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

NWeid1
Messages
81
Reaction score
0

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

 
Physics news on Phys.org
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.
 
Back
Top