Question

If the value of the integral $\int\limits_{0}^{5} \frac{x+[x]}{e^{x-[x]}} d x=\alpha e^{-1}+\beta$, 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)

$I=\int\limits_{0}^{5} \frac{x+[x]}{e^{x-[x]}} d x$

$\int\limits_{0}^{1} \frac{t+2}{e^{t}} d t+\int\limits_{0}^{1} \frac{z+4}{e^{z}} d z+\ldots . .+\int\limits_{0}^{1} \frac{y+8}{e^{y}} d x$
$\Rightarrow \int\limits_{0}^{5} \frac{5 x+20}{e^{x}} d t=5 \int\limits_{0}^{1} \frac{x+4}{e^{x}} d x$
$\Rightarrow 5 \int\limits_{0}^{1}(x+4) e^{-x} d x$
$\left.\Rightarrow 5 e ^{- x }(- x -5)\right|_{0} ^{1} $
$\Rightarrow -\frac{30}{ e }+25$
$\alpha=-30$
$\beta=25 $
$\Rightarrow 5 \alpha+6 \beta=0$
$(\alpha+\beta)^{2}=5^{2}=25$