# Proving the definition of the integral

• Feb 7th 2010, 07:59 AM
paupsers
Proving the definition of the integral
Suppose $f \in R[0, \pi]$. Prove that the sequence

$a_{n}=\frac{\pi}{n}\sum f(\frac{k\pi}{n})$

converges to $\int f$.

Note that the sum is from k=1 to n, and the integral is from 0 to pi.

Can anyone at LEAST give me a starting point for this? I have no idea where to even begin this...
• Feb 7th 2010, 08:23 AM
felper
Note that the sum you have, is a Riemann sum of the form:

$\dfrac{b-a}{n}\displaystyle\sum_{i=0}^nf\left(a+i\dfrac{b-a}{n}\right)$

Where $\dfrac{b-a}{n}$ is a partition of the interval, $a+i\dfrac{b-a}{n}$ is a selection of points of each interval defined by the partition. The norm of the partition is regular, that means that every interval defined by the partition has the same lenght. This Riemann sum is defined by a aritmetic progression. In more abstract sence, a Riemann sum just have to look like this:

$\sum_{i=0}^nf(a_i)\Delta x_i$

The Riemann integration tell us that if a Riemann sum converges when the norm of the partition tends to 0, the function is Riemann-integrable. So, can you tell me which is the partition in this particular case? can you express your sum like a Riemann sum?
• Feb 7th 2010, 08:25 AM
Plato
I have no idea what you mean by prove.
I assume the by $f\in R[0,\pi]$ you mean the function is Riemann integrable of the interval.
Then observe that each $a_n$ is a right-hand approximating sum over a regular partition of the interval.
• Feb 7th 2010, 10:05 AM
paupsers
Quote:

Originally Posted by felper
Note that the sum you have, is a Riemann sum of the form:

$\dfrac{b-a}{n}\displaystyle\sum_{i=0}^nf\left(a+i\dfrac{b-a}{n}\right)$

Where $\dfrac{b-a}{n}$ is a partition of the interval, $a+i\dfrac{b-a}{n}$ is a selection of points of each interval defined by the partition. The norm of the partition is regular, that means that every interval defined by the partition has the same lenght. This Riemann sum is defined by a aritmetic progression. In more abstract sence, a Riemann sum just have to look like this:

$\sum_{i=0}^nf(a_i)\Delta x_i$

The Riemann integration tell us that if a Riemann sum converges when the norm of the partition tends to 0, the function is Riemann-integrable. So, can you tell me which is the partition in this particular case? can you express your sum like a Riemann sum?

So the partition is $\frac{\pi}{n}$
So $\sum_{i=0}^nf(a_i)\Delta x_i=\frac{\pi}{n}\sum_{i=0}^nf(a_i)$? But this is where I started...