Question

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

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

Answer 1

The correct option is A $45(e-1)$

$I=\underset{0}{\overset{10}{\int }}\left[x\right].e^{\left[x\right]+1-x}dx$
$=\underset{1}{\overset{2}{\int }}{e}^{2-x}dx+\underset{2}{\overset{3}{\int }}2\cdot e^{3-x}dx+\underset{3}{\overset{4}{\int }}3\cdot e^{4-x}dx+\cdots +\underset{9}{\overset{10}{\int }}9\cdot e^{10-x}dx$
$=-\left\{\right(1-e)+2(1-e)+(1-e)+\cdots +9(1-e\left)\right\}$
$=45(e-1)$