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Proof of the fundamental theorem of calculus

  1. May 23, 2012 #1
    1. The problem statement, all variables and given/known data

    This is supposed to be a proof of the fundamental theorem of calculus.



    I'm not really sure what that proves, but to me at least it does not prove that the area under a curve is the antiderivative of the function and then inserting the upper x value and subtracting from it the lower x value.

    I'm sort of convinced that to find the area beneath a function you take the antiderivative. For instance if your velocity is constant at 50 mph and you travel 2 hours, then you take the antiderivative of that, 50x, then plug in 2 for x, you get 100 miles and subtract that from zero and you still get 100. Are they just proving by analogy that since the antiderivative works for simple geometric shapes for which the area is known that it must work for other more complex functions?

    I also understand how the infinitesimal sections of an area cancel in the middle. If you have an interval from a to z, and you keep adding in subintervals, say,

    (a-b) + (c-b) + (d - c) + (z - d)

    I understand how those cancel except z - a.

    However, I'm still not fully convinced that to find the area beneath a curve you take the antiderivative of the function and subtract the upper x interval from the lower x interval. I don't see how the proof above shows that.
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  3. May 23, 2012 #2


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    Consider the graph of function y= f(x). Define F(x) to be "the area under the graph y= f(x) between some value, a, and x. For some small h, we have F(x+h) as the area under y= f(x) so F(x+h)- F(x) is the area under the curve between x and x+h. That area lies between [itex]f(x_*)h[/itex] and [itex]f(x^*)h[/itex] where [itex]x_*[/itex] is the x value which gives the minimum value of f and [itex]x^*[/itex] is the x value which gives the maximum value of f.

    That is
    [tex]f(x_*)h\le F(x+h)- F(x)\le f(x^*)h[/tex]
    Dividing through by h,
    [tex]f(x_*)\le \frac{F(x+h)- F(x)}{h}\le f(x^*)[/tex]

    Taking the limit as h goes to 0 gives F'(x) in the center while both left and right sides go to f(x).
  4. May 23, 2012 #3

    D H

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    It doesn't show that. The fundamental theorem of calculus has two parts. This proof was of one of the parts,
    [tex]\frac d {dx} \int_a^x f(t) dt = f(x)[/tex]

    You are essentially asking for the relationship between the indefinite integral and the definite integral. What is it that allows us to use the same symbol [itex]\int[/itex] for both? This is the subject of second part of the fundamental theorem of calculus (or the first part, depending on which book you read). In short, the second fundamental theorem of calculus says that if [itex]F(x)[/itex] is an antiderivative of the function [itex]f(x)[/itex], then
    [tex]\int_a^b f(x) dx = F(b) - F(a)[/tex]
    The proof is fairly simple given the proof of the first. Let [itex]G(x)[/itex] be the definite integral
    [tex]G(x) = \int_a^x f(t)dt[/tex]
    By the first fundamental theorem of calculus, [itex]\frac{dG(x)}{dx} = f(x) \equiv F'(x)[/itex] for all x in [itex][a,b][/itex]. Thus [itex]G(x)[/itex] and [itex]F(x)[/itex] differ at most a constant [itex]c[/itex]:
    [tex]G(x) = F(x) + c[/tex]
    By definition, [itex]G(a) = 0[/itex], and thus [itex]c = -F(a)[/itex]. In other words,
    [tex]G(x) = F(x) - F(a)[/tex]
    Now look at [itex]G(b)[/itex]. From above,
    [tex]G(b) = F(b) - F(a)[/tex]
    By definition,
    [tex]G(b) = \int_a^b f(t)dt[/tex]
    and thus
    [tex]\int_a^b f(t)dt = F(b) - F(a)[/tex]
  5. May 23, 2012 #4

    That's what I want proof of, that the area under a curve is the antiderivative of the function. I already understand the part about b - a. I sort of understand how the area under the curve is the antiderivative of the formula, after all, for a rectangle with points (2,2) and (4,2). The formula is = c = 2. So the antiderivative of 2 is 2x. And I'm already convinced that subtracting 2(4) - 2(2) will give you the area under the line, c = 2. So I sort of understand it, I just think there should be a clearer way to express that truth then the one the mathematicians are using.
    Last edited: May 23, 2012
  6. May 23, 2012 #5
    The Riemann integral ##\int_a^bf(x)dx## is defined to be the limit of Riemann sums, if the limit exists. You have a text. I'm sure you find find the full definition there.
    Note that the definition of the definite integral makes no mention of areas under curves.

    The Fundamental Theorem of Calculus (at least the version that you're talking about) states the following:

    If ##f## is continuous on ##[a,b]## and ##F## is any antiderivative of ##F## (that is ##F'=f##), then ##\int_a^bf(x)dx=F(b)-F(a)##.

    Note that the statement of the Fundamental Theorem of Calculus makes no mention of area under curves.

    If we have a region ##R## in ##\mathbb{R}^2## bounded below by the ##x##-axis, on the left by the line ##x=a##, on the right by ##x=b##, and above by the continuous graph of ##y=f(x)## (so we're assuming ##a<b## and ##f(x)\geq0## for all ##x\in[a,b]##), then we can define the area ##A(R)## of the region in multiple ways. In most undergraduate calculus texts, the area under the curve is defined by [itex]A(R)=\int_a^bf(x)dx.[/itex] If you prefer Lebesgue, Jordan, Hausdorff, or any other well-defined measure, then it can be shown that the integral definition is equivalent for this particular case. But that is the subject of graduate-level analysis. If you want a proof of that fact, you'll need to first teach yourself (or have someone teach you) some measure theory.
  7. May 24, 2012 #6

    D H

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    The Riemann integral is all about about calculating the area under a curve. Any reasonable text will make this very clear.

    No, it doesn't. In the original post, robertjford80 wrote about the First Fundamental Theorem of Calculus, which describes the relationship between the derivative and the definite integral. Robert didn't show the definition of the function F(x), but by context (and by the nature of the proof) it is a definite integral, not an antiderivative. The relationship between the antiderivative (indefinite integral) and the definite integral is the subject of the Second Fundamental Theorem of Calculus.
  8. May 24, 2012 #7
    Perhaps you should go back and read your text. Integrals are motivated by the "area under the curve" problem. But when they finally get around to defining them, it is not about areas. Integrals can be interpreted in terms of areas. But integrals are not areas. Integrals are so much more than areas. Integrals are about chopping up a problem into many pieces, approximating on each piece, adding it all back up, and taking a limit.

    Give me a citation from any "reasonable" text that includes the word "area" in its definition of a Riemann integral.

    You're right. The proof that he posted was for the First Fundamental Theorem of Calculus. But he is very clearly talking about wanting a proof for the Second Fundamental Theorem of calculus. In fact he wants a special proof that just handles the situation when the integral in question is being used to compute an area.

    Last edited: May 24, 2012
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