Question

ConsiderConsider the integral
$I=\underset{0}{\overset{10}{\int }}\frac{[x]e^{[x]}}{e^{x-1}}dx$
where $[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $I$ is equal to:

• A. 9( e -1)
• B. 45( e +1)
• C. 45( e -1)
• D. 9( e +1) the integral

Answer 1

Answer: (C)

$I=\underset{0}{\overset{10}{\int }}[x]\cdot e^{[x]-x+1}$
$I=\underset{0}{\overset{1}{\int }}0dx+\underset{1}{\overset{2}{\int }}1\cdot e^{2-x}+\underset{2}{\overset{3}{\int }}2\cdot e^{3-x}+\dots .+\underset{9}{\overset{10}{\int }}9\cdot e^{10-x}dx$
$\Rightarrow I=\sum _{n=0}^9\underset{n}{\overset{n+1}{\int }}n\cdot e^{n+1-x}dx$
$=-\sum _{n=0}^{9}n{\left(e^{n+1-x}\right)}_n^{n+1}$
$=-\sum _{n=0}^{9}n\cdot \left(e^0-e^1\right)$
$=(e-1)\sum _{n=0}^{9}n$
$=(e-1)\cdot \frac{9\cdot 10}{2}$
$=45(e-1)$