Question

Let $f : R \rightarrow R$ be defined as $f ( x )= e ^{- x } \sin x$. If $F :[ 0 , 1] \rightarrow R$ is a differentiable function such that $F ( x )=\int\limits_{0}^{ x } f ( t ) dt$, then the value of $\int\limits_{0}^{1}\left( F ^{\prime}( x )+ f ( x )\right) e ^{ x } dx$ lies in the interval

• A. $\left[\frac{327}{360}, \frac{329}{360}\right]$
• B. $\left[\frac{330}{360}, \frac{331}{360}\right]$
• C. $\left[\frac{331}{360}, \frac{334}{360}\right]$
• D. $\left[\frac{335}{360}, \frac{336}{360}\right]$

Answer 1

Answer: (B)

$f ( x )= e ^{-x} \sin x$
Now, $F ( x )=\int\limits_{0}^{ x } f ( t ) dt $
$\Rightarrow F ^{\prime}( x )= f ( x )$
$I =\int\limits_{0}^{1}\left( F ^{\prime}( x )+ f ( x )\right) e ^{ x } dx =\int\limits_{0}^{1}( f ( x )+ f ( x )) \cdot e ^{ x } dx$
$=2 \int\limits_{0}^{1} f ( x ) \cdot e ^{ x } dx =2 \int\limits_{0}^{1} e ^{- x } \sin x \cdot e ^{ x } dx$
$=2 \int\limits_{0}^{1} \sin d d x$
$=2(1-\cos 1)$
$I =2\left\{1-\left(1-\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{8} \ldots \ldots \ldots .\right)\right\}$
$I =1-\frac{2}{4}+\frac{2}{16}-\frac{2}{19}+\ldots . .$
$1-\frac{2}{4}< I <1-\frac{2}{4}+\frac{2}{16}$
$\frac{11}{12}< I <\frac{331}{360}$
$\Rightarrow I \in\left[\frac{11}{12}, \frac{331}{360}\right]$
$\Rightarrow I \in\left[\frac{330}{360}, \frac{331}{360}\right]$