Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Any proof for the definition of the definite integral

  1. Mar 30, 2004 #1
    Hello all, new poster here.

    When learning about the definition of the definite integral, a few books that I have read through first define anitdifferentiation, then explains some Riemann sums and then

    POOF

    [tex]\int_{a}^{b} f(x) dx = \lim_{||\Delta||\rightarrow\0}\sum_{i=1}^n f(\xi_i) \Delta_i x[/tex]

    So is there a proof for this or do I just gotta accept it? Please help with the integral definition.
     
    Last edited by a moderator: May 3, 2018
  2. jcsd
  3. Mar 30, 2004 #2

    HallsofIvy

    User Avatar
    Science Advisor

    Depends upon exactly what you mean. One doesn't have to "prove" definitions: they are given. Of course, one might want to prove that the definition actually gives you something worth while! In this case, it doesn't, necessarily: there exist many functions that are N0T "integrable" and for which the definition of the definite integral doesn't make sense.

    Probably you will have to get a lot of calculus under your belt and then take either an "Advanced Calculus" or a "Mathematical Analysis" course to see a real proof that continuous functions are always integrable. (But there exist non-continous functions that are also integrable. I didn't see a theorem stating exactly what functions are integrable until I was in graduate school.)

    Roughly speaking- it goes like this: divide your x-axis (from a to b) in to segments and, on each segment choose x* so that f(x*) is the largest possible on that segment. Turn around and choose x* so that f(x*) is the smallest possible on that segment. If x is any point on the segment, then by definition f(x*)<= f(x)<= f(x*) so,taking the length of the segment to be &Delta; x, f(x*)&Delta; x<= f(x)&Delta; x<= f(x*&Delta; x and then &Sigma;f(x*)&Delta; x<= &Sigma;f(x)&Delta; x<= &Sigma;f(x*&Delta; x .

    IF f is continuous then you can show that, in the limit, x* goes to x* so the two end sums go to the same thing. Since the sum for ANY choice of x is "trapped" between them, it must also give the same result: the definite integral.

    By the way, if you think of this in terms of "area under the curve", notice that x* always gives rectangles that include the area while x* always gives rectangles that are included in the area so that the "area" always lies between the two. Of course, the area is a constant so if the two limits are the same then that common value must BE the area. That is, one can show that the definite integral gives the area under the curve without having to give a precise definition of "area"!

    The proof that one can get the definite integral by evaluating the anti-derivative at the endpoints is a different matter and is quite often given in calculus books.
     
  4. Mar 30, 2004 #3
    The whole is equal to the sum of it's parts.
     
  5. Mar 30, 2004 #4
    And that makes perfect sence for the Riemann sums, but since when did taking a function and increasing its exponent value by one and divide that by the value in the final exponent equate to the area under the curve (simple case). That is where there is a leap of faith for me.

    With Simpson's rule, one uses some algebra and calculus to prove that the area under the curve is

    1/3 h (y0 + 4y1 + y2)
     
  6. Mar 30, 2004 #5
    This page seems to provide some interesting historical information about the derivative and finally the integral. It's a bit heavy depending on your level of experience, but very wordy, so stick with it until the end.

    http://www.math.wpi.edu/IQP/BVCalcHist/calc2.html

    Section 2.5 The Ellusive Inverses – the Integral and Differential is what caught my eye...seems to target your curiosity.
     
  7. Mar 30, 2004 #6
    I'll leave it to someone more qualified then I am to give you rigorous proof.

    Take a right triangle with two legs of 5 inches each. Take the length of one side(5 inches) and " increase it's exponent by one and divide that by the value in the final exponent" to find it's area. Now stop and look at what you have done, and spend a little time thinking about it. Is this a big leap of faith?
    Would you rather use your Riemann sum here?

    If you are talking about a position function, there are two ways to arrive at the total distance traveled:
    1.) The summation of an infinite number of infinitesimal distances.
    2.) subtracting the intial position from the final position.

    They both result in the same answer.

    The Riemann sum with yield the same result as the definite integral, if it is carried to completion(When delta x = dx). This will take you a long, long, time...

    -Mike
     
    Last edited by a moderator: Mar 30, 2004
  8. Mar 30, 2004 #7

    HallsofIvy

    User Avatar
    Science Advisor

    If this is what you really meant to say then you are way off. You can't "prove" that because, in general, it isn't true. Except for curves of polynomials of degree 2 or less, Simpson's rule gives only an approximation to the area.

    The idea that the area under the curve is the inverse of finding the derivative (which is what I guess you mean "increase it's exponent by one and divide that by the value in the final exponent") is not a leap of faith- the proof is given in every calculus book- it's a fairly straightforward application of the mean value theorem.

    I wondered for a moment what "increase it's exponent by one and divide that by the value in the final exponent" had to do with triangles- it's interesting that it DOES work in the case you give! If you have right triangle with legs of length 5, then I guess the "exponent" of the length of one side is 1: increasing it by 1 means you have 52 and then you divide by 2: yes, by Jove, the area is (1/2)(52)!
     
  9. Mar 30, 2004 #8

    NateTG

    User Avatar
    Science Advisor
    Homework Helper

    You mean the fundemental theorem of calculus.

    Here's an informal version:

    Let's say that we have a function [tex]F(x)[/tex] with derivative [tex]f(x)[/tex] on [tex][a,b][/tex].

    Now, if we partitiion [tex][a,b][/tex] with [tex]x_i[/tex] so that [tex]a=x_0<x_1<x_2...<x_n=b[/tex] then [tex]F(x_1)-F(x_0) \approx (x_1-x_0) f(x_0)[/tex]

    Now, the left sum of [tex]f(x)[/tex] on the same partition is:
    [tex]\sum_{i=0}^{n-1}(x_{i+1}-x_i)f(x_i) \approx \sum_{i=0}^{n-1} F(x_{i+1})-F(x_i)[/tex]

    But it's easy to see that the sum on the right hand side collapses into:
    [tex]F(x_n)-F(x_0)[/tex] which is the result that you would expect from the fundemental theorem of calculus.

    There's a bit of work involved making this formal, and some more work showing that the limit exists, but this should help a bit.
     
  10. Mar 31, 2004 #9
    OK, a bit simplified, but here is an example where using a couple of different numbers can be misleading.

    0 + 0 = 0

    0 x 0 = 0

    2 + 2 = 4

    2 x 2 = 4

    [sarcasm]I have just shown per example that addition and multiplication are the same operators. [/sarcasm] I believe that we all know that addition and multiplication are different operators, but some examples can lead one to believe that.

    There are many more complex funtion(s) that have ugly solution(s) after integration. It still seems like pulling a rabbit out of a hat when saying the integral is equal to the area under the curve. Riemann sums is the only way to back it up.

    As for the Simpson's rule, I did overstate. I was trying to state that the area under the curve for a parabola is equal to

    1/3 h (y0 + 4y1 + y2) and can be algebraically proven.
     
  11. Mar 31, 2004 #10
    I like that one because it kind of sneaks up on you.

    Triangle:
    If f(x) = x
    then F(x) = 1/2x^2

    Pyramid:
    if f(x) = x^2
    then F(x) = 1/3x^3

    mikie1,
    As you work with calculus over a period of time, these things get much easier to visualize.
    -Mike
     
    Last edited by a moderator: Mar 31, 2004
  12. Mar 31, 2004 #11
    mikie1,
    I hope your attitude about calculus changes. When you say that calculus is in any way "ugly", it sounds like you're trying to fight with it. Calculus is truly beautiful. I hope that you see this soon, it will make calculus much easier and more enjoyable for you.

    My little parlor trick, "or pulling a rabbit out of a hat" as you put it, shows you that you have already been using calculus for years, and you didn't even know it. Take the formula for the circumference of a circle(2 pi r) and integrate it. You now have the formula for the area of a circle(pi r^2).

    Take the surface area of a sphere(4 pi r^2) integrate that and you have the formula for the volume of a sphere (4/3 pi r^3). Does any of this look "ugly"?

    I think that pulling a rabbit out of a hat is a wonderful solution! The fact that you can see calculus in this way is a great start! -Mike
     
    Last edited by a moderator: Mar 31, 2004
  13. Mar 31, 2004 #12

    HallsofIvy

    User Avatar
    Science Advisor

    Now, he didn't actually say that calculus was "ugly"- he said that integrals have "ugly" solutions, by which he seems to mean "not trivial".

    This is generally true of "inverse" problems: If you are given some complicated polynomial, f(x), given a and asked to find y= f(a), that's straightforward calculation. If you are asked to find x such that f(x)= a, that's an extremely difficult problem.

    The derivative is defined through a simple formula. The "anti-derivative" is defined as its inverse. That's why anti-derivatives tend to be complicated.

    Once again, the "fundamental theorem of calculus"- that the definite integral, defined as "area", can be derived through the anti-derivative, is NOT "pulled out of a hat" but has a simple proof given in any calculus book.
     
  14. Mar 31, 2004 #13
    Thanks for all the replies!

    A little bit about me. I'm 35, took some calc in high school (1986) and graduated nontraditionally in 1995 with a BS in electrical engineering. I have some extra time on my hands and I'd like to clear up some things from my educational past that seemed vague to me or the prof didn't have a good explanation.

    One was the Simpson's rule. In a problem solving / computer apps class one of the profs said that with Simpson's rule, we are estimating a curve with parabolas and the coefficients of the parabola must be determined in order to calculate the area under the curve. This is untrue since Simpson's rule just uses three points and a formula. Another one is paint cans that can't be painted.

    HallsofIvy - Thanks, I do find that calc can be pretty kewl and U R exactly right on my interpretation of "ugly". I need to be a little more clear since I'm dealing with "purest" on this board.

    I'll read through some of the links to see if I can "see the light".
     
  15. Apr 2, 2004 #14

    HallsofIvy

    User Avatar
    Science Advisor

    "In a problem solving / computer apps class one of the profs said that with Simpson's rule, we are estimating a curve with parabolas and the coefficients of the parabola must be determined in order to calculate the area under the curve. This is untrue since Simpson's rule just uses three points and a formula."

    No, believe it or not, what your professor told you is not untrue. His/her point was that the "formula" you are referring to is derived by approximating the curve by the parabola that goes through those three points.

    What you are saying could be interpreted as "No, it's untrue that I have to learn anything- I just have to memorize formulas."
     
  16. Apr 4, 2004 #15
    hi.. i'm facing a trouble in finding a proof for the derivation of the volume of a sphere, a cylinder, and a cone using integration.. so please provide me with the solution for this assignment that i have & i can't answer.. with my appreciation & thanks a lot for your support
    nermeena
     
  17. Apr 4, 2004 #16
    Let height of the cylinder to be L and the area of the bottom A. Then the volume will be --> [tex]\lim_{||\Delta L||\rightarrow\0}\sum_{i=1}^n A \Delta L = \int_{0}^{L} A dL[/tex] = AL.
     
  18. Apr 5, 2004 #17

    HallsofIvy

    User Avatar
    Science Advisor


    I. You are trying to "hijack" a thread on a completely different subject. If you are not responding to something said in this thread start a new one.

    II. Since you say this is an "assignment", it should be in the "homework" section and you should show what you have tried. One reason for that is to let us know what kind of help you need. I can think of several ways of doing these but I don't know what methods you already know and are allowed to use.
     
  19. Apr 6, 2004 #18
    You are reading into that way too much! :wink:
     
  20. Apr 12, 2004 #19
    Maybe this helps

    To Mike1 original question:

    What I always have in mind, when I think of definite integrals is this (though someone already told me that I am wrong, but I don't care)

    I define the integral sign to mean

    b/
    |f(x)=f(a)+f(a+dx)+f(a+2dx)+.....+f(b-2dx)+f(b-dx)+f(b)
    a/

    in direct analogy with ordinary the summation sign.

    (INTERMEZZO (skip it if you like):
    The ordinary summation sign uses a discreet index n, with n=0,1,2,.....
    The integral sign uses an index x, which is not a discreet index, but a continuous index. How do we generalize from a discreet index to a continuous index? Well the best way we can do is to invent a "dx" which is supersmall by definition, and this way we can interpret my previous definition of the integral sign almost "as if" it "slides" through the continuum. When you think about it, this is kind of weird, didn't Cantor proof that there we more natural numbers than real numbers? Anyway, I am not smart enough for that stuff.END OF INTERMEZZO)

    The "dx" here is Leibniz's and Newton's infinitesimal.

    Now, you may think that this is useless, because you never see

    b/
    |f(x)
    a/

    without the "dx", that is, you always see

    b/
    |f(x)dx
    a/

    Okay, let's do that then.

    b/
    |f(x)dx=f(a)dx+f(a+dx)dx+f(a+2dx)dx+.....+f(b-2dx)dx+f(b-dx)dx+f(b)dx
    a/

    An infinite number of terms to add, it looks hopeless. But hey, Leibniz found a trick. What if somehow we know that for some F(x) we know that

    dF(x)/dx=(F(x+dx)-F(x))/dx=f(x)

    You might wonder how that could possibly simplify things, but I am gonna do it anyway: replacing f(x) by (F(x+dx)-F(x))/dx we get

    b/
    |f(x)dx=
    a/

    [(F(a+dx)-F(a))/dx]dx+[(F(a+2dx)-F(a+dx))/dx]dx+[(F(a+3dx)-F(a+2dx))/dx
    dx..............[(F(b-dx)-F(b-2dx))/dx]dx+[(F(b)-F(b-dx))/dx]dx+[(F(b+dx)-F(b))/dx]dx

    Now because the "dx" in

    b/
    |f(x)dx
    a/

    is chosen by me (and if you are smart you choose the same) to be exactly the same as the "dx" in

    (F(x+dx)-F(x))/dx

    the two "dx's" cancel each other out, so that

    b/
    |f(x)dx=
    a/

    F(a+dx)-F(a)+F(a+2dx)-F(a+dx)+F(a+3dx)-F(a+2dx).......F(b-dx)-F(b-2dx)+
    F(b)-F(b-dx)+F(b+dx)-F(b)

    The F(a+dx) term at the beginning can be found again three positions further up, only now with a minus sign, so they add up to zero. So it is with most of the terms, and as you can find out for yourselves, only two terms are left

    b/
    |f(x)dx=F(b+dx)-F(a)
    a/

    A miracle has happened right in front of your eyes! No need to add together an infinite number of terms, all we have to do is know the anti-derative of f(x).

    You might be bothered by the fact that instead of F(b)-F(a) my derivation actually gives F(b+dx)-F(a), but the difference is only infinitesimal.

    I warned you, somebody "authorotive" already told me that my derivation is wrong, somehow, in a way I do not understand, it has to to with the mathematical rigidity of me using the infinitesimal "dx", but I do know that the great great Riemann has used his brains to crack this one, so yeah, probably I am oversimplifying things to much, like always, but when I am gonna have kids someday, this is the way I am gonna explain it to him.

    So what do you think of my definition? I have a lot more stupid stuff. For instance, I define the dirac-delta function to be

    dirac(x-a)= 1/dx if x=a
    0 if x=not(a)
     
  21. Apr 12, 2004 #20

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    As you got Cantor's cardinals the wrong way round, can I beseech you on behalf of the mathematical community not to pass on this knowledge for so many reasons (misuse of infinitesimals, presumption that adding up an uncountable number of non-zero objects is remotely rigorous, writing a 'function' as equal to the reciprocal of a 1-form). Whatever the suggestive nature of things might be (and notations have evolved to be deliberately suggestive) doesn't justify this level of abuse. If you want proof the why you shouldn't do this as it leads you to get the wrong answers consider:

    suppose F(x,y,z)=0 defines implictly x,y, and z as a funtion of each other then what is:

    [tex]\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}[/tex]

    naively, according to the above, it probably is 1, when it isn't, it's -1. Try working out why

    [tex]\frac{d^2y}{dx^2} \neq \frac{1}{\frac{d^2x}{dy^2}}[/tex]

    as well.
     
  22. Apr 14, 2004 #21
    Help me get my undestanding of infinitesimals right?

    To M. Grime:

    Yes, having switched the real numbers with the natural numbers is pretty embarrassing.

    I have a question. You say that my understanding of the infinitesimal is totally wrong. Maybe you can help me get it right. Let me explain my thoughts on the subject and then maybe you could correct me, or anybody.

    Let

    Ndx=h

    with N some finite integer, dx some finite real number (it is not a infinitesimal yet but is will become one later) and h also some finite real number.
    Now let's shrink dx, but I want to keep h a constant.This means that N must become larger, and not just that, I want to keep N an integer too, which forms a restriction on what values dx can take. So h is the only really arbitrary number here, as long as it is finite.

    Now let us assume that my shrinking powers are such that I can shrink dx so much that it becomes appropiate to call dx a infinitesimal. What is the value of N to satisfy the equation Ndx=h? Well, I think that N must equal h/dx. Is this absurd, N=h/dx?

    Now what I think is that even with dx being a infinitesimal, one will not be able to go through ALL of the real numbers between the interval 0 and h. I do NOT think that dx means the difference between nearest-neighbours on the real number system continuum. The infinitesimal is just a limit. So to me, your earlier objection that I am trying to count an uncountable infinity does not apply. By using dx I was not using all the real numbers of the uncountable infinity, I only was merely using a countable subset of that uncountable infinity, which is ALMOST like going through all the real numbers.

    But I am still worried about N=h/dx. I have "discovered" a proof of the natural number e, which I feel is very elegant, but it merrily uses expressions like N=h/dx. This proof of e, which I regard as my greatest personal triumph in trying to understand infinitesimals, will collapse if you can explain to me why N=h/dx is such nonsense.

    Did I mention that this is the N I am using in my definition of the integral? Let's rewrite Ndx= h to (h/dx)dx=h. Here the dx in the denominator cancels the dx in the numerator. I CHOOSE both of the dx's to be exactly the same, because I WANT h on the right handside of the equation.

    Now you also have a argument involving partial deratives. I am still thinking about that. But "partial" deratives may be beside the point. Because when you write (dx/dy) you implicitly also mean to keep to keep the variable z a constant, and when you write (dz/dx) you implicitly mean to keep the variable y a constant. That may be why the dx in (dx/dy) does not simply cancel the dx in (dz/dx), they are both used in different contexts, only the notation does not reflect that. Well. Maybe. Like I said, I am still thinking about that.

    I know very well that most of you (probably all of you) do not agree with me. That is why it is so important that, despite my limited brainpower, one of you with your scary monstertalent for maths tries to help me get it right. (But don't use words like 1-form, I have no idea what you mean, and besides, as you may have noticed, I am not a real mathematician)

    ydnef
     
    Last edited: Apr 14, 2004
  23. Apr 14, 2004 #22

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    If all you are doing is saying that 'infinitesimal' means some suitably small real number, then you're doing a numerical method of integration, and you're using something akin to the uniform subdivision of the interval (fixing N to be a natrual number). Moreover, when there is an anti-derivative it drops out quite nicely. That is nothing special, and as almost every function doesn't posses an antiderivative it's of only partial importance, but good to know anyway at highschool. Thus you've just proved, albeit without any rigour, the fundamental theorem of calculus

    There is a difference between the infinitesimal idea of approximation, where one ought to write \deltax, in calculus and the rigorous manipulation of dx which is not even a real number. So what you write whilst being suggesive is not correct. So I suppose the first thing to say to you is dx is not a real number. The second is that you're just justifying the idea that integration is the reverse of differentiation and can be done to a reasonable numerical degree. Sort of moving the furntiure around. And as long as you stop using dx as a number when it isnt' one you're fine. If you treat them like numbers you will eventually get the wrong answers becuase things happen that are counter intuitive.
     
  24. Apr 14, 2004 #23
    You lost me when you said that dx is not a real number. Let me try to explain why I think it is real number by using an example.

    Does there exist a f(x) such that (df(x)/dx)=f(x)? Well suppose there is? (We already know there is but let's pretend we do not know yet)

    Written full out this equation is

    (f(x+dx)-f(x))/dx=f(x)

    For sure the dx here represents an infinitesimal, there can be no doubt at all what dx is here. We can rewrite the equation to

    f(x+dx)=(1+dx)f(x)

    According to this shifting by dx on the x-line just means multiplying f(x) by (1+dx). So

    f(x+2dx)=((1+dx)^2)f(x)=...=(1+2dx+dx^2)f(x)

    f(x+3dx)=((1+dx)^3)f(x)=...=(1+3dx+3dx^2+dx^3)f(x)

    f(x+4dx)=((1+dx)^4)f(x)=...=(1+4dx+6dx^2+4dx^3+dx^4)f(x)

    What do we see appearing as the coefficients? Pascal's triangle
    1
    11
    121
    1331
    14641
    .........

    We can replace the expressions by the binomialcoefficient (I think it is calles that) and make a general formula for

    f(x+Ndx)=.....(some formula which contains the binomialcoefficient)

    Now I take Ndx=h in the way like I explained before and ...........(I shortcut the thing now, the internetcafe where I am in is closing).......and in the end I get the series

    1+1/2!+1/3!....

    the series one generally refers to as e.
    So what I did is, I took the infinitesimal dx and treated it like a real number and voila, I get e. So why didn't things go wrong then.

    I'll check the net again tomorrow, hopefully you're not tired of me.

    ydnef
     
  25. Apr 14, 2004 #24

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    All you're doing is estimating the integral with a numerical approximation, which if there is an antiderivative yields the obvious answer.

    dx is not a real number, it is not a number at all, if you're saying it is then you don't mean dx, you mean [tex]\delta x[/tex] which is tiny. Look up these things. You are just doing the numerical integral with equal intervals of length dx (when you shouldn't be) and then saying as dx goes to zero you get the anti-derivative. And? That is rather obvious and taught to every undergrad, with the proper notation.

    Treating it as a proper number can reveal some obvious answers, and the way you generate e is exactly the way we were taught as callow 16 year olds when attempting to find a function whose derivative is itself.

    So, in some sense, congratulations, you've got what we already know and teach. It's just that you're using the wrong names for things. Just use a delta x, do the manupulation, let it tend to zero and voila the answer, just as is supposed to be.

    edit actually we do know when diff eqns have solutions and when they are unique.
     
    Last edited: Apr 14, 2004
  26. Apr 15, 2004 #25
    Okay, for me this could be the last time for me because we can't seem to convince each other.

    What I think you are saying is this:
    Manipulating dx is a tricky business. Sometimes they cancel each other out and we are fine. But there many more instances when they do not simply cancel each other out and you have to be careful.

    Maybe you mean when you say that dx is not a number, you mean that dx does not represent the same number all the of time, it depends on the context, and in that sense it is just a symbol.

    Let me elaborate. When you say that I have to be careful cancelling the one infinitesimal dx by the other infinitesimal dx, you mean that in the same way as you would say that one has to be careful cancelling the one infinity 8 by the other infinity 8 (My keyboard doesn't have a the correct symbol for infinity, so I'll make do with the 8, which is the same only turned 90 degrees)
    Infinities come in many forms, they can go to infinity logarithmic, quadratic or linear,etc. The result of the division infinity/infinty=8/8 depends totally upon what the symbols actually represent. Naively by just looking at the symbols, you would say that they cancel and the result of the division is one, because they look the same. But only when you know for a fact that the two dx's represent the same thing then you can cancel them. If you do not know that then be careful, because the symbols do not keep track of what type of infinity is really meant.

    And so it is with the dx's. The dx's are just symbols. Off course one has to be careful in cancelling the one dx by the other dx, because they might not represent the same thing, they are just symbols, and you have to check what they represent. But in my case, when I define the integral, I happen to know for a fact that they are the same. What I am gonna show you now may not be rigourous enough for you and I hope you won't have an heartattack:

    Let's define the integral like I wanted to:

    b/
    |f(x)dx=f(a)dx+f(a+dx)dx+f(a+2dx)dx+.....+f(b-2dx)dx+f(b-dx)dx+f(b)dx
    a/

    How do I know that the dx inside f(a+dx) is the same dx as which I multiply it with? The answer: Because I want to give the integral an interpretation, namely I want it to mean the following

    b/
    |f(x)dx="area underneath the curve f(x)"
    a/

    (for simplicity take f(x)>0)

    Take time to convince yourself that the integral can only take on that interpretation if and only if the dx inside f(a+dx) is the same as the dx you multiply it with.
    Now let's substitute f(x)=dF(x)/dx into f(a+dx) (wether that is possible is like saying wether a function is integrable or not).
    We get

    [f(a+2dx)-f(a+dx)]/dx

    This says that the dx in the denominator is the same as the dx in f(a+dx), which as I showed is the same as the dx we multiply it with, and thus the two dx's cancel out. That is

    ([f(a+2dx)-f(a+dx)]/dx)dx=f(a+2dx)-f(a+dx)

    What I think I have shown is, that in order to get the interpretation right, namely that of "area under a curve", I have to choose all the dx's, in this special case, to be the same. My desire to get a certain interpretation for the integral forces me to make all the infinitesimals to be the same thing. If I do not, the integral could not be interpreted like that. In this special case I do not need to find out what type of infinitesimals the dx's actually represent because I discovered that to get the interpretation right they must be the same and they cancel out and we don't have to worry about them.
    This does NOT AT ALL quench your thirst for rigour, does it?

    Why is the fact that I represent the integral as a discreet sum so disturbing to you (not only you I suppose). Maybe it is because you think that all a discreet sum can lead to is an approximation and could never define the integral perfectly. But you forget that I am using the infinitesimal dx. It does not matter that the sum is discreet, by using the infinitesimal dx we ensure that we do not get an approximation but an exact answer. That is the power of our beloved infinitesimal. At first sight you would not think that such a small thing could be such a powerful tool. Numerical integration does not make use of this wonderful tool. It is in this sense that I can say that what I do in doing integrals is not like numerical integration at all, because numerical integration does not make use of that one special ingredient.

    So to me dx as a symbol represents a real number. I know dx/dx does not automatically mean 1, because the dx in the denominator might mean
    (x(z+dz)-x(z)), whereas the other dx might mean (x(y+dy)-x(y)), one has to check, but in my specific case, there are no worries. But the fact that one has to be extremely careful in other circumstances does not mean that dx is not an element of the real numbers. It just means that you have to careful. Well, to me at least.

    But hey, I do appreciate it that you are trying to teach me something. Nobody else seems to care that I am wrong.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook