Let $f:R\to R$ be defined as $f\left(x\right)=e^{-x}\sin x.$ If $F:\left[0, 1\right]\to R$ is a differentiable function such that $F\left(x\right)={\int }_0^{x}f\left(t\right)dt,$ then the value of ${\int }_0^1\left(F^{\text{'}}\left(x\right)+f\left(x\right)\right)e^{x}dx$ lies in the interval

Given $f\left(x\right)=e^{-x}\sin x$

Now, $F\left(x\right)={\int }_0^{x}f\left(t\right)dt$

Using Newton Leibnitz rule i.e. $\frac{d}{dx}{\int }_{u\left(x\right)}^{v\left(x\right)}f\left(t\right)dt=f\left(v\left(x\right)\right)·\left(v^{\text{'}}\left(x\right)\right)-f\left(u\left(x\right)\right)·\left(u^{\text{'}}\left(x\right)\right),$

$\Rightarrow F^{\text{'}}\left(x\right)=f\left(x\right)$

Now, $I={\int }_0^1\left(F^{\text{'}}\left(x\right)+f\left(x\right)\right){\mathrm{e}}^x\mathrm{dx}$

$\Rightarrow I={\int }_0^1\left(f\right(x)+f(x\left)\right)·{\mathrm{e}}^x\mathrm{dx}$

$\Rightarrow I=2{\int }_0^{1}f\left(x\right)·e^{x}dx$

$\Rightarrow I=2{\int }_0^1{e}^{-x}\sin x·e^{x}dx$

$\Rightarrow I=2{\int }_0^1\sin xdx$

$\Rightarrow I=2{\left[-\cos x\right]}_0^1$

$\Rightarrow I=2(1-\cos 1)$

Using the expansion of $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$

$\Rightarrow I=2\left\{1-\left(1-\frac{1}{2}+\frac{1}{4!}+\frac{1}{6!}+...\right)\right\}$

$\Rightarrow I=1-\frac{2}{4!}+\frac{2}{6!}-\frac{2}{9!}...$

$\Rightarrow 1-\frac{2}{4!}<I<1-\frac{2}{4!}+\frac{2}{6!}$

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

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

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