# A Fresh Start

#### Cyrus

Ok guys, about two weeks ago I posted asking about the fundamental theroem of calculus and the use of the dummy variable. Ive had some time to clear my head. I want to take a fresh new approach to it. I have reread what you guys put and I will just write down what I think. Please be patcient and thanks in advance for putting up with my B.S.

Ok, so first things first. Lets start with a standard integral evaluated from a to b.

$$\int^b_a f(x)dx$$

In order to do this, we plug in values between a and b that vary, and sum it over the interval. When we do this, we start with the lower bound, a, and plug it in and evalute it. We multiply this buy dx, which is a small piece of the parameter domain, and repeat the process for the next increment. Notice that in this process, dx was the value of a small increment in the parameter domain. It is usually the limit of delta x, as x goes to zero. No where in this definition of dx, did the value of x go anywhere in the x part of dx. For the first value of a, we did not put f(a)da. Because that is now how we defined the use of the symbol dx. dx has a very specific meaning. Although there is an x in the dx, it does NOT mean put in a value where you see the x in the term dx.

Now lets move on to the FTC: Here we have:

$$F(x)= \int^x_a f(x)dx$$

Hopefully at this point, we all agree that this notation is wrong. Lets see why. Lets use crosson and mattgrime's point of evaluating this function at the point x=3. Here we have a FUNCTION. When we normally see a FUNCTION , our standard procedure says to put in a value everywhere we see an x. If we choose to evaluate the point x=3, then we would have to do the following:

$$F(3)=\int^3_a f(3)d3$$

Notice that we had to put in a 3, for the term d3. But now we have an integral where the function ONLY ever takes on one value, f(3). ALSO, we have a term called d3. What is d3? d3 is nothing, its pure garbage. In fact, the way we defined an integral in the first example tells us that the dx part should NEVER EVER have a value in that x part of dx. Even though dx has an x in it, its defined such that you dont PUT a value in that x, PEROID, EVEN IN STANDARD INTEGRALS THAT DO WORK!

Ok, so now what. Since im a smarta$$. Lets say, ok, I know what an integral is and what dx REALLY means, and based on its definition, I have to leave that x in the dx alone. So lets say I CHOOSE to do that. $$F(3) = \int^3_a f(3)dx$$ But now look what I have done. I STILL get an integral, which is now something I CAN POSSIBLY evalute, BUT what does the integral represent? Now it just means im summing f(3), (3-a) times. AGAIN, this is NOT what I had set out to do in the fist place. So EVEN IF I try to be smart about saying, wait a minute, d3 violates the definition of an integral, leave that part alone as dx, I STILL get something that makes no sense. The only way to to satisify ALL THREE of these conditions, is to introduce a separate dummy variable. THEN and ONLY then does it give me what I want, an integral that can be though of as an area up to here, where 'here' is the value of x. (Not to be rude, but this post is really a quesiton for MattGrime, so please let him respond first if you dont mind) Last edited: #### matt grime Science Advisor Homework Helper cyrusabdollahi said: Ok, so first things first. Lets start with a standard integral evaluated from a to b. $$\int^b_a f(x)dx$$ In order to do this, we plug in values between a and b that vary, and sum it over the interval. i'd rather you didn't use the word sum since you are only "summing by analogy" When we do this, we start with the lower bound, a, and plug it in and evalute it. We multiply this buy dx, which is a small piece of the parameter domain, and repeat the process for the next increment. no that is not what we do. Notice that in this process, dx was the value of a small increment in the parameter domain. no it isn't. It is usually the limit of delta x, as x goes to zero. No where in this definition of dx, did the value of x go anywhere in the x part of dx. For the first value of a, we did not put f(a)da. Because that is now how we defined the use of the symbol dx. dx has a very specific meaning. yes it does though you don't appear to know what it is Although there is an x in the dx, it does NOT mean put in a value where you see the x in the term dx. Now lets move on to the FTC: Here we have: $$F(x)= \int^x_a f(x)dx$$ no we do not. you think writing this is meaningful, we do not Hopefully at this point, we all agree that this notation is wrong. Lets see why. Lets use crosson and mattgrime's point of evaluating this function at the point x=3. that isn't "my point". My point is that you are just randomly misassigning meaning to various symbols and hoping you new meaning can be interpreted. if you want to make sense of that set of symbols feel free to do so, we can think of no way of doing so. the notation is wrong in the sense that it corresponds to nothing we use and has nothig to do with the FTC. if you want to give it a meaning feel free. Here we have a FUNCTION. When we normally see a FUNCTION , our standard procedure says to put in a value everywhere we see an x. If we choose to evaluate the point x=3, then we would have to do the following: $$F(3)=\int^3_a f(3)d3$$ Notice that we had to put in a 3, for the term d3. But now we have an integral where the function ONLY ever takes on one value, f(3). ALSO, we have a term called d3. What is d3? d3 is nothing, its pure garbage. In fact, the way we defined an integral in the first example tells us that the dx part should NEVER EVER have a value in that x part of dx. Even though dx has an x in it, its defined such that you dont PUT a value in that x, PEROID, EVEN IN STANDARD INTEGRALS THAT DO WORK! if you understood what the dx signified this would be unnecessary. Ok, so now what. Since im a smarta$$. Lets say, ok, I know what an integral is and what dx REALLY means,
you don't

and based on its definition, I have to leave that x in the dx alone. So lets say I CHOOSE to do that.

$$F(3) = \int^3_a f(3)dx$$

But now look what I have done. I STILL get an integral, which is now something I CAN POSSIBLY evalute, BUT what does the integral represent?
why must it represent anything?

Now it just means im summing f(3), (3-a) times.
no you are not adding anything up 3-a times

AGAIN, this is NOT what I had set out to do in the fist place. So EVEN IF I try to be smart about saying, wait a minute, d3 violates the definition of an integral, leave that part alone as dx, I STILL get something that makes no sense.
the integral now makes sense but is nothing special

[quoteThe only way to to satisify ALL THREE of these conditions,
what 3 conditions?

is to introduce a separate dummy variable.[/quote

but this is how integrals are defined. all you are doing is using the notation properly.

THEN and ONLY then does it give me what I want, an integral that can be though of as an area up to here, where 'here' is the value of x.

(Not to be rude, but this post is really a quesiton for MattGrime, so please let him respond first if you dont mind)

what is the point of this?

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#### Galileo

Homework Helper
Cyrus, you are causing unnecessary confusion. You should interpret
$$\int_a^b f(x)dx$$
as one symbol, as a whole and it's simply a number. There's a reason for this notation, but it may be best not to assign a specific meaning to dx.

As I'm sure you know:
$$\int_a^b f(x)dx=\int_a^b f(t)dt=\int_a^b f(\Omega)d\Omega=\int_a^b f(\heartsuit)d\heartsuit$$
It's just a dummy variable.

When you write:
$$F(x)=\int_a^x f(t)dt$$
F is a function of x. x is the independent variable. The t is a dummy variable, I could use whatever symbol I want instead of t, EXCEPT x! Because x already HAS a meaning. If you write x instead of t, you are assigning 2 different meanings to the same symbol and that is where your confusion started, the rest followed.

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#### Cyrus

Ok Matt, if you do not mind, I would like to step through each of your complaints to understand each problem.

First, you said

i'd rather you didn't use the word sum since you are only "summing by analogy"
By this, do you mean we do a limit of the riemann summation? I thought that is how math texts defined the integral, a limit of the riemann summation, I guess I was wrong.

Next you say,

no that is not what we do.
Again, I thought you evalute the function at a point, and multiply it by a small piece of 'delta x', add it to the sum, then increment by delta x, and repeat the process. But you let the limit of delta x approach zero while doing this process. I guess I was wrong there too.

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#### CrankFan

cyrusabdollahi said:
Again, I thought you evalute the function at a point, and multiply it by a small piece of 'delta x', add it to the sum, then increment by delta x, and repeat the process. But you let the limit of delta x approach zero while doing this process. I guess I was wrong there too.
I'm guessing that the iterative process you're thinking of (evaluate, add, increment, repeat) is related to a source that defines integrals as Riemann Sums and shows that the area under the curve can be approximated by the sum of rectangular regions. It might also show examples of approximate area calculations with a fixed number of rectangles.

The purpose of those descriptions aren't to teach you how to calculate a particular integral but to show you an idea that will lead to a means of calculating integrals. It's also there to help convince you that the approach being considered will lead to results that agree with your intuition.

Thus, the "process" being described isn't carried out (except maybe for a few cases with very few rectangles just so that you get the idea behind it). You're supposed to become convinced that if it did get carried out, that is if the number of approximating rectangles increased and their widths got smaller, then the area of all of those infinitely thin rectangles of height f(x) would be the best possible approximation of the area under the curve.

Calculating a Riemann sum requires some basic algebraic manipulation, no iterative summing process is needed. Various techniques exist which allow us to calculate (elementary) integrals directly, without considering rectangles or Riemann sums

#### Cyrus

Thus, the "process" being described isn't carried out
If an integral is not really a limit on the riemann sum, then what IS actually taking place?

Calculating a Riemann sum requires some basic algebraic manipulation, no iterative summing process is needed.
But it is an iterative process, in a sense. Each point you evalute it is deltaX away. I.e [f(x)+f(x+deltax)+...f(n)] delta X. The value of f(x) is incremented in steps by deltax, and added.

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#### CrankFan

cyrusabdollahi said:
If an integral is not really a limit on the riemann sum, then what IS actually taking place?
I'm not saying that an integral can't be thought to be the limit of the Riemann sum. I'm saying that it doesn't require an iterative process to evaluate the limit of that Riemann sum.

cyrusabdollahi said:
But it is an iterative process, in a sense. Each point you evalute it is deltaX away. I.e [f(x)+f(x+deltax)+...f(n)] delta X. The value of f(x) is incremented in steps by deltax, and added.
* The process of approximating the area under the curve that you're referring to is carried out when deltaX is fixed. As an exercise, people can compute approximations in that tedious fashion.

* The observation that the approximations get better as deltaX gets smaller, isn't a process to calculate an integral. It's an argument to convince you that what we agree to define the integral as gives the best possible approximation. Once we accept that definition, evaluating an integral is no more tedious than evaluating the limit of a Riemann sum. Evaluating that limit does not entail performing an iterative process of the type you alluded to (in post #4).

#### Cyrus

Right, to be clear. You are saying that the integral is equal to the LIMIT of the riemann sum? If so, I do not disagree with you.

#### Crosson

After reading Cyrus's first post in this thread, I am convinced that he understands the point about dummy variables that we have been trying to show him.

As a side note, I started that whole business about "d3" to make a point in what I hoped would be a funny way. It has been quite surprising to see the responses, in general people seem to have a viscious attitude towards this style of teaching. Maybe it is just because it is difficult to convey sarcasm over the internet.

#### Cyrus

Thank you crosson. Sometimes I do get carried away, and I need smart people like you to smack me silly! But im glad we are on the same page :-) Matt and I still seem to have some disagreement, but I look forward to hearing his input in helping me out.

#### CrankFan

cyrusabdollahi said:
Right, to be clear. You are saying that the integral is equal to the LIMIT of the riemann sum? If so, I do not disagree with you.
I was also saying that statements like the following, in reference to a definite integral, indicate confusion.

"When we do this, we start with the lower bound, a, and plug it in and evalute it. We multiply this buy dx, which is a small piece of the parameter domain, and repeat the process for the next increment"

Which isn't a bad thing (not knowing), I just thought it might help to point that out .... in the case that you that felt Matt's "no that is not what we do.", and "no it isn't." weren't verbose enough.

#### Cyrus

Ok, does it make you happy if i say, we can think of it that way? Because, after all, the integral is DEFINED EXACTLY as the limit of this process. I was just trying to convey my idea by describing the process. I know that you dont ACTUALLY do that when you do the integration, i.e. an iterrative process.

#### matt grime

Homework Helper
There is a difference between a (numerical) integral being the limit of a sum (which it is) and it actually being a sum itself (which it is not in this sense), and dx is not a little piece of the domain. in this case it is simply a reminder to tell you what you're integrating and from whence the idea came. you are not mutlplying anything by dx's and adding things up.

I believe Crosson (and his post here reinforces that view) was attempting to show you simply that your notation was bad by putting '3 in for x' in a nice informal way. I don't see the point of all this. NOtation is simply notation use it and understand it, that is all.

the riemann integral of a function, when it exists, is the limit of the upper (and lower) riemann sums, note we are not multiplying values of f by little bits dx and adding them up. i am certainly happy to think of it as being *like* a sum, a continuous version of a discrete object, but i don't think it necessarily helps for you to describe it *as* a sum and talk about it as if it is a discrete thing.

anyway, amazingly this is sufficient to prove the fundamental theorem of calculus.

one thing i don't think you appreciate is that when we say

$$int_a^xf(x)dx$$

doesn't mean anything we simply mean that it doesn't mean anything. this is different from 'it can never mean anything'. we mean that you are making a clash with our conventions. we cannot interpret it, we don't understand it. we have notation that means definite integral and indefinite integral and you are mixing them and we don't know what the result is supposed to mean. if you want to tell us you have a meaning for it fine, what is it?

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#### Cyrus

Yes matt. I agree with you. It is meaningless, but as crosson was saying. I was being sarcastic by giving it meaning to help prove to myself that it STILL wont work out if you TRY to give it meaning. I agree, int^x_a f(x)dx is trash. EVEN if you TRY to make it have meaning it STILL WONT WORK PEROID. See what Im doing? I said ok, leave the dx alone, now all you have is the integral of a constant function f(3)dx, from a to 3, which is equal to f(3) *(3-a), but you said no its not. Sure it is. Its the integral of a constant funciton. But thats not what we set out to do in the first place. So by me TRYING to give it meaning, I still proved to myself it wont work, end of story, no matter how you try to fudge evaluating it. Does that help?

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#### matt grime

Homework Helper
i didn't say that int_a^3 f(3)dx was not f(3)(3-a) indeed i said it was a long time ago. i did take issue with you stating that you were adding up f(3) 3-a times since that is nonsense. your last post only serves more to make me ask what the point of this thread is.

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#### Cyrus

Crosson got my point, and ill settle with that. No hard feelings matt, thanks for your help.

#### matt grime

Homework Helper
and the point of this thread was....?

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