Proving the Fundamental Theorems of CalculusDate: 02/02/2001 at 10:22:32 From: Junior Wellington Subject: The Fundamental Theorem of Calculus Good morning to you all. There appears to be some confusion in regard to the first and second fundamental theorems of calculus. In some books, the first theorem is given, and in others, the second is. What is the correct order? Also, could you please provide a proof of the second fundamental theorem? Yours sincerely, Junior Date: 02/12/2001 at 16:04:10 From: Doctor Jordi Subject: Re: The Fundamental Theorem of Calculus Hi, Junior - thank you for writing to Dr. Math. There is often a trade-off to be made between rigor and clarity, which is one of the reasons that different authors present the same material differently. I will try to give you a rigorous proof. Before reading it, you should consult the following explanation from our archives: Riemann Sums and the Integral http://mathforum.org/dr.math/problems/brian1.18.99.html This is not a rigorous proof, but only a motivation for the proof that I'm going to provide. In particular, the explanation states that "by definition, the inverse of integration is differentiation," something that is true only for a specific kind of integration. Historically, indefinite integration has always been defined to be the inverse of differentiation, i.e. the indefinite integral of f(x) is the collection of all possible antiderivatives of f(x), which happen to differ only by a constant. But definite integration, motivated by the problem of finding areas under curves, was originally defined as a limit of Riemann sums. Only later was it discovered that the limits of these Riemann sums can actually be computed with antiderivatives, leading to our modern FTC. The Fundamental Theorem of Calculus (FTC) is usually presented as a pair of statements, sometimes with one as a corollary of the other: IF: f(t) is continuous on an open interval I; a and b are constants contained in I; x and t are in I; and F(t) is any antiderivative of f(t); THEN: 1) Int[a,x](f(t) dt) = F(x); and 2) Int[a,b](f(t) dt) = F(b) - F(a) Statement (2) is sometimes called the "Evaluation Theorem" because it allows us to evaluate the definite integral. (Note that I've used "Int[a,b](f(t) dt)" to mean "The definite integral of f(t) from a to b." Sometimes I will only write Int[a,b] if I'm only concerned with the limits of integration and the function is implied or understood.) So, how do we prove these statements? Statement 1: ------------ Let's give a definition of A(x). We will define, for constant a, A(x) = Int[a, x](f(t) dt) You can think of this function A(x) as the area under the curve from a up to x, for fixed a and variable x. Look at the diagram here to see more clearly what I mean: The main motivation for this proof is that dA/dx = f(x). Is this true? Well, if so, then by the definition of the derivative, dA A(x + h) - A(x) -- = lim[h -> 0] --------------- dx h which we think is equal to f(x). The general idea of the proof I am about to show you is to think of h as an increment in x, make that increment infinitesimal, and see what implications follow from that. Let's look in the numerator of the definition above. For any h > 0 such that x + h is in I, A(x + h) = Int[a, x + h](f(t) dt) and thus A(x + h) - A(x) = Int[a, x + h](f(t) dt) - Int[a, x](f(t) dt) = Int[x, a](f(t) dt) + Int[a, x + h](f(t) dt) (true, since Int[r, s] = - Int[s, r]) = Int[x, x + h](f(t) dt) (*) (true, since Int[r, d] + Int[d, s] = Int[r, s]) Now, can you picture what that last integral looks like? Remember, we are about to make h infinitely small (h -> 0), so you can think of the integral Int[x, x + h] as a very thin rectangle under the graph of f(t) of width h and height somewhere between f(x) and f(x + h). Look at the above diagram again to see what I'm talking about. The integral (*) is nothing more than a small increment in the area function, A(x). What else can we say about that little integral, that slim rectangle? Just how narrow is it, and more importantly, how tall is it? Well, we can definitely give some restrictions on how tall it can be, and thus we can also give some restrictions on its area. We can do this by using a theorem from calculus that places limits on the value of a definite integral. Let m be the minimum value of f(t) in the interval [x, x + h] and M be the corresponding maximum. Thus, the area of our integral Int[x, x + h](f(t) dt) must be greater than or equal to hm and less than or equal to hM. That is, hm <= Int[x, x + h](f(t) dt) <= hM Divide through by h and get: Int[x, x + h](f(t) dt) m <= ---------------------- <= M h Hey, this is really getting somewhere! Do you recognize that little fellow, bounded by m and M? We certainly have met him before, but he looked a little different, something more like A(x + h) - A(x) m <= --------------- <= M h Great - we're almost there. We now finally let h -> 0 and see what happens. The bounded quantity above will be our derivative, dA/dx. What about m and M, though? What values will they take now? Well, remember they were defined as the minimum and maximum values of f(t) in the interval [x, x + h]. As h gets smaller and smaller, m and M have fewer and fewer values to choose from, until all that's left is f(x). So, m and M both tend to f(x) as h tends to 0. Using the Squeeze Theorem for limits, since both m and M tend to the same limit, f(x), the quantity bounded by them must also tend to that limit. Thus, dA/dx = f(x). Therefore, an antiderivative of f(x) is A(x). Thus Int[a,x](f(t)dt) = F(x) This completes the proof of statement 1. The exchange of F(x) for A(x) in that last step is justified because we specified that they are both antiderivatives of f(x), although they may differ by a constant. The value of this constant, in fact, will be crucial for proving statement 2. Statement 2: ------------ The second part of the FTC follows immediately from the first part, to the point where it is practically just a corollary. We want to know what Int[a, b](f(t) dt) evaluates to. The natural way to do this is by noting that we already know two possible ways to express the integral Int[a, x](f(t) dt) with its antiderivatives A(x) and F(x). Since A(x) and F(x) have the same derivative, f(x), they must differ only by a constant. That is, A(x) = F(x) + C (**) We want to now know what that constant is. We do a little trick and let x = a, so that A(a) = 0, because A(a) = Int[a, a](f(t) dt) = 0. We see that: 0 = F(a) + C so C = - F(a) Substitute back in (**). A(x) = F(x) - F(a) We now replace x by b and get the conclusion of Statement 2: A(b) = Int[a,b](f(t)dt) = F(b) - F(a) And we have proved everything! **************************************** One more comment. It is possible to prove Statement 2 without relying on Statement 1, by using the definition of the integral and the mean value theorem for derivatives. The general flavor of this alternate proof is as follows: 1) For each subinterval in the partition used for the Riemann sum in the definition of the integral, use the mean value theorem for derivatives. 2) Each subinterval will have an expression of the form F(j) - F(k). When you add up all those expressions, you will ultimately end up with F(b) - F(a) where a and b are your limits of integration. 3) You will find that for any partition there exists a Riemann sum that is equal to F(b) - F(a), so that, therefore, the integral that is defined as the limit of these Riemann sums will also have that value. Perhaps this is the alternate stream of logic you have seen in your textbooks, and it is also perfectly okay. Which one is correct? It depends on what you want it for. The second possible proof I outlined for proving statement 2 is in a sense stronger than the complete proof I gave before. It is very sly to note this, but this second proof says nothing about the continuity of the function we are integrating, since it goes directly into the definition of the integral. Notice how I had to include the continuity of f in the hypothesis to the FTC as I presented it to you. That is why, if we use the second proof, we have a stronger form of statement 2 that only requires the function to be integrable (all continuous functions in an interval can be integrated, but not all integrable functions are continuous). As a final note, I will say that in mathematics there is not always a perfectly "correct" answer. Sometimes we do things one way to emphasize one point; sometimes we do them another way, concentrating and being very careful and pedantic about another point. It is all part of the game. Write back if you have questions about anything. I hope I addressed all your questions adequately. Best regards, - Doctor Jordi, The Math Forum http://mathforum.org/dr.math/ |
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