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

In summary: I just cannot figure out how to make the connection between the limit and the integral.In summary, the limit \lim_{n\rightarrow inf} \frac{1}{n}(\sin(\pi/2) + \sin(\pi/2) + ... + \sin(\pi)), where n is approaching infinity, can be converted to an integral by dividing the interval [0, π] into n equal sub-intervals and approximating the value of sin(x) in each sub-interval by its value at the right endpoint. The limit of this sum is equal to the area under the graph of y=sin(x), which can be written as the integral \int_{0}^{1}sin(x)\,dx. The
  • #1
NWeid1
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0

Homework Statement


My professor gave us this problem:

Convert the limit to an integral:


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

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

[tex] = \frac{1}{\pi} \int\limits_{0}^{\pi}\sin{x}\, dx[/tex]

[tex] = \int\limits_{0}^{1}\sin{x}\, dx [/tex]

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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  • #2
Hi NWeid1! :smile:


NWeid1 said:

Homework Statement


My professor gave us this problem:

Convert the limit to an integral:


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

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

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 [itex]\int\limits_{0}^{\pi}\sin{x}\, dx[/itex]?
 
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  • #3
Well this is exactly what I have in my notes, and I know the answer is [tex] = \int\limits_{0}^{1}\sin{x}\, dx [/tex]

But I was just writing it down when he was teaching it, I never really understood the process.
 
  • #5
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.
 
  • #6
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.
 
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  • #7
NWeid1 said:

Homework Statement


My professor gave us this problem:

Convert the limit to an integral:


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

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

then

[tex] \lim_{n\rightarrow inf} \frac{1}{\pi}(\frac{\pi}{n} \sum_{i=1}^{n} \sin(\frac{i\pi}{n}))[/tex]
So you are approximating the value of sin(x) inside the "i"th sub-interval by its value at the right endpoint, [itex]sin(i\pi/n)[/itex]. 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 [itex]sin(i\pi/n)[/itex] and base [itex]\pi/n[/itex]

[tex][tex] = \frac{1}{\pi} \int\limits_{0}^{\pi}\sin{x}\, dx[/tex]
Now you make that approximation exact by taking the limit- this limit of the sum will be equal to the area under y= sin(x).

[tex] = \int\limits_{0}^{1}\sin{x}\, dx [/tex]
and, of course, the area under the graph is the integral.

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.
The sum goes from [itex]0\pi/n= 0[/itex] to [itex]n\pi/n= \pi[/itex]. That's how he got the limits of integration. And the difference bertween two consecutive value, say [itex]2\pi/n- \pi/n[/itex], or [itex]n\pi/n- (m-1)\pi/n)[/itex] is [itex]\Delta x= \pi/n[/itex] so you are dividing the length, [itex]\pi[/itex] into n equal parts.

Homework Equations





The Attempt at a Solution

 
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  • #8
Heh! What HoI said! :smile:
(@Hoi: you have way too much spare time! ;)


Just one addition.
You write:
[tex] = \frac{1}{\pi} \int\limits_{0}^{\pi}\sin{x}\, dx
= \int\limits_{0}^{1}\sin{x}\, dx[/tex]
But this is not true, which you will see if you evaluate the integral.
 
  • #9
Oh, yeah I learned all the riemann sums.
 

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

What is the purpose of expressing limits as an integral?

Expressing limits as an integral is a mathematical process used to find the area under a curve, also known as the definite integral. It is an important tool in calculus and is used to solve a variety of real-world problems.

How do you express limits as an integral?

To express limits as an integral, you first need to determine the function or curve that you want to find the area under. Then, you need to set up the integral by choosing the appropriate limits of integration and integrating the function with respect to the variable of integration.

What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, which means you are finding the area under the curve between those two limits. An indefinite integral does not have any limits of integration, and therefore, represents the general antiderivative of a function.

Can you explain the process of finding the limits of integration?

The limits of integration represent the boundaries of the area under the curve that you want to find. To determine these limits, you can use the given problem or graph to identify the points of intersection of the function and the x-axis. These points will be your upper and lower limits of integration.

How can expressing limits as an integral be applied in real life?

Expressing limits as an integral has various real-world applications, such as calculating the volume of irregularly shaped objects, finding the work done by a variable force, and determining the average value of a function over a given interval. It is also used in fields like physics, engineering, and economics to solve complex problems.

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