Question

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

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

Answer 1

The correct option is A $[\frac{330}{360},\frac{331}{360}]$

$F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)d\left(t\right)$
Applying Leibnitz theorem,$F^{\prime }\left(x\right)=f\left(x\right)$
$I=\underset{0}{\overset{1}{\int }}(F^{\prime }\left(x\right)+f\left(x\right))e^{x}dx=\underset{0}{\overset{1}{\int }}2f\left(x\right)e^{x}dx$
$\Rightarrow I=\underset{0}{\overset{1}{\int }}2\sin xdx$
$\Rightarrow I=2(1-\cos 1)$
$=[1-(1-\frac{1^2}{2!}+\frac{1^4}{4!}-\frac{1^6}{6!}\cdots )]$

Now,
$2[1-(1-\frac{1}{2}+\frac{1}{24})]<2(1-\cos 1)<2[1-(1-\frac{1}{2}+\frac{1}{24}-\frac{1}{720})]$
$\Rightarrow \frac{330}{360}<2(1-\cos 1)<\frac{331}{360}$
$\Rightarrow \frac{330}{360}<I<\frac{331}{360}$