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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


    [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?
  2. jcsd
  3. Mar 30, 2004 #2


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    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.


    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...

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


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    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


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    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.

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

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

    As you work with calculus over a period of time, these things get much easier to visualize.
    Last edited by a moderator: Mar 31, 2004
  12. Mar 31, 2004 #11
    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


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    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


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    "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
  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


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    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


    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


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


    Okay, let's do that then.


    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


    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



    Now because the "dx" in


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


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



    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


    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

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    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.
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