Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined as $\mathrm{f}(\mathrm{x})=\mathrm{e}^{-\mathrm{x}} \sin \mathrm{x}$. If
$\mathrm{F}:[0,1] \rightarrow \mathrm{R}$ is a differentiable function
such that $\mathrm{F}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}$, then the value of $\int_{0}^{1}\left(\mathrm{~F}^{\prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})\right) \mathrm{e}^{\mathrm{x}} \mathrm{dx}$ lies in the interval
Correct Option: 2,
$f(x)=e^{-x} \sin x$
Now, $\mathrm{F}(\mathrm{x})=\int_{0}^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt} \quad \Rightarrow \mathrm{F}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x})$
$\mathrm{I}=\int_{0}^{1}\left(\mathrm{~F}^{\prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})\right) \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\int_{0}^{1}(\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x})) \cdot \mathrm{e}^{\mathrm{x}} \mathrm{dx}$
$=2 \int_{0}^{1} \mathrm{f}(\mathrm{x}) \cdot \mathrm{e}^{\mathrm{x}} \mathrm{dx}=2 \int_{0}^{1} \mathrm{e}^{-\mathrm{x}} \sin \mathrm{x} \cdot \mathrm{e}^{\mathrm{x}} \mathrm{d} \mathrm{x}$
$=2 \int_{0}^{1} \sin \mathrm{xdx}$
$=2(1-\cos 1)$
$\mathrm{I}=2\left\{1-\left(1-\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{[8} \ldots \ldots \cdots\right)\right\}$
$\mathrm{I}=1-\frac{2}{4}+\frac{2}{16}-\frac{2}{[9}+\ldots \ldots$
$1-\frac{2}{4}<\mathrm{I}<1-\frac{2}{4}+\frac{2}{6}$
$\frac{11}{12}<\mathrm{I}<\frac{331}{360}$
$\Rightarrow \mathrm{I} \in\left[\frac{11}{12}, \frac{331}{360}\right]$
$\Rightarrow \mathrm{I} \in\left[\frac{330}{360}, \frac{331}{360}\right]$
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