## Friday, July 18, 2008

### Definite Integrals

Surprisingly, it is not easy to define what such an integral is. We want it to be equal to the area under the graph of a function - but to get it we need to define a specific algorithm that will produce it. One of such definitions is given by Riemann - it is called the Riemann Integral.
To properly define the Riemann Integral, we need to define a few other things firstly.

Lets suppose that we want a definition that will give us the area under the graph of some function f(x), which goes from the interval [a,b] to the real line (it is not necessary to assume that f is defined in all of [a,b], but it must be bounded in [a,b]). For this interval lets define:

P={a=$x_{0}}.

As you see from the definition P is a group that has n+1 different points and it includes both a and b. Note that according to the definition P is finite. Now for any such group lets define w=max|$x_{i}-x_{i-1}$|. Obviously w>0. Now we can write the following sum:

$^n_{i=1}\sum f(t_{i})$($x_{i}-x_{i-1}$)

In this sum $t_{i}$ is any number such that $x_{i-1}<=t_{i}<=x_{i}$.
The next step would be to define a limit "according to w". We will define it in the following way:
We will say that the limit =A if and only if:

$\forall\epsilon>0 \exists \delta>o$$\forall P$ $w(P)<\delta \rightarrow$|$^n_{i=1}\sum f(t_{i})$($x_{i}-x_{i-1}$)-A|<$\epsilon$ $\forall t$

Basically, it says that if for any division of the interval such that the maximum distance between two points in the division is less that delta, the difference between the sum and A is less than epsilon, and such delta exists for any given epsilon (and for any selections of the points $t_{i}$) , than the limit is A.
The limit, if it exists, is the integral according to Riemann. If the limit doesn't exists the function has no integral according to Riemann.

Since we have a definition, lets use it to calculate an integral. For example, lets calculate the integral of f(x)=2 in the interval [0,1]. f($t_{i}$)=2 for any part of the interval, so we can get it out of the sum. The rest of the sum is the length of the interval:

lim$^n_{i=1}\sum f(t_{i})$($x_{i}-x_{i-1}$)=2(0--1)=2

Unsurprisingly the answer is two. Notice that this is indeed the answer only because the calculation here is independent from both P and the points t we selected. But what if the function is slightly more complex, for example f(x)=x^2? It is still possible to calculate the integral using this definition, but it is better to use another definition.

The Riemann definition is a very good definition, but sometimes it is easier to use a different definition. There is more than one other definition, but I want to talk only about one of them - the definition according to Darbo.

In this definition we again make use of P, which is defined in exactly the same way as before. We also need to define two other things:

$m_{i}$=inf(f(x)) , $M_{i}$=sup(f(x)) , $x \in[x_{i-1},x_{i}]$

Using this we can define the Upper Darbo sum as following:

$^n_{i=1}\sum M_{i}(x_{i}-x_{i-1})$

And also the Lower Darbo sum:

$^n_{i=1}\sum m_{i}(x_{i}-x_{i-1})$

The next step is to define the lower and upper integral. The upper integral is the infimum of the group of the upper sums, and the lower integral is the supremum of the lower Darbo sums. If both of these integrals are equal, we will say that the function has an integral according to Darbo.

In the next posts I will show that these two ways to define the integral are equivalent, and will also discuss ways to calculate the integral.