Question

Let $f:R\to R$ be defined as $f\left(x\right)=e^{-x}\sin x$. If $F:\left[0,1\right]R\to$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.

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

Answer 1

The correct option is A

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

Determine the value of ${\int }_0^1\left[F\text{'}\left(x\right)+f\left(x\right)\right]e^{x}dx$

We have, $f\left(x\right)=e^{-x}\sin x$ and $F\left(x\right)={\int }_0^{x}f\left(t\right)dt$

Differentiating using liebnitz rule:

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

Now,

$$\begin{gathered} {\int }_0^1\left[F\text{'}\left(x\right)+f\left(x\right)\right]e^{x}dx \\ I={\int }_0^1\left[f\left(x\right)+f\left(x\right)\right]e^{x}dx \\ I={\int }_0^{1}2f\left(x\right)e^{x}dxI={\int }_0^{1}2e^{-x}\sin x{e}^{x}dx \\ I={\int }_0^{1}2\sin xdx \\ I=2{{\left[-\cos x\right]}^1}_0 \\ I=-2\left[\cos 1-\cos 0\right] \\ I=2\left[1-\cos 1\right] \\ I=2\left[1-\left\{1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}...\right\}\right] \\ I=1-\frac{2}{4!}+\frac{2}{6!}.... \\ \therefore 1-\frac{2}{4!}<I<1-\frac{2}{4!}+\frac{2}{6!} \\ \Rightarrow 1-\frac{2}{24}<I<1-\frac{2}{24}+\frac{2}{360} \\ \Rightarrow \frac{11}{12}<I<\frac{331}{360} \\ \Rightarrow \frac{330}{360}<I<\frac{331}{360} \end{gathered}$$

Hence, the correct answer is Option (A).