# If the integral

Question:

If the integral $\int_{0}^{10} \frac{[\sin 2 \pi x]}{e^{x-[x]}} d x=\alpha e^{-1}+\beta e^{-\frac{1}{2}}+\gamma$,

where $\alpha, \beta, \gamma$ are integers and $[\mathrm{x}]$ denotes the greatest integer less than or equal to $x$, then the value of $\alpha+\beta+\gamma$ is equal to :

1. 0

2. 20

3. 25

4. 10

Correct Option: 1

Solution:

Let $\mathrm{I}=\int_{0}^{10} \frac{[\sin 2 \pi \mathrm{x}]}{\mathrm{e}^{\mathrm{x}-[\mathrm{x}]}} \mathrm{dx}=\int_{0}^{10} \frac{[\sin 2 \pi \mathrm{x}]}{\mathrm{e}^{|\mathrm{x}|}} \mathrm{dx}$

Function $f(x)=\frac{[\sin 2 \pi x]}{e^{[x]}}$

is periodic with

period '1'

Therefore

$I=10 \int_{0}^{1} \frac{[\sin 2 \pi x]}{e^{|x|}} d x$

$=10 \int_{0}^{1} \frac{[\sin 2 \pi x]}{e^{x}} d x$

$=10\left(\int_{0}^{1 / 2} \frac{[\sin 2 \pi x]}{e^{x}} d x+\int_{1 / 2}^{1} \frac{[\sin 2 \pi x]}{e^{x}} d x\right)$

$=10\left(0+\int_{1 / 2}^{1} \frac{(-1)}{\mathrm{e}^{\mathrm{x}}} \mathrm{dx}\right)$

$=-10 \int_{1 / 2}^{1} \mathrm{e}^{-\mathrm{x}} \mathrm{dx}$

$=10\left(\mathrm{e}^{-1}-\mathrm{e}^{-1 / 2}\right)$

Now,

$10 \cdot \mathrm{e}^{-1}-10 \cdot \mathrm{e}^{-1 / 2}=\alpha \mathrm{e}^{-1}+\beta \mathrm{e}^{-1 / 2}+\gamma$ (given)

$\Rightarrow \alpha=10, \beta=-10, \gamma=0$

$\Rightarrow \alpha+\beta+\gamma=0$