Let f : R

$f(x)=e^{-x} \sin x$

Now, $\mathrm{F}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt} \quad \Rightarrow \mathrm{F}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x})$

$=2 \int_{0}^{1} f(x) \cdot e^{x} d x=2 \int_{0}^{1} e^{-x} \sin x \cdot e^{x} d x$

$=2 \int_{0}^{1} \sin x d x$

$=2(1-\cos 1)$

$I=2\left\{1-\left(1-\frac{1}{2}+\frac{1}{\lfloor 4}+\frac{1}{\lfloor 6}+\frac{1}{\lfloor 8} \ldots \ldots . . .\right)\right\}$

$I=1-\frac{2}{4}+\frac{2}{16}-\frac{2}{19}+\ldots \ldots$

$1-\frac{2}{\lfloor 4}

$\frac{11}{12}<\mathrm{I}<\frac{331}{360}$

$\Rightarrow \mathrm{I} \in\left[\frac{11}{12}, \frac{331}{360}\right]$

$\Rightarrow I \in\left[\frac{330}{360}, \frac{331}{360}\right]$

Ans. (2)