Let f : R $\to$ R be defined as f(x) = e^$-$xsinx. If F : [0, 1] $\to$ R is a differentiable function with that F(x) = $\int_0^x {f(t)dt}$, then the value of $\int_0^1 {(F'(x) + f(x)){e^x}dx}$ lies in the interval

F(x) = $\int_0^x {f(t)dt}$ <br><br>$\Rightarrow$ F'(x) = f(x) by Leibnitz theorem<br><br>I = $\int\limits_0^1 {(F'(x) + f(x)){e^x}dx = \int\limits_0^1 {2f(x){e^x}dx} }$<br><br>$I = \int\limits_0^1 {2\sin x\,dx}$<br><br>$I = 2(1 - \cos 1)$<br><br>$$ = 2\left\{ {1 - \left( {1 - {{{1^2}} \over {2!}} + {{{1^4}} \over {4!}} - {1 \over {6!}} + ...} \right)} \right\}$$<br><br>$$ = 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}}} \right)} \right\} &lt; 2(1 - \cos 1) &lt; 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}} - {1 \over {720}}} \right)} \right\}$$<br><br>${{330} \over {360}} &lt; 2(1 - \cos 1) &lt; {{331} \over {360}}$<br><br>${{330} \over {360}} &lt; 1 &lt; {{331} \over {360}}$