Question

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

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

Answer 1

Answer: (C)

$I =\int\limits_{0}^{10}[ x ] \cdot e ^{[ x ]- x +1}$
$I =\int\limits_{0}^{1} 0 d x +\int\limits_{1}^{2} 1 \cdot e ^{2- x }+\int\limits_{2}^{3} 2 \cdot e ^{3- x }+\ldots .+\int\limits_{9}^{10} 9 \cdot e ^{10- x } dx$
$\Rightarrow I = \displaystyle\sum_{n=0}^{9} \int\limits_{n}^{n+1} n \cdot e ^{ n +1- x } d x$
$=-\displaystyle\sum_{n=0}^{9} n \left( e ^{n+1- x }\right)_{ n }^{ n +1}$
$=-\displaystyle\sum_{n=0}^{9} n \cdot\left( e ^{0}- e ^{1}\right)$
$=( e -1) \displaystyle\sum_{ n =0}^{9} n$
$=( e -1) \cdot \frac{9 \cdot 10}{2}$
$=45( e -1)$