Consider the integral
Question:

Consider the integral

$I=\int_{0}^{10} \frac{[x] e^{[x]}}{e^{x-1}} d x$

where $[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $I$ is equal to:

  1. $9(\mathrm{e}-1)$

  2. $45(e+1)$

  3. $45(\mathrm{e}-1)$

  4. $9(e+1)$


Correct Option: , 3

Solution:

$I=\int_{0}^{10}[x] \cdot e^{[x]-x+1}$

$I=\int_{0}^{1} 0 d x+\int_{1}^{2} 1 \cdot e^{2-x}+\int_{2}^{3} 2 \cdot e^{3-x}+\ldots .+\int_{9}^{10} 9 \cdot e^{10-x} d x$

$\Rightarrow \quad I=\sum_{n=0}^{9} \int_{n}^{n+1} n \cdot e^{n+1-x} d x$

$=-\sum_{n=0}^{9} n\left(e^{n+1-x}\right)_{n}^{n+1}$

$=(e-1) \sum_{n=0}^{9} n$

$=(\mathrm{e}-1) \cdot \frac{9 \cdot 10}{2}$

$=45(\mathrm{e}-1)$

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