) = e^{x-[x]} = e^{\{x\}},$ where $\{x\}$ is the fractional part of $x.$Since the function $f(x) = e^{\{x\}}$ is periodic with period $1,$

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$100(e-1)$

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(A) Let $f(x) = e^{x-[x]} = e^{\{x\}},$ where $\{x\}$ is the fractional part of $x.$Since the function $f(x) = e^{\{x\}}$ is periodic with period $1,$ we have $\int_{n-1}^{n} e^{\{x\}} dx = \int_{0}^{1} e^{\{x\}} dx$ for any integer $n.$In the interval $[0, 1],$ the fractional part $\{x\} = x.$Therefore,$\int_{0}^{1} e^{\{x\}} dx = \int_{0}^{1} e^x dx = [e^x]_0^1 = e^1 - e^0 = e - 1.$Now,the summation is $\sum_{n=1}^{100} \int_{n-1}^{n} e^{\{x\}} dx = \sum_{n=1}^{100} (e-1) = 100(e-1).$

The value of $\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]} dx,$ where $[x]$ is the greatest integer $\leq x,$ is

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