# The UNDEFINED integral.

• Apr 15th 2009, 06:28 PM
VonNemo19
The UNDEFINED integral.
Is the epsilon-delta definition of the integral totally useless, or can it be applied in the same way as the definition of the ordinary limit? Thinking about it just twists my noodle. I've tried researching but, answers to this question aren't forthcomming. And please, if you want to tell me anything about this thing being intuitively clear, save it. I want strict proof. HELP ME!!(Worried)
• May 23rd 2009, 06:37 PM
Media_Man
Example?
Would you be referring to the definition of the integral as $\int_a^b f(x) dx=\lim_{\Delta x\rightarrow 0}\sum f(x_i)\Delta x$ ?
• May 23rd 2009, 07:41 PM
VonNemo19
Quote:

Originally Posted by Media_Man
Would you be referring to the definition of the integral as $\int_a^b f(x) dx=\lim_{\Delta x\rightarrow\infty}\sum f(x_i)\Delta x$ ?

Well, no. However, if you think that you can justify this statement in terms of first principals, go for it.

The definition I speak of is a little more precise:

Let $f$ be a function defined on an interval $[a,b]$ then, if for each $\epsilon>0$ there exists a $\delta>0$ such that

$\mid\sum_{i=1}^{n}f(x_i)\Delta_i{x}-A\mid<\epsilon$ whenever $\parallel\Delta\parallel<\delta$

we say that $f$ is integrable on the interval $[a,b]$ and we write

$\int_a^b{f(x)}dx=A$

This is the definition I'm talking about. I'd like to hear your thoughts. You can see that this definition parallels the definition of the ordinary limit in many ways. My beef here is that, in the the definition of the ordinary limit the number $L$ is either known, or can be found by analytic methods. But, in this definition, you can't show that a function is integrable without knowing the number $A$ which, in turn, pressuposes that you have integrated the the thing already! Do you see the circular logic here. Now I now that it can be shown that the summ is approaching a number $A$ by some other method (sequencing?), but I have not studied these methods yet. Help me a little bit here and you've definitely got a thanks coming'.

Peace(Hi)
• May 23rd 2009, 08:00 PM
matheagle
Quote:

Originally Posted by Media_Man
Would you be referring to the definition of the integral as $\int_a^b f(x) dx=\lim_{\Delta x\rightarrow\infty}\sum f(x_i)\Delta x$ ?

I don't think you want $\Delta x\rightarrow\infty$
In fact letting n go to infinity isn't good either.
You need the mesh, the largest subinterval, to go to zero.
• May 23rd 2009, 08:09 PM
Media_Man
It's a visual
See animation first: Integration and Riemann Sums

To find the area under a curve $f(x)$ between $x=a$ and $x=b$, we chop the area up into space-filling rectangles. Define the width of the rectangle to be an arbitrarily small number, $\Delta$, and the number of rectangles used an arbitrarily large number, $n$ (notice that $n*\Delta=b-a$). Then the area of the entire space is the same as the sum of the areas of all the little rectangles. Notice that the height of each rectangle is going to be the value of the function at the x-value that rectangle is sitting on, so the area of an individual rectangle is length X width, $f(x_i)*\Delta$. Since we can get any level of accuracy here, it doesn't really matter where precisely this value of x lies.

So given $\Delta$ and $n$, the area $A\approx\sum_{i=0}^n f(x_i)*\Delta$ , where $x_i\approx a+\Delta*i$ . Now it should follow intuitively that the more rectangles you use, the more accurate the approximation will be. Therefore, we formally define the area by $\int_a^b f(x)dx=\lim_{\Delta\rightarrow 0}\sum_{i=0}^n f(x_i)\Delta$ . Interesting trivia for you: Sir Isaac Newton used a capital "S" to denote the sum of these rectangles, which got stretched so much we invented a new symbol for it, taking the strict $S_a^b$ to the new $\int_a^b$ . Likewise, we refer to the tiny widths or "differentials" by "dx" instead of $\Delta$.

Now, going back to our "intuition." You can see by the animation that the more rectangles we use, the less white space we get, therefore increasing our accuracy. So, let's start with an assumption: the area does exist and it is unique, call it A. Not saying we know what it is yet, just that it exists.

Now, let's say you want to find A with a precision of $\pm \epsilon$. That is, you seek to set up your system of rectangles such that $|\sum_{i=0}^n f(x_i)\Delta-A|<\epsilon$. Our intuition tells us that any level of accuracy can be achieved, given enough rectangles. To translate this into math language, for any given $\epsilon$, a $\delta$ can be found such that for all $0<\Delta<\delta$, the desired level of accuracy is achieved by this approximating sum.

The proof of this concept is quite visual, and the notation is never really consistent. Hope this helps, though.
• May 24th 2009, 09:49 AM
Scott H
You will find in mathematics that we sometimes talk about numbers without knowing what they are. :)
• May 24th 2009, 02:00 PM
VonNemo19
Could you go into a little more detail on how the assumption that the value $A$ exists, and how it is that by making this assumption, we can later prove that it, in fact, does exist. Also, how would you ever be able to choose an $\epsilon$ if you don't choose the correct value of $A$?

Perhaps all of my questions would be answered with an example. Here's something easy.

Use the definition of the riemann integral to show that $f(x)=2x$ is integrable on the closed interval $[0,1]$.

I would like this to be in the form of an $\epsilon-\delta$ proof. If that's possible.

Quote:

Originally Posted by Scott H
You will find in mathematics that we sometimes talk about numbers without knowing what they are. :)

Ok. Im not seeing it. We still have to choose an $\epsilon$ and we can't do that without KNOWING the value of $A$ How could we ever show that the thing is approaching A, without knowing what A is? That is my question. 'Round and 'round we goooooooo............(Dance)

So you're saying

Integrate first, ask questions later. If that is the only way, I'll accept it. But could someone please give me an example in terms of epsilon delta. Books are entirely devoid of this stuff.
• May 24th 2009, 05:44 PM
Media_Man
Getting a bit philosophical on us, here...
Quote:

Could you go into a little more detail on how the assumption that the value exists, and how it is that by making this assumption, we can later prove that it, in fact, does exist.
Take a bucket outside and fill it with an arbitrary amount of water. How many liters is it? Even though you don't know exactly, certainly this figure exists, right? Since it exists, give it a name, A. Now take a measuring cup and pour the water into the cup, exactly one cup at a time, counting how many cups it holds. This simple measuring system gives you 1 cup, 2 cups, 3 cups, etc., and with each "iteration" as a mathematician would call it, the number gets closer and closer to A. Voila, when the iteration is complete, we have calculated A exactly.

This is not so different than what we are doing here. We look at a curve and we know that there is area under it, and that the amount of area is an unquestionably unique number. We're coming up with an iterative system of approximating this area which after a sufficient number of iterations, gives us the answer to any level of accuracy. So no, we are not assuming A exists and using this assumption to prove it does indeed exist. We know it exists, and we are attempting to calculate it to an arbitrary level of accuracy.

Quote:

Also, how would you ever be able to choose an $\epsilon$ if you don't choose the correct value of A?
Here you are correct. In general, given a $\delta$ it is not possible to determine how many rectangles are necessary to find A with an accuracy within $\pm \epsilon$. This changes for different functions.
• May 24th 2009, 06:06 PM
Media_Man
A "tangible" example, whatever that may mean
$

\int_a^b f(x) dx=\lim_{\Delta\rightarrow 0}\sum f(x_i)\Delta
$

Let $f(x)=2x$, $a=0$, $b=1$, and $x_i=i*\Delta$. (Keep in mind the sum goes from $0$ to $n$, where $n\Delta=b-a=1$.)

$

\int_0^1 2x dx=\lim_{\Delta\rightarrow 0}\sum_{i=0}^{n} 2(i*\Delta)\Delta
$
= $\lim_{\Delta\rightarrow 0}2\Delta^2\sum_{i=0}^{n} i$

Now recall the formula we all learned in discrete math, the story that little Gauss discovered a shortcut to adding all the numbers from 1 to 100 with this formula: $\sum_{i=0}^{n} i = \frac{1}{2}n(n+1)$

So, $\lim_{\Delta\rightarrow 0}2\Delta^2\sum_{i=0}^{n} i$ = $\lim_{\Delta\rightarrow 0}2\Delta^2\frac{1}{2}n(n+1)$ = $\lim_{\Delta\rightarrow 0}\Delta^2(n^2+n)$

Now, remember $n\Delta=1$, so...

$\lim_{\Delta\rightarrow 0}\Delta^2(n^2+n)$ = $\lim_{\Delta\rightarrow 0}\Delta^2(\frac{1}{\Delta^2}+\frac{1}{\Delta})$= $\lim_{\Delta\rightarrow 0}(1+\Delta)$ = 1

Ergo, A=1 QED.

*Note: the invention of integrals and calculus was intended as a cohesive system of legal shortcuts to allow us $not$ to have to do these kinds of calculations.

**Notice the second to last step, that $A=\lim_{\Delta\rightarrow 0}(1+\Delta)$ . So if we want to know A within an accuracy of $\pm\epsilon$, well, $|1+\Delta-1|=|\Delta|<\epsilon$ . So in this case, for this function, $\delta=\epsilon$.

They say a picture is worth a thousand words. Well this post is getting a little convoluted even for me. (Giggle)
• May 25th 2009, 03:42 AM
HallsofIvy
You don't "assume that A" exists. And you certainly do not "later prove that it does exist".

You define a function to be "integrable" if and only if that A exists and then show how to find A if f is integrable- that is, how to find A if it exists.
• May 25th 2009, 04:23 PM
VonNemo19
Quote:

Originally Posted by HallsofIvy
You don't "assume that A" exists. And you certainly do not "later prove that it does exist".

You define a function to be "integrable" if and only if that A exists and then show how to find A if f is integrable- that is, how to find A if it exists.

You seem frustrated. You obviously find fault with Media Man's argument. It seemed sound to me. Could you justify your statement. I just want to understand All of the ins and outs. Yes, of course it stands to good reason that the integral is nothing more than the invers of differentiation, but I seldom rely on good reason. I like strict proof. So again, how 'bout this $A$ business......................?(Wondering)
• May 25th 2009, 06:47 PM
Media_Man
If you are asking for an elementary epsilon delta proof of the Fundamental Theorem of Calculus (that integration is the inverse of differentiation), that is a tall order. The proof is quite lengthy and detailed.

Here is a visual argument though. Consider $F(x)=\int_0^x f(t)dt$. According to the FTC, $F'(x)=f(x)$. Remember, $F(x)$ is a function telling you how much area is contained under the curve $f(t)$ on the interval $[0,x]$. Now picture what you get by $F(x+h)-F(x)$ for some small $h$: You get a narrow rectangular sliver of area of width $h$ and height $\approx f(x)$. The smaller $h$ is, the more accurate: $F(x+h)-F(x)\approx h*f(x)$. Hence, $f(x)=\lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}h$. Look familiar? (Surprised)

Quote:

You define a function to be "integrable" if and only if that A exists and then show how to find A if f is integrable- that is, how to find A if it exists.
Apologies. HallsofIvy is correct. Not all functions are integrable. But if a function is continuous on an interval, it is indeed integrable on that interval (the converse is not true). Forgive me, I have a knack for getting twisted in technicalities. A visual rule of thumb is that a function is integrable on an interval if and only if the function traps a finite amount of area under its curve inside that interval. Obviously, your example $f(x)=2x$ fills the bill, so we are safe in assuming A exists and we are not chasing a wild goose.
• May 25th 2009, 06:54 PM
Media_Man
Backing up through the proof, remember the last few steps, where $n\Delta=b-a=1$? Well, on the interval $[0,x],~~ n\Delta=x$. Observe:

$

\lim_{\Delta\rightarrow 0}\Delta^2(n^2+n)
$
= $

\lim_{\Delta\rightarrow 0}\Delta^2(\frac{x^2}{\Delta^2}+\frac{x}{\Delta})
$
= $

\lim_{\Delta\rightarrow 0}(x^2+\Delta x)
$
= $x^2$

So $F(x)=\int_0^x 2tdt=x^2$

This slightly more technical definition is shorthanded by $F(x)=\int 2xdx=x^2$. Again, the proofs are more technical than you think.
• May 25th 2009, 07:05 PM
VonNemo19
(Clapping)I'm liking you more and more, Media Man. I really appreciate your help here. Generalizing the integral of 2x from 0 to x was a good show indeed. However, you spoke of a tall order regarding a proof. Well, I simply love math, and one day, I like to pursue a doctorate in the subject, so tall orders are the only orders I like to give.

But, don't worry about it know. You've done enough. Thanks.

By the way, if you are any good at $\epsilon-\delta$ Proofs. I will be posting a lot of threads on the topic. It's just one of those things that I feel that I need to master.

Peace out.(Hi)