The value of


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

  1. (1) $100(e-1)$

  2. (2) $100 e$

  3. (3) $100(1-e)$

  4. (4) $100(1+\mathrm{e})$

Correct Option: 1


$\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]} d x$

$=\int_{0}^{1} e^{[x\}} d x+\int_{1}^{2} e^{\{x\}} d x+\int_{2}^{3} e^{\{x\}} d x+\ldots \ldots \int_{99}^{100} e^{\{x\}} d x \quad(\because\{x\}=x-[x])$

$=\left.e^{x}\right|_{0} ^{1}+\left.e^{(x-1)}\right|_{1} ^{2}+\left.e^{(x-2)}\right|_{2} ^{3}+\ldots \ldots+\left.e^{(x-99)}\right|_{99} ^{100}$

$=(e-1)+(e-1)+(e-1)+\ldots \ldots(e-1)$


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