Talk:Integral

Needs section on AUC, or split AUC to its own page

"Area under the curve" redirects here, but this page does not define an AUC in terms of its use in statistics or give the reader an indication of how they should interpret an AUC when they first encounter one. — Preceding unsigned comment added by 145.117.146.70 (talk)

(Integral == area under the curve); True?

The following text in the introduction to this article got me wondering:

[...] the definite integral [...] is defined informally to be the net signed area of the region in the xy-plane bounded by the graph of ƒ, the x-axis, and the vertical lines x = a and x = b.

Given a function $f: \mathbb{R} \rightarrow \mathbb{R}_+, \ f \in C^{\infty} \$, is "the area under the graph" not also the formal definition? As far as I know, the motivation behind both the Riemann and Lebesgue integrals is measuring areas under curves and irregular volumes in a meaningful way. Furthermore, the ways I've seen Lebesgue/Riemann integrals developed and motivated usually emphasizes that definitions are consistent with areas or volumes.

Bottom line: let $f: x \mapsto f(x) \$ have the properties as above. Is "the area under the graph of $f \$ between $a$ and $b$" not a valid, formal definition of $\int_a^b f(x) dx \$? If not, why not?

Cheers! Trolle3000 [talk] 05:32, 23 June 2011 (UTC)

A single definite integral is *always* the area between the graph and the axis wrt integration.

Another way you could think of the definite integral is as a product of two averages: one is the average length of the infinitely many vertical lines in the region and the other is the interval width (infinitely many horizontal lines in a rectangle representing the area of the region).

A hardly known fact is that all integrals are indeed *line* or *path* integrals. As for Lebesgue theory - it is not required in any form or shape.

71.132.128.219 (talk) 21:16, 23 August 2011 (UTC)

I can see some problems with using "the area under the graph" as part of a formal definition of integration:

1. We only have a direct (i.e. non-calculus) way of calculating area for simple geometric shapes such as rectangles and triangles. To formally define the concept of the "area" enclosed by a general curve, you would have to approximate the region under the curve by a set of rectangles (or some other simple shape), add up the areas of the rectangles, then see if there is a limit as the width of these rectangles tends to zero. In effect, this is the formal definition of the Riemann integral - so this just introduces "area" as an intermediate concept into the standard definition.
2. Using "the area under the graph" as a formal definition for both Riemann and Lebesgue integrals does not explain why there are functions that are integrable under the Lebesgue definition but not under the Riemann definition - unles you say that "area" means different things in the two definitions, which is somewhat confusing.
3. It is not obvious how "the area under the graph" definition generalises to related concepts such as arc length integrals and contour integrals. Gandalf61 (talk) 08:05, 23 June 2011 (UTC)

I realize this simple definition will not hold when talking about integrals from $-\infty$ to $\infty$ or when we're dealing with limits of sequences of functions.

However, I think we can agree that a simple, closed, non-pathological curve encloses some well-defined property that could be called area, and that a curve homeomorphic to the real line has some well-defined property that could be called arc length? After all, those properties can be measured with either

1. a piece of string and a ruler, or
2. cardboard, scissors and good kitchen scales.

As long as we're talking about functions on compact intervals in $\mathbb{R}^3$, why shouldn't we be able to assign physical meaning to the integrals? Trolle3000 [talk] 07:10, 23 June 2011 (UTC)

That's what the informal bit in the lead is about. It is no way to go around a formal definition of an integral though, there is more than one definition and the area would be defined by the integral so it is a bit of a tautology. For instance in one definition the area can't be defined if all the points on the graph are one except for the rationals where it is zero. In others there's problems with the function going off to infinity or dealing with a path integral round a point in the complex plane. What on earth is the area under a complex number function? Dmcq (talk) 08:55, 23 June 2011 (UTC)

What I think the original poster wants to use is the following statement: If S is the region under the curve and μ is Lebesgue measure, then $\mu(S) = \int_a^b f(x)\,dx$. That's true. But the usual way of proving it is to prove that both sides are equal to $\iint_S d\mu$. That points out a philosophical difficulty with this approach, namely that you have to go up one dimension. So if you want to define surface integrals, then you interpret them as volumes of appropriate regions (specifically, regions in the normal bundle to the surface). If you want to define volume integrals, then you have to measure hypervolumes. Etc. One can do this in principle—Lebesgue n-measure is defined for any n, so the definition is not circular—but it would be, I think, messier than the usual treatment. And one would still want to have a description of which functions are integrable, limit theorems, and so on, and I think that would be more difficult in this framework because the definition of the integral is so indirect. Ozob (talk) 10:36, 23 June 2011 (UTC)

I guess the lead could mention that measure is a mathematical generalization of the concept of area and that might make some of the rest more accessible. Dmcq (talk) 17:12, 23 June 2011 (UTC)

I think what really bugs me is the word "informally" - it leaves the reader thinking that there is something more to it than area, and there really isn't - apart from all the math, of course! In the article about integrals at Mathworld it says:

An integral is a mathematical object that can be interpreted as an area or a generalization of area.

I like that statement. It is precise, and doesn't leave the reader wanting information. So I propose we edit the text in this article to:

Given a function ƒ of a real variable x and an interval [a, b] of the real line, the definite integral [...] can be interpreted as the net signed area of the region in the xy-plane bounded by the graph of ƒ, the x-axis, and the vertical lines x = a and x = b.

What do you say? Trolle3000 [talk] 17:27, 23 June 2011 (UTC)

Mathworld tends to be informal anyway. Even so you left one very important thing out from what Mathworld said - 'or generalization of area'. Without that you are left with the informal bit. There is more to it than area. When a mathematician generalizes a teacup can turn into a doughnut. Dmcq (talk) 17:51, 23 June 2011 (UTC)

@Dmcq, what you write is one of the things I love about math, but also the reason I think the formal/informal discussion should be left out of the introduction - no need to complicate things further. When dealing with a function $f \in C^{\infty}, f: [a,b] \in \mathbb{R} \rightarrow \mathbb{R}_+$ and an integral $\int_a^b f(x) dx$, why make it messier than necessary? In this case, how can you interpret the integral if not as the area below the graph? And why not tell that out loud? Trolle3000 [talk] 22:47, 23 June 2011 (UTC)

If f represents velocity, then I am much more inclined to interpret its integral as displacement. To my mind, displacement is a more accurate description of an integral than area. (For example, the interpretation as displacement is true even if f is not non-negative.) But even that is only valid in the one-dimensional case. Ozob (talk) 23:45, 23 June 2011 (UTC)

Now we are talking physics. If "f(t)" represents (non-negative) acceleration and "t" represents time, then the integral as well as the area below the graph represents speed. If "f(s)" represents (non-negative) force and "s" represents displacement, then both the integral the area below the graph represents work. But we are still dealing with an area, only with other units than "length x length" Trolle3000 [talk] 23:57, 23 June 2011 (UTC)

No, once you make those interpretations, the integral is no longer an area. The integral is only an area if f is interpreted as height above the x-axis. Ozob (talk) 10:27, 24 June 2011 (UTC)

History

I have removed the paragraph

That same century, the Indian mathematician Aryabhata used a similar method in order to find the volume of a cube.

since it has little in common with what the cited article states:

The formulas for the sums of the squares and cubes were stated even earlier. The one for squares was stated by Archimedes around 250 B.C. in connection with his quadrature of the parabola, while the one for cubes, although it was probably known to the Greeks, was first explicitly written down by Aryabhata in India around 500

Sasha (talk) 22:57, 2 January 2012 (UTC)

This article has been edited by a user who is known to have misused sources to unduly promote certain views edits (see WP:Jagged 85 cleanup). I searched the page history, and found 3 edits by Jagged 85. Examination of the sources used by this editor often reveals that the sources have been selectively interpreted or blatantly misrepresented, going beyond any reasonable interpretation of the authors' intent. Tobby72 (talk) 19:22, 19 January 2012 (UTC)

integration

integration of cos x/sin²x — Preceding unsigned comment added by 41.221.159.84 (talk) 15:58, 18 February 2012 (UTC)