I<0? Evaluate New Year Challenge Integral

  • Context: MHB 
  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Challenge Year
Click For Summary
SUMMARY

The integral $$I=\int_{2013}^{2014} \frac{\sin x}{x}\,dx$$ is evaluated to determine its sign. The discussion concludes that since the function $\frac{\sin x}{x}$ is positive in the interval from 2013 to 2014, the integral $$I$$ is greater than zero, i.e., $$I>0$$. This conclusion is supported by the properties of the sine function and its behavior over the specified range.

PREREQUISITES
  • Understanding of definite integrals
  • Knowledge of the sine function and its properties
  • Familiarity with the concept of improper integrals
  • Basic calculus skills
NEXT STEPS
  • Study the properties of the sinc function, specifically $$\frac{\sin x}{x}$$
  • Explore numerical integration techniques for evaluating definite integrals
  • Learn about the behavior of oscillatory integrals
  • Investigate the applications of integrals in real-world scenarios
USEFUL FOR

Mathematics students, calculus instructors, and anyone interested in integral calculus and the analysis of oscillatory functions.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
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):)
 
Physics news on Phys.org
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.:)
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K