Formal definition of the integral?

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I've just realized that often, especially is physics, integrals are used merely as syntactic sugar representing an infinite sum of the form: lim(dx->0) sum(f(x)*dx) where dx is delta-x. Take for example the simplified equation for center of mass: cm = I(p dm)/tm where I is the integral symbol, p is the position of a point mass relative to the centroid of the body, dm is the mass of the individual point masses (and so p dm is the first moment), and tm is the total mass. The integral in that equation, as far I know, isn't really calculating an area, volume, etc. It's merely a fancy way of writing an infinite sum of the above form. What I'm wondering is how the integral is formally defined. I've looked it up on the internet and found it to always be described as the area under the curve, volume, curl, etc. Is it correct to say that, in a very broad sense, the integral is just fancy syntax for infinite sums of the above form?

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I'm not a math expert but this is the way I look at it.

The indefinite integral is the infinite sum: lim(dx->0) sum(f(x)*dx) = F(x).
The definite integeral is F(b)-F(a) which also "happens" to represents the area under f(x) from x=a to x=b.

I think it's important to not forget that the integral is an infinite sum. I've seen many students distribute roots inside the integral. If they would have known the definition they would have known that this is a no-no.

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Well, the integral is the anti-derivative. That formula is the Riemann sum.

I think a mathematician would most likely be annoyed by making very broad generalizations.

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Quote:
 Original post by nilknI've just realized that often, especially is physics, integrals are used merely as syntactic sugar representing an infinite sum of the form:lim(dx->0) sum(f(x)*dx)where dx is delta-x. Take for example the simplified equation for center of mass:cm = I(p dm)/tmwhere I is the integral symbol, p is the position of a point mass relative to the centroid of the body, dm is the mass of the individual point masses (and so p dm is the first moment), and tm is the total mass.The integral in that equation, as far I know, isn't really calculating an area, volume, etc. It's merely a fancy way of writing an infinite sum of the above form.What I'm wondering is how the integral is formally defined. I've looked it up on the internet and found it to always be described as the area under the curve, volume, curl, etc.Is it correct to say that, in a very broad sense, the integral is just fancy syntax for infinite sums of the above form?

Pretty much. What you are describing is called the Reimann integral. There are other definitions of integrals and that leads to measure theory. But integration is all about adding stuff up.

Perhaps try the definition in wikipedia

-Josh

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 Original post by Name_UnknownI think a mathematician would most likely be annoyed by making very broad generalizations.

20th Century mathematics was mostly about generalization [smile]

-Josh

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Thanks for clearing that up.

jjd: Of all the many places I looked, wikipedia just wasn't among them! The definition provided there satisifed my curiosity. Thanks for the link.

Now that this great burden has been lifted I can continue with my life. [grin]

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Quote:
Original post by jjd
Quote:
 Original post by Name_UnknownI think a mathematician would most likely be annoyed by making very broad generalizations.

20th Century mathematics was mostly about generalization [smile]

-Josh

Ok, I think my Differential Equations professor would be very annoyed by broad generalizations ;-)

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Quote:
 Original post by Name_UnknownOk, I think my Differential Equations professor would be very annoyed by broad generalizations ;-)

Haha! [smile] I don't doubt... although I'm sure he finds the Sturm-Louiville theorem most satisfying [wink][wink]

-Josh

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Quote:
 Original post by nilknThanks for clearing that up.jjd: Of all the many places I looked, wikipedia just wasn't among them! The definition provided there satisifed my curiosity. Thanks for the link.Now that this great burden has been lifted I can continue with my life. [grin]

You're welcome! Calculus is actually a pet peeve of mine. I really don't like the way it is taught in most universities. One thing that I would love to see in a calculus course is historical material used to introduce and motivate the topics covered. The fundamental theorem of calculus is often presented after teaching techniques of integration and I think its a real anticlimax for the student. I mean, it's the FUNDAMENTAL THEOREM!!! But the fact that the area under a curve and the slope of the curve are so tightly related is truly weird! I mean, why would you expect this?!

OK, I'll hold my rant in... [wink]

-Josh

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I think it would be strange if the shape of the curve had nothing to do with the area under it.

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Quote:
 Original post by LilBudyWizerI think it would be strange if the shape of the curve had nothing to do with the area under it.

Sorry, I think you're missing the point [smile]

-Josh

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Perhaps you are missing mine which is that the slope defines the shape. I think that is a concept that could be introduced pretty much as soon as the slope-intercept form of the equation for a line is introduced.

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Have a look at this:
http://mathworld.wolfram.com/RiemannIntegral.html

I think the moderator should put a link to mathworld from this forum.

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Quote:
 Original post by LilBudyWizerI think it would be strange if the shape of the curve had nothing to do with the area under it.

Quote:
 Original post by LilBudyWizerPerhaps you are missing mine which is that the slope defines the shape. I think that is a concept that could be introduced pretty much as soon as the slope-intercept form of the equation for a line is introduced.

Well, actually you're just proving my original point. It is easy to be told that this is the way things are and believe them. That was not the case for Newton or Leibnitz.

The first person known to have integrated a function was Abu Ali al-Hasan ibn al-Haytham (also know as Alhazen) some time around 970 BC. It would be remiss to not point out that Archimedes (287 - 212 BC) integrated all sorts of spirals and curves, and that he also showed how to calculate the slopes of tangents to spirals.

Calculating derivatives and integrals were not new to Newton or Leibnitz. The tools you are talking about have been around for a very long time. So to suggest that the concept of a slope, and how it relates to a curves shape, should easily lead to calculus, is pretty naive to say the least.

However, the calculations preceeding Newton and Leibnitz only applied to special cases and limited classes of functions. What Newton and Leibnitz showed was a way to make the same calculations for almost any function conceivable at the time using a single method, which was actually far simpler than some of the other, more limited, techniques. That is why their calculus is hailed as such a phenomenal achievement and why you do not learn the "method of exhaustion" that Archimedes used. And that is why I say that the fact that slope and area of a curve are so tightly, and perhaps I should also say "simply", related is incredible! Furthermore, I did not suggest that the slope of a curve (or shape if you prefer) would not be related to the area under the curve, but that the relationship can expressed so succinctly (tightly) and with such an enormous amount of generality is, IMO, mind-blowing.

-Josh

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Quote:
Original post by jjd
Quote:
 Original post by LilBudyWizerI think it would be strange if the shape of the curve had nothing to do with the area under it.

Quote:
 Original post by LilBudyWizerPerhaps you are missing mine which is that the slope defines the shape. I think that is a concept that could be introduced pretty much as soon as the slope-intercept form of the equation for a line is introduced.

Well, actually you're just proving my original point. It is easy to be told that this is the way things are and believe them. That was not the case for Newton or Leibnitz.

The first person known to have integrated a function was Abu Ali al-Hasan ibn al-Haytham (also know as Alhazen) some time around 970 BC. It would be remiss to not point out that Archimedes (287 - 212 BC) integrated all sorts of spirals and curves, and that he also showed how to calculate the slopes of tangents to spirals.

Calculating derivatives and integrals were not new to Newton or Leibnitz. The tools you are talking about have been around for a very long time. So to suggest that the concept of a slope, and how it relates to a curves shape, should easily lead to calculus, is pretty naive to say the least.

However, the calculations preceeding Newton and Leibnitz only applied to special cases and limited classes of functions. What Newton and Leibnitz showed was a way to make the same calculations for almost any function conceivable at the time using a single method, which was actually far simpler than some of the other, more limited, techniques. That is why their calculus is hailed as such a phenomenal achievement and why you do not learn the "method of exhaustion" that Archimedes used. And that is why I say that the fact that slope and area of a curve are so tightly, and perhaps I should also say "simply", related is incredible! Furthermore, I did not suggest that the slope of a curve (or shape if you prefer) would not be related to the area under the curve, but that the relationship can expressed so succinctly (tightly) and with such an enormous amount of generality is, IMO, mind-blowing.

-Josh

Perhaps I have a basic missunderstanding, but I dont understand why you think it is so important to relate the slope to the integral.
The definition of the rieman integral does not include the slope or derivative in it, and I think for a reason.
You can have functions, not only they do not have a slope, but they also are not continuos. Which you can calculate their rieman integral.
I believe there are also such functions with infinity amount of points(Countable, and in a finity fragment) with no continuity, which you can calculate their rieman integral.

-josh

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"Perhaps I have a basic missunderstanding, but I dont understand why you think it is so important to relate the slope to the integral."

Yup, you have an incredibly complete and basic misunderstanding.

The briefest look at the equation d = r * t makes the misunderstanding clear. Rate of change and area are obviously and necessarily intimately related.

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Quote:
 Original post by The C modest godPerhaps I have a basic missunderstanding, but I dont understand why you think it is so important to relate the slope to the integral.The definition of the rieman integral does not include the slope or derivative in it, and I think for a reason.You can have functions, not only they do not have a slope, but they also are not continuos. Which you can calculate their rieman integral.
Correct. But the Riemann Integral was invented in the 19th century, while the Fundamental Theorem of Calculus was first observed by Newton and Leibnitz in the 17th century, and the proof of it relied on an informal geometrical argument which did not consider the pathological curves which you mention here.

Still, the Fundamental Theorem of Calculus is enormously important. Many of the functions we are interested in have antiderivatives on some interval, and for these curves the FTC connects the problem of finding areas to the problem of finding slopes.

The FTC is also part of the reason why Calculus is called what it is. The FTC showed that there is a systematic method for solving many area problems. The Method of Exhaustion which jjd mentions above does nothing of the sort: It required that you had already solved the area problem before you started, and just gave you a technique for proving that your solution was correct (though the logic behind the Method of Exhaustion was virtually watertight; Newton and Leibnitz' calculus was sloppy in comparison).

Quote:
 I believe there are also such functions with infinity amount of points(Countable, and in a finity fragment) with no continuity, which you can calculate their rieman integral.
The definition of a Riemann Integral requires a function to be defined at all points on an interval.

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I think it's important to note here that even though the finite integral of a function happens to coincide with the area under the graph of the function, integration has nothing to do with geometry in general. Geometry is merely the field where we first stumbled upon integraion and differentiation from the beginning.

Integration and differentiation are mathematical operations that exists on their own outside specific fields in mathematics, much like multiplication and division. In fact it is very useful to view integration and differentiation as continous analogues to multiplication and division.

The fundamental theorem of calculus is remarkable indeed, but not more remarkable than the fact that multiplication and division are each other's inverse operation, or addition and subtraction.

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"The definition of a Riemann Integral requires a function to be defined at all points on an interval."

False. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.

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Quote:
Quote:
 I believe there are also such functions with infinity amount of points(Countable, and in a finity fragment) with no continuity, which you can calculate their rieman integral.
The definition of a Riemann Integral requires a function to be defined at all points on an interval.

Sigh. It is the case that Riemann-integrable functions may be discontinuous at an infinite number of points. A correct statement is: A Riemann-integrable function is continuous except possibly on a set of measure zero.

Intelligent discussion of the limits of Riemann integration requires an understanding of the Lesbeque and the Stieltjes integrals (they "solve" different "problems" with Riemann's version).

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Quote:
 Original post by sherifffruitflyFalse. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.
It's a degenerate case, where definitions often disagree. Still, I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

The theorem you are thinking to justify otherwise will state that any function continuous on a closed interval is integrable on that interval, not that a function which is everywhere continuous is everywhere Riemann-integrable.

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Original post by NotAnAnonymousPoster
Quote:
 Original post by sherifffruitflyFalse. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.
It's a degenerate case, where definitions often disagree. Still, I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

The theorem you are thinking to justify otherwise will state that any function continuous on a closed interval is integrable on that interval, not that a function which is everywhere continuous is everywhere Riemann-integrable.

You can say that all you want - it's just false. Definitions do not vary on this to any substantial degree (and hence neither to the theorems). ROFL - you think there's substantial variance in exactly *what* the Riemann integral is? LMAO.

And there's nothing "degenerate" about my example. It's a perfectly fine continuous, differentiable, and Riemann-integrable function.

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Quote:
 Original post by sherifffruitflyYou can say that all you want - it's just false. Definitions do not vary on this to any substantial degree (and hence neither to the theorems). ROFL - you think there's substantial variance in exactly *what* the Riemann integral is? LMAO.
Not for the Riemann integral, no. Which is why I said:
Quote:
 Original post by MeI think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.
Quote:
 Original post by sherifffruitflyAnd there's nothing "degenerate" about my example. It's a perfectly fine continuous, differentiable, and Riemann-integrable function.
I wasn't clear there. I meant that your function f is defined on the degenerate interval [0,0].

But we won't get anywhere here without actually supplying some definitions. Try this one at PlanetMath:

Riemann Integral

The definition requires that the function in question be defined on a non-degenerate interval.

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Quote:
Original post by NotAnAnonymousPoster
Quote:
 Original post by sherifffruitflyFalse. A Riemann-integrable function's domain does not have to include an interval. Consider the function f, defined only at 0, such that f(0)=0. This is Riemann-integrable (everywhere continuous and differentiable even). But its domain does not include any interval.
It's a degenerate case, where definitions often disagree. Still, I think most definitions of Riemann-integral will entail that f in this case is not Riemann-integrable.

The theorem you are thinking to justify otherwise will state that any function continuous on a closed interval is integrable on that interval, not that a function which is everywhere continuous is everywhere Riemann-integrable.

Moreover, the theorem I was thinking of was not the one you tried to putin my mouth. The one I was after was the one I actually said - that the Riemann-integrable functions are those which are continuous except on a set of measure zero.