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:
Correct Option: , 3
Solution:
$I=\int_{0}^{10}[x] \cdot e^{[x]-x+1}$
$I=\int_{0}^{1} 0 d x+\int_{1}^{2} 1 \cdot e^{2-x}+\int_{2}^{3} 2 \cdot e^{3-x}+\ldots .+\int_{9}^{10} 9 \cdot e^{10-x} d x$
$\Rightarrow \quad I=\sum_{n=0}^{9} \int_{n}^{n+1} n \cdot e^{n+1-x} d x$
$=-\sum_{n=0}^{9} n\left(e^{n+1-x}\right)_{n}^{n+1}$
$=(e-1) \sum_{n=0}^{9} n$
$=(\mathrm{e}-1) \cdot \frac{9 \cdot 10}{2}$
$=45(\mathrm{e}-1)$