# Let f

Question:

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

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

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

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

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

Correct Option: 2,

Solution:

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

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

$\mathrm{I}=\int_{0}^{1}\left(\mathrm{~F}^{\prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})\right) \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\int_{0}^{1}(\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x})) \cdot \mathrm{e}^{\mathrm{x}} \mathrm{dx}$

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

$=2 \int_{0}^{1} \sin \mathrm{xdx}$

$=2(1-\cos 1)$

$\mathrm{I}=2\left\{1-\left(1-\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{[8} \ldots \ldots \cdots\right)\right\}$

$\mathrm{I}=1-\frac{2}{4}+\frac{2}{16}-\frac{2}{[9}+\ldots \ldots$

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

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

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

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

Leave a comment

Free Study Material