# Let f : R

Question:

Let $f: R \rightarrow R$ be defined as $f(x)=e^{-x} \sin x$. 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 $\int_{0}^{1}\left(F^{\prime}(x)+f(x)\right) e^{x} d x$ lies in the interval

1. $\left[\frac{327}{360}, \frac{329}{360}\right]$

2. $\left[\frac{330}{360}, \frac{331}{360}\right]$

3. $\left[\frac{331}{360}, \frac{334}{360}\right]$

4. $\left[\frac{335}{360}, \frac{336}{360}\right]$

Correct Option: , 2

Solution:

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

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)