Let f:RR be defined as f(x)=e^-x x.

Question

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

Accepted Answer

F(x) = $$\int_0^x {f(t)dt} $$

$$ \Rightarrow $$ F'(x) = f(x) by Leibnitz theorem

I = $$\int\limits_0^1 {(F'(x) + f(x)){e^x}dx = \int\limits_0^1 {2f(x){e^x}dx} } $$

$$I = \int\limits_0^1 {2\sin x\,dx} $$

$$I = 2(1 - \cos 1)$$

$$ = 2\left\{ {1 - \left( {1 - {{{1^2}} \over {2!}} + {{{1^4}} \over {4!}} - {1 \over {6!}} + ...} \right)} \right\}$$

$$ = 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}}} \right)} \right\} < 2(1 - \cos 1) < 2\left\{ {1 - \left( {1 - {1 \over 2} + {1 \over {24}} - {1 \over {720}}} \right)} \right\}$$

$${{330} \over {360}} < 2(1 - \cos 1) < {{331} \over {360}}$$

$${{330} \over {360}} < 1 < {{331} \over {360}}$$

Solutions · 1