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

1. Dec 8, 2011

### NWeid1

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution

2. Dec 8, 2011

### I like Serena

Hi NWeid1!

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: Dec 8, 2011
3. Dec 8, 2011

### NWeid1

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.

4. Dec 8, 2011

5. Dec 8, 2011

### NWeid1

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. Dec 9, 2011

### I like Serena

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: Dec 9, 2011
7. Dec 9, 2011

### HallsofIvy

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

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.

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 $\pi/n$

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).

and, of course, the area under the graph is the integral.

The sum goes from $0\pi/n= 0$ to $n\pi/n= \pi$. That'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.

Last edited by a moderator: Dec 9, 2011
8. Dec 9, 2011

### I like Serena

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

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