I<0? Evaluate New Year Challenge Integral

[anemone]

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

Spoiler

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

 

[kaliprasad]

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

Spoiler

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

Spoiler

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 $I \gt 0$

 

[anemone]

Thanks kaliprasad for participating.

Spoiler

Your solution is quite ingenious, and please note that the value $136.90^\circ$ should be $136.40^\circ$.

Solution of other:

Spoiler

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

\(\displaystyle \int_{2013}^{a} \frac{\sin x}{x}\,dx>\int_{2013}^{a} \frac{\sin x}{2014}\,dx\)

and

\(\displaystyle \int_{a}^{2014} \frac{\sin x}{x}\,dx>\int_{a}^{2014} \frac{\sin x}{2013}\,dx\)

So adding them up yields

\(\displaystyle \begin{align*}I=\int_{2013}^{a} \frac{\sin x}{x}\,dx+\int_{a}^{2014} \frac{\sin x}{x}\,dx&>\int_{2013}^{a} \frac{\sin x}{2014}\,dx+\int_{a}^{2014} \frac{\sin x}{2013}\,dx\\&>\frac{\cos 2013}{2014}-\frac{\cos 2014}{2013}+\frac{\cos a}{2013}-\frac{\cos a}{2014}\\&>0\end{align*}\)

 

[kaliprasad]

Thanks kaliprasad for participating.

Spoiler

Your solution is quite ingenious, and please note that the value $136.90^\circ$ should be $136.40^\circ$.

Spoiler

oops my mistake. false start in 2016.

 

[anemone]

Spoiler

oops my mistake. false start in 2016.

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