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I don't know if this is an appropriate question for this site. If not, I apologize.

A colleague of mine maintains that the Second Fundamental Theorem of Calculus shouldn't be taught as such (i.e. that it is "fundamental") but rather be thought of, and referred to as The Evaluation Theorem or The Net Change Theorem.

My question isn't really about the "should" or "shouldn't" but rather "How did with end up with two FTC's?" What is the reasoning behind referring to both as Fundamental?

Again, if this is off-topic, I apologize and please close down the question!

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    $\begingroup$ As a side note: Quite a few theorems in mathematics have weird names, e.g. the "fundamental theorem of algebra". At least from todays perspective, this is a terrible name. As a professor I know once said: "It's neither fundamental, nor is it about [abstract] algebra." (in German the 'abstract' is redundant). $\endgroup$ Feb 22, 2017 at 11:14
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    $\begingroup$ Tangential comment, but I think "Net change theorem" is a good name because it emphasizes the intuition that the total change $f(b) - f(a)$ is the sum of all the little changes $f'(x) \, dx$. $\endgroup$
    – littleO
    Feb 22, 2017 at 11:25

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The Fundamental Theorem of Calculus, very loosely stated, is "differentiation is the inverse of integration".

Recall that, in general, to show two operations $S$ and $T$ are inverses, you must show that both $ST$ and $TS$ are identities. That's what the two parts are: loosely stated,

  • The first part shows that differentiating an integral gives the original function
  • The second part shows that integrating a derivative gives the original function
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  • $\begingroup$ (disclaimer: I do not know the details of the historical evolution of the name) $\endgroup$
    – user14972
    Feb 22, 2017 at 10:50
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The reason for the name "fundamental" is probably because it gives a very direct and non-obvious relation between the two central concepts of calculus namely integration and differentiation. While a version of integration was known from the time of ancient Greeks, no one thought of differentiation till Newton arrived on the scene. And because of the Fundamental Theorem of Calculus the process of integration was made considerably easier. Before FTC (time of ancient Greeks) the only way to integrate was to add lot of small terms. Newton changed that to subtraction of two terms via FTC.

Note further that the two Fundamental Theorems of calculus are different from each other and we do need two of them. Only when functions involved are continuous we can combine two theorems into one. See this answer for more details.

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  • $\begingroup$ Both tangent-finding and area-finding methods were known to the ancient Greeks, and algebraic methods for both were known, in special cases and special cases only, before Newton (see Fermat's method of adequality, for example). I don't think you can really justify claiming that integration predates Newton, but differentiation does not. $\endgroup$
    – Chappers
    Feb 22, 2017 at 12:07
  • $\begingroup$ @Chappers: I don't think people viewed tangents in terms of difference quotient, but instead tangents were handled by their geometric properties. At least that's what I gather from works of Apollonius on conics. Areas on the other hand were always handled by thinking as sum of a lot of small areas. Thus the summation aspect of integration was known, but non thought of differences, rate of change etc which is the central aspect of differentiation. $\endgroup$
    – Paramanand Singh
    Feb 22, 2017 at 12:24
  • $\begingroup$ "Areas on the other hand were always handled by thinking as sum of a lot of small areas." Not by Archimedes, e.g., if you look at what he's actually doing: he doesn't sum areas to find the area of a region at all. The heuristic weighs slices of the area, and shows that a certain property holds for all such slices (and hence the whole area), whereby we obtain the value of $A$. Actual proof is done by approximating area with inscribed and circumscribed polygons, yes, but no limit is taken: instead, it is shown that if the area is $B\neq A$, one finds a polygon that has area between $A$ and $B$. $\endgroup$
    – Chappers
    Feb 22, 2017 at 12:46
  • $\begingroup$ This is why when the heuristic does not come to Europe in the Renaissance, it takes so long to produce an algorithm that finds the value of the area: when the only method you have to prove areas is proof by contradiction, you need a way to find the value of the area in the first place. If they had had the heuristic, they wouldn't have needed to invent calculus. $\endgroup$
    – Chappers
    Feb 22, 2017 at 12:50
  • $\begingroup$ @Chappers : I think you need to look into method of exhaustion as well as Cavalieri's principle which is more in line with integral calculus. The method of exhaustion is more related to the idea of supremum and infimum which corresponds to the integral as sup and inf of upper lower Darboux sums. $\endgroup$
    – Paramanand Singh
    Feb 22, 2017 at 12:52
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The first theorem states that if $f:[a,b] \to \mathbb{R}$ is continuous, then the function $g: [a,b] \to \mathbb{R}$ defined by $g(x) = \int_a^x f(t) \ dt$ is differentiable and satisfies $g'(x) = f(x)$ for all $x \in [a,b]$.

In a sense, one might be inclined to think that the second theorem is not a theorem in its own light. If $F$ is any function defined on $[a,b]$ with $F' = f$ we have $F(x) = g(x) + C$ and so $F(x) - F(a) = (g(x) + C) - (g(a) + C) = g(x) - g(a) = g(x) = \int_a^x f(t) \ dt$. It's an immediate corollary of the first theorem, and therefore does not deserve a name.

There's two counterpoints to this:

$(1)$ This issue depends on how the theorems are presented. The second fundamental theorem is presented in different ways. In introductory calculus books, it is often presented the way I describe above: if $f$ is continuous on $[a,b]$ and $F' = f$ on $[a,b]$, then $F(x) - F(a) = \int_a^x f$ for all $x \in [a,b]$. However, we can actually weaken a hypothesis of the second theorem so that the continuity of $f$ is not required, only Riemann integrability. This is stronger and requires a fundamentally different proof.

$(2)$ Pedagogically, it's better to have two separate theorems. The first theorem enables to us to differentiate integrals. For example, if $f(x) = \int_{\cos x}^{x^2} e^t \ dt$, what is $f'(x)$? Conversely, the second theorem allows us to compute integrals. What is $\int_{-1}^{1} \frac{2x+2}{3x^2 + 6x - 25} \ dx$? We compute the antiderivative and solve. The problem domains for both theorems are different. If we have one "umbrella" theorem called the fundamental theorem of calculus, it's not the most convenient and may not be as intuitive to students.

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I would agree with your colleague with respect to the following development.

We define the concept of the definite integral $\int_{a}^{b}f$ of a function $f$ in terms of that of limit, the Riemannian partition thing. Then, logically independently, we can define the concept of a primitive $F$ of $f$, i.e. $F$ being such that its derivative $DF = f$. Then a question naturally comes: How do we find a primitive of a suitable function? The FTC answers this question; i.e. the function $x \mapsto \int_{c}^{x}f$ (where $c \in [a,b]$ and $x$ runs through a suitable subset of $[a,b]$ so that $\int_{c}^{x}f$ exists, of course) is a handy choice because $D\int_{c}^{x}f = f$. This is the theorem usually referred to as the First FTC. By this theorem and the property that any constant shift of a primitive is still a primitive, which follows directly from the definition, we have got the theorem that $\int_{a}^{b}f = F(b) - F(a)$ and then the problem of computing a definite integral of a function became that of finding a primitive of the function.

So I think the FTC is fundamentally one theorem. However, if we are to have beginning readers in mind, then, for the sake of application ease, it would be reasonable to give the two theorems two titles. The way how the mind of a mathematically literate person operates is usually different from a mathematically less literate person.

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First you get to know the antiderivatives, with the idea is the area covered under a curve. And it can be obtain for some easy integrand f(x) (like some power formulas) that you gonna integrate by using the limits of a Riemann sum by using some algebraic proof over difference of squares.

Then you get to know the FTC 1 'evaluation' that is used to proof the area under a curve is actually the antiderivative evaluated at the end points, more on that you learn some properties of integrals and how you can swap limits of integration, split integrals, the average value has to cover the same area, and the more important the integral from '$a$' to '$a$' is zero.

The FTC 2 'integral formulas' tell you that you can express a function expresed with the integral formulas (that is you don't get a number anymore; you get different value depending on where is you final point).

$$F(x)=\int_a^x f(t)dt$$

That has lower limit of integration as initial point or initial condition as $(x,F(x))=(a,0)$; and the upper limit has a change of variables so the computer don't get confused. The lower limit is very important: if you want to have a initial point with value of 5 then you arrange as $$F(x) =\int _a^x f(t) dt+5;\quad F(a)=\int_a^a f(t)dt+5=5$$ and you can use more properties of integrals like this.

Then you address the reality that not all the function that you want to integrate has an actual antiderivative that you can evaluate at the endpoints; but you can take its linear approximation, quadratic approximation; you just have to get used to the notation of this formulas and how to obtain its derivatives.

Then you pass to composite function, maybe some piece wise. Then numerical integration and some method to find antiderivatives.

All that I relearn in edx courses for free.

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  • $\begingroup$ I think that extend to multivariable in the theorem for line integrals for the Green's theorem and Gauss' theorem but I'm still learning those subjects. $\endgroup$
    – JorgeLRG
    Aug 29, 2022 at 21:34

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