Can someone explain this proccess for expressing limits as an integral?

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Homework Help Overview

The discussion revolves around converting a limit expression involving a summation of sine functions into an integral. The original poster presents a limit that appears to be related to Riemann sums and seeks clarification on the process of expressing it as an integral, specifically questioning the choice of Δx and the manipulation of the formula to derive the integral limits.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the definition of Riemann integrals and how they relate to the limit and summation presented. There are inquiries about the reasoning behind the choice of Δx as π/n and how the limits of integration are determined. Some participants also express confusion about the process of converting the limit to an integral.

Discussion Status

The discussion is ongoing, with participants providing insights into the definition of integrals and Riemann sums. Some guidance has been offered regarding the interpretation of the limit and the approximation of areas under curves, but there is no explicit consensus on the understanding of the original problem.

Contextual Notes

Participants note potential typos in the original limit expression and question the accuracy of the provided formulas. There is a recognition of the need for clarity in the definitions and processes involved in converting limits to integrals.

NWeid1
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Homework Statement


My professor gave us this problem:

Convert the limit to an integral:


\lim_{n\rightarrow inf} \frac{1}{n}(\sin(\pi/2) + \sin(\pi/2) + ... + \sin(\pi))
he said it was right endpoint from [0,π]
then he set Δx = \frac{\pi}{n}
then

\lim_{n\rightarrow inf} \frac{1}{\pi}(\frac{\pi}{n} \sum_{i=1}^{n} \sin(\frac{i\pi}{n}))

= \frac{1}{\pi} \int\limits_{0}^{\pi}\sin{x}\, dx

= \int\limits_{0}^{1}\sin{x}\, dx

I just don't understand how he found Δx as being π/n or how he munipulated the formula to get the limits for the inetgral.

Homework Equations





The Attempt at a Solution

 
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Hi NWeid1! :smile:


NWeid1 said:

Homework Statement


My professor gave us this problem:

Convert the limit to an integral:


\lim_{n\rightarrow inf} \frac{1}{n}(\sin(\pi/2) + \sin(\pi/2) + ... + \sin(\pi))
he said it was right endpoint from [0,π]
then he set Δx = \frac{\pi}{n}
then

\lim_{n\rightarrow inf} \frac{1}{\pi}(\frac{\pi}{n} \sum_{i=1}^{n} \sin(\frac{i\pi}{n}))
= \frac{1}{\pi} \int\limits_{0}^{\pi}\sin{x}\, dx

It appears that your first formula contains a number of typos, since I think it should match the summation.


Do you know how a (Riemann) integral is actually defined?
In this case \int\limits_{0}^{\pi}\sin{x}\, dx?
 
Last edited:
Well this is exactly what I have in my notes, and I know the answer is = \int\limits_{0}^{1}\sin{x}\, dx

But I was just writing it down when he was teaching it, I never really understood the process.
 
You mean familiar with the process of converting an integral to a limit? I learned it but that is kind of what I'm asking. The only notes I have on this is what I have above, which I can't understand by the example.
 
Well, do you know that an integral is approximated by inscribed rectangles?

Sorry for not explaining more, but I have no clue what you know and do not know.
 
Last edited:
NWeid1 said:

Homework Statement


My professor gave us this problem:

Convert the limit to an integral:


\lim_{n\rightarrow inf} \frac{1}{n}(\sin(\pi/2) + \sin(\pi/2) + ... + \sin(\pi))
No. Based on what you say below, you want
\lim_{n\to \inf} (\frac{1}{n} sin(\pi/n)+ sin(2\pi/n)+ ...+ sin(n\pi/n))

he said it was right endpoint from [0,π]
then he set Δx = \frac{\pi}{n}
Yes, you are dividing the interval from 0 to \pi into n equal length sub-intervals so each has length \pi/n. The two endpoints of the first sub-interval are 0 and \pi/n, the two endpoinst of the second sub-interval are \pi/n and 2\pi/n, etc.

then

\lim_{n\rightarrow inf} \frac{1}{\pi}(\frac{\pi}{n} \sum_{i=1}^{n} \sin(\frac{i\pi}{n}))
So you are approximating the value of sin(x) inside the "i"th sub-interval by its value at the right endpoint, sin(i\pi/n). And you are approximating the area under the graph of y= sin(x) inside each sub-interval by the area of the rectangle having height sin(i\pi/n) and base \pi/n

= \frac{1}{\pi} \int\limits_{0}^{\pi}\sin{x}\, dx
<br /> Now you make that approximation exact by taking the limit- this limit of the sum will be <b>equal</b> to the area under y= sin(x).<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> = \int\limits_{0}^{1}\sin{x}\, dx </div> </div> </blockquote>and, of course, the area under the graph is the integral.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I just don&#039;t understand how he found Δx as being π/n or how he munipulated the formula to get the limits for the inetgral. </div> </div> </blockquote> The sum goes from 0\pi/n= 0 to n\pi/n= \pi. That&#039;s how he got the limits of integration. And the difference bertween two consecutive value, say 2\pi/n- \pi/n, or n\pi/n- (m-1)\pi/n) is \Delta x= \pi/n so you are dividing the length, \pi into n equal parts.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2> </div> </div> </blockquote>
 
Last edited by a moderator:
Heh! What HoI said! :smile:
(@Hoi: you have way too much spare time! ;)


Just one addition.
You write:
= \frac{1}{\pi} \int\limits_{0}^{\pi}\sin{x}\, dx<br /> = \int\limits_{0}^{1}\sin{x}\, dx
But this is not true, which you will see if you evaluate the integral.
 
Oh, yeah I learned all the riemann sums.
 

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