I<0? Evaluate New Year Challenge Integral

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Discussion Overview

The discussion revolves around evaluating the integral $$I=\int_{2013}^{2014} \frac{\sin x}{x}\,dx$$ and determining whether it is less than, equal to, or greater than zero. The focus appears to be on reasoning and justification for the evaluation of this integral.

Discussion Character

  • Exploratory, Debate/contested

Main Points Raised

  • Some participants propose evaluating the integral based on the behavior of the function $$\frac{\sin x}{x}$$ over the interval from 2013 to 2014.
  • Others argue that the oscillatory nature of the sine function may influence the value of the integral, suggesting that it could be positive, negative, or zero.
  • A later reply acknowledges the contributions of participants but does not provide any further analysis or conclusions regarding the value of the integral.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the integral is less than, equal to, or greater than zero, indicating that multiple competing views remain.

Contextual Notes

The discussion lacks detailed mathematical analysis or specific reasoning steps that could clarify the evaluation of the integral, leaving assumptions and dependencies on the behavior of the sine function unresolved.

anemone
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Let $$I=\int_{2013}^{2014} \frac{\sin x}{x}\,dx$$. Determine with reason if $I<0,\,I=0$ or $I>0$?

This challenge is one of my top favorite problems that can be cracked using purely elementary method! (Sun):)
 
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anemone said:
Let $$I=\int_{2013}^{2014} \frac{\sin x}{x}\,dx$$. Determine with reason if $I<0,\,I=0$ or $I>0$?

This challenge is one of my top favorite problems that can be cracked using purely elementary method! (Sun):)

converting to degree we have lower limit around $136.90^\circ$ and upper limit around $193.70^\circ$
now from $136.90^\circ$ to $180^\circ$ degrees $\sin$ is $\ge 0$ and from $180^\circ$ to 1$93.70^\circ$ . it is $\le 0$. They come in the same cycle as difference is 1 radian
integral from $(180-13.70)^\circ$ i.e $166.30^\circ$ to $193.70^\circ$ shall be zero provided denominator is constant. as denominator is decreasing the integral from $166.30^\circ$ to $193.70^\circ$ is positive and adding another positive quantity that is integral from $136.90^\circ$ to $166.30^\circ$ which is positive so sum $I \gt 0$
 
Thanks kaliprasad for participating.
Your solution is quite ingenious, and please note that the value $136.90^\circ$ should be $136.40^\circ$.

Solution of other:
Split the definite integral into two part, with $a$ being the zero at about $2013.75$, note that we have:

$$\int_{2013}^{a} \frac{\sin x}{x}\,dx>\int_{2013}^{a} \frac{\sin x}{2014}\,dx$$

and

$$\int_{a}^{2014} \frac{\sin x}{x}\,dx>\int_{a}^{2014} \frac{\sin x}{2013}\,dx$$

So adding them up yields

$$\begin{align*}I=\int_{2013}^{a} \frac{\sin x}{x}\,dx+\int_{a}^{2014} \frac{\sin x}{x}\,dx&>\int_{2013}^{a} \frac{\sin x}{2014}\,dx+\int_{a}^{2014} \frac{\sin x}{2013}\,dx\\&>\frac{\cos 2013}{2014}-\frac{\cos 2014}{2013}+\frac{\cos a}{2013}-\frac{\cos a}{2014}\\&>0\end{align*}$$
 
anemone said:
Thanks kaliprasad for participating.
Your solution is quite ingenious, and please note that the value $136.90^\circ$ should be $136.40^\circ$.

oops my mistake. false start in 2016.
 
kaliprasad said:
oops my mistake. false start in 2016.

Please don't worry about it...and it was after all an honest mistake, I understand it completely.:)
 

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