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Dec 12th, 2002, 03:57 PM
#1
Thread Starter
Hyperactive Member
Integral Definition
Give me a clear definition for someone that does not know anything about Calculus.
What is that curve thing?
This thing:
. _
..\
_/
53323737 15 743 313402 05 740313063. 17 15 4150 743 313402 05 140393403437 5203 743 30210.

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Dec 12th, 2002, 04:26 PM
#2
What kind of riddle is that? I can only see a number of lines and points, no curve thing.
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Dec 12th, 2002, 05:05 PM
#3
New Member
Sorry but my english is not very good, and I am not sure about giving you this explanation correctly in english, so, i will answer you in spanish!!
Aunque no se ve muy claro, por el título de tu mensaje, deduzco que se trata del símbolo de la integral que es algo así como una "S" bastante estirada verticalmente. Bien, esto es un operador matemático que, básicamente, denota el area comprendida entre la función que sigue al símbolo de integral y el eje de coodenadas. (Area, si es una función real; pero si fuera una función con llegada en RxR sería el volumen y así sucesivamente)
Esta sería la idea gráfica, pero algebaicamente hablando se trata de la operación contraria a la derivación (del mismo modo que la división es la operación contraria a la multiplicación)
Sin embargo existen varios matices importantísimos en el tema de integrales que hacen el tema muy extenso, pero a modo intuitivo te puedo decir que las hay "definidas", "indefinidas", "iteradas", "reales", etc... Además no siempre tienen solución analítica, aunque sí numérica. Pero para el calculo numérico de éstas hacen falta técnicas que sólo suelen estar al alcance de los matemáticos.
Entiendo que esto no es más que una introducción y que puede parecer lioso, pero espero servirte de ayuda, aunque debo decir que es un tema MUUY extenso y que no está al alcance de todos.
(Para entender integración debes tener muy bien controlado lo referente a derivación)
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Dec 13th, 2002, 03:04 AM
#4
Addicted Member
Well let me try in English but in non-calculus terms.
Consider a sheet of graph paper and draw your self an origin and an x & y coordinate system. Now draw an arbitrary line lets say slanting up to the right at about 45 degrees. Make that as long as you like then drop a perpendicular to the x axis from its end.
So now you have a triangle and what we need- for whatever reason- is its area. The easiest way since you've got graph paper is to count the squares. Think of it as lots of columns- each column is as wide as the small squares on your paper, perhaps 1mm or maybe 1/10", whatever. So you can say that each column is that wide, and as high as the slanting line at that x-distance from the origin.
Since the height of the column is a function of the equation of the line (in this case y=mx+c 'cos it's a straight line) you can calculate the height as opposed to having to measure it, and the area is the sum of the heights of all columns times the width.
In that case, with a discrete width (eg the 1mm or whatever) we would consider the area to be the capital sigma. We'd call the column width delta-x.
Problem is , if you draw this, is that at the top of the column you've got a triangle and we're actually not 100% accurate 'cos we've counted as area, a small piece actually outside.
So make the column infinitesimal ; then our delta-x is written as dx and the exact height at any x value, times dx gives us the exact area of that skinny column. When we add them up, we get the exact area, and that's when we use the elongated 's' which you failed so misreably to draw earlier.
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Dec 13th, 2002, 03:36 PM
#5
Thread Starter
Hyperactive Member
Okay that's understandable. But how could you times dx with x and get an area? Does dx=exact hieght?
53323737 15 743 313402 05 740313063. 17 15 4150 743 313402 05 140393403437 5203 743 30210.

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Dec 13th, 2002, 10:44 PM
#6
transcendental analytic
let dx be just a number. troughout the function f you with an interval D measured y=f(x) for x, and sum upp the rectangles with side dx and height y=f(x), this is called the undersum, an approximation of the area below the function, (negative area if below 0) now similarly you can start at y=f(D+x) and get the oversum, which are rectangles above the function when it raises. Now if you decrease D the areas will get more accurate, and when dx is infinitely small, you can prove that there's exactly one value I for the following equation
S1(D)>= I >= S2(D)
where S1 and S2 are the over respective under sums of the rectangles with width D and interval D, when D is an infinitely small number. We call this dx, and I we call the intergral for f.
Use  
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To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Dec 13th, 2002, 10:56 PM
#7
Fanatic Member
A spanish post cannot be appreciated well enough simply because no one could understand it! well i took the trouble to translate the above spanish paragraph to english -- the translation isn't very good but it'll do:
Although it is not seen very sure by the title of your message, I deduce that one is the symbol of the integral that is something as well as a "S" stretched enough vertically. Well, this is a mathematical operator who, basically, denotes the area between the function that follows the symbol of coodenadas integral and the axis of. (Area, if it is a real function; but if outside a function with arrival in RxR would be the volume and so on) This would be the graphical idea, but algebaically speaking it is the opposite operation to the differentiation (in the same way that the division is the opposite of multiplication) Nevertheless the several most important points in the subject of integrals make intergration a very extensive topic, but intuitively I can say to you that there are these "defined", "indefinite", "iteration", "real", etc... In addition they do not always have analytical solutions, although they are numerical. But from numerical calculations of these lacks technicality that usually are only within reach of mathematicians. I understand that this is not more than an introduction and that can seem vague, but I hope it serves you as aid, although I must say that it is an extensive subject and that it is not within reach of all. (to understand integration you must have to know differentiation very well).
I (Bugz) disagree with the last line. Mrs Romiens once told me that sometimes it is also a good idea to teach intergration before differentiation.
Massey RuleZ! ^-^__  Cheers!  __^-^ Massey RuleZ!
Did you know that...
The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!
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Dec 20th, 2002, 04:45 PM
#8
Addicted Member
So you will take Mrs. Romiens' word over Mr. White's?
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Dec 20th, 2002, 04:59 PM
#9
Fanatic Member
yawn i do whatever i like and no one is going to dictate what I do or how I feel.
Massey RuleZ! ^-^__  Cheers!  __^-^ Massey RuleZ!
Did you know that...
The probability that a random rational number has an even denominator is 1/3 (Salamin and Gosper 1972)? This result is independently verified by me (2002)!
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