# Consider the integral

Question:

Consider the integral

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

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

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

2. (2) $45(\mathrm{e}+1)$

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

4. (4) $9(\mathrm{e}+1)$

Correct Option: , 3

Solution:

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

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

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

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

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

$=-\sum_{\mathrm{n}=0}^{9} \mathrm{n} \cdot\left(\mathrm{e}^{0}-\mathrm{e}^{1}\right)$

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

Leave a comment