Question

Let $f:R\to R$ be defined as $f(x)=e^{-x}\sin x$. If $F:[0,1]\to R$ is a differentiable function such that $F(x)=\underset{0}{\overset{x}{\int }}f(t)dt$, then the value of $\underset{0}{\overset{1}{\int }}\left(F^{\prime }(x)+f(x)\right)e^{x}dx$ 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, $F(x)=\underset{0}{\overset{x}{\int }}f(t)dt$
$\Rightarrow F^{\prime }(x)=f(x)$
$I=\underset{0}{\overset{1}{\int }}\left(F^{\prime }(x)+f(x)\right)e^{x}dx=\underset{0}{\overset{1}{\int }}(f(x)+f(x))\cdot e^{x}dx$
$=2\underset{0}{\overset{1}{\int }}f(x)\cdot e^{x}dx=2\underset{0}{\overset{1}{\int }}{e}^{-x}\sin x\cdot e^{x}dx$
$=2\underset{0}{\overset{1}{\int }}\sin ddx$
$=2(1-\cos 1)$
$I=2\left\{1-\left(1-\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{8}\dots \dots \dots .\right)\right\}$
$I=1-\frac{2}{4}+\frac{2}{16}-\frac{2}{19}+\dots ..$
$1-\frac{2}{4}<I<1-\frac{2}{4}+\frac{2}{16}$
$\frac{11}{12}<I<\frac{331}{360}$
$\Rightarrow I\in \left[\frac{11}{12},\frac{331}{360}\right]$
$\Rightarrow I\in \left[\frac{330}{360},\frac{331}{360}\right]$