Question

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

• A. 100
• B. 25
• C. 16
• D. 36

Answer 1

Answer: (B)

$I=\underset{0}{\overset{5}{\int }}\frac{x+[x]}{e^{x-[x]}}dx$

$\underset{0}{\overset{1}{\int }}\frac{t+2}{e^t}dt+\underset{0}{\overset{1}{\int }}\frac{z+4}{e^z}dz+\dots ..+\underset{0}{\overset{1}{\int }}\frac{y+8}{e^y}dx$
$\Rightarrow \underset{0}{\overset{5}{\int }}\frac{5x+20}{e^x}dt=5\underset{0}{\overset{1}{\int }}\frac{x+4}{e^x}dx$
$\Rightarrow 5\underset{0}{\overset{1}{\int }}(x+4)e^{-x}dx$
${\Rightarrow 5e^{-x}(-x-5)|}_0^1$
$\Rightarrow -\frac{30}{e}+25$
$\alpha =-30$
$\beta =25$
$\Rightarrow 5\alpha +6\beta =0$
$(\alpha +\beta {)}^2=5^2=25$