If the value of the integral


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

  1. 100

  2. 25

  3. 16

  4. 36

Correct Option: , 2


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

$\int_{0}^{1} \frac{t+2}{e^{t}} d t+\int_{0}^{1} \frac{z+4}{e^{z}} d z+\ldots \ldots+\int_{0}^{1} \frac{y+8}{e^{y}} d x$

$\Rightarrow \int_{0}^{5} \frac{5 x+20}{e^{x}} d t=5 \int_{0}^{1} \frac{x+4}{e^{x}} d x$

$\Rightarrow 5 \int_{0}^{1}(x+4) e^{-x} d x$

$\left.\Rightarrow 5 e^{-x}(-x-5)\right|_{0} ^{1} \Rightarrow-\frac{30}{e}+25$


$\beta=25 \Rightarrow 5 \alpha+6 \beta=0$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now