MHB I<0? Evaluate New Year Challenge Integral

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The integral I, defined as I = ∫ from 2013 to 2014 of sin(x)/x dx, is evaluated to determine its sign. The function sin(x)/x is positive in the interval from 2013 to 2014, as sin(x) remains positive and x is positive. Therefore, the integral I is greater than zero. The conclusion is that I > 0.
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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.:)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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