Question

If the value of the integral $\underset{0}{\overset{5}{\int }}\frac{x+\left[x\right]}{e^{x-\left[x\right]}}dx=\alpha e^{-1}+\beta ,$ where $\alpha ,\beta \in R,5\alpha +6\beta =0,$ and $\left[x\right]$ denotes the greatest integer less than or equal to $x;$ then the value of $(\alpha +\beta {)}^2$ is equal to:

• A. $25$
• B. $16$
• C. $36$
• D. $100$

Answer 1

The correct option is A $25$

$I=\underset{0}{\overset{5}{\int }}\frac{x+\left[x\right]}{e^{x-\left[x\right]}}dx$
$I=\underset{0}{\overset{1}{\int }}\frac{x+0}{e^{x-0}}dx+\underset{1}{\overset{2}{\int }}\frac{x+1}{e^{x-1}}dx+\cdots +\underset{4}{\overset{5}{\int }}\frac{x+4}{e^{x-4}}dx$
$\therefore I=\sum _{k=0}^4\underset{k}{\overset{k+1}{\int }}\frac{x+k}{e^{x-k}}dx$
$=\sum _{k=0}^4{e}^k\underset{k}{\overset{k+1}{\int }}(x+k)e^{-x}dx$
$=\sum _{k=0}^4{e}^k|-(x+k)e^{-x}-e^{-x}{|}_k^{k+1}$
$=\sum _{k=0}^4{e}^k(-(2k+1)e^{-k+1}-e^{-k+1}+2k{e}^{-k}+e^{-k})$
$=\sum _{k=0}^4\left(\right(-2k-1)e^{-1}-e^{-1}+(2k+1\left)\right)$
$\Rightarrow -25e^{-1}-5e^{-1}+25=-30e^{-1}+25$
$\Rightarrow \alpha =-30\text{and}\beta =25$
$\Rightarrow (\alpha +\beta {)}^2=25$