A student friend of mine recently gave me a proof of the fundamental theorem of calculus which does not correspond to any I can find in the textbooks. It starts by considering an increasing continuous function and taking the area of a thin rectangle topped with a triangle created by adding to the $x$ value (like the wikipedia entry for the theorem):

$$ \begin{align} A(x + h) & = A(x) + hA'(x)\\ hA'(x)& = A(x + h) - A(x)\\& = hf(x) + \frac12h\cdot hf'(x) \end{align} $$

where $A$ is the area function, $h$ is the added value, the first term on the RHS is the area of the rectangle and the second term on the RHS is the area of the triangle. The $hf'(x)$ term is the projection that forms the vertical side of the triangle. Apparently the curve can be composed of small straight lines making the triangle possible. The next step is:

$$\begin{align} hA'(x) &= hf(x)\\ A'(x) & = f(x) \end{align}$$

RHS 2 is discarded because $h^2$ is negligible. Is this proof legitimate? If so why isn't it in the textbooks? It seems much simpler than the alternatives.

  • $\begingroup$ You seem to assume that $A'$ exists, whereas I would have taken that to be part of what was to be proved. $\endgroup$ – Michael Hardy Jul 25 '14 at 16:25

It's a good start. But there are three problems.

  1. There's an assumption that integrals and areas are the same. It's true that the integral was defined to try to generalize the standard area computations, but are you certain that it succeeds? If not, then proving things about areas doesn't prove things about integrals (which involve lower and upper sums, etc.).

  2. The claim that A(x + h) = A(x) + hA'(x) is false. What's true is that the right hand side is a very good approximation of the left, under certain assumptions, and it gets better and better as $h$ gets smaller.

  3. That same claim assumes implicitly that the area function $A$ is differentiable. While most uses of the FTC depend on the fact that the derivative of $A$ is $f$, the deeper part of the theorem is that $A$ is differentiable at all. Once you know that, actually computing the derivative isn't so hard. :)

The details that are hidden in these two things are precisely what all those "book" proofs actually handle.

On the other hand, a picture and proof-sketch like this is a great idea for motivating the claim of the theorem, and for guiding the student to a more complete and correct proof.

  • $\begingroup$ Regarding point 2, might I draw the OPs attention to Taylor's theorem: en.wikipedia.org/wiki/Taylor's_theorem $\endgroup$ – lemon Jul 25 '14 at 14:28
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    $\begingroup$ Agreed...although Taylor's theorem wants your function to be differentiable, and part of the conclusion of the FTC is that $A$ actually is differentiable. I'll add that to my answer. $\endgroup$ – John Hughes Jul 25 '14 at 14:29
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    $\begingroup$ "It's not proving it's true, but proving that it's plausible" is a nice slogan for this. $\endgroup$ – Semiclassical Jul 25 '14 at 14:30
  • $\begingroup$ That's a great way of putting it. And it's not a bad first step for any mathematical question: before you try to write a full proof, at least convince yourself that the thing you're trying to prove is plausible. $\endgroup$ – John Hughes Jul 25 '14 at 14:32
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    $\begingroup$ An alternative (jokey) version: "Could you convince a physicist with it?" $\endgroup$ – Semiclassical Jul 25 '14 at 16:32

The theorem asserts that $A'$ exists and equals $f$. The argument given appears to assume $A'$ exists, rather than treating that as a part of what was to be proved.

You can argue as follows.

Suppose $f$ is continuous. $$ \frac{A(x+h)-A(x)}{h} = \frac 1 h \int_x^{x+h} f(t)\,dt \tag 1 $$ Given $\varepsilon>0$, we can say that $f(t)$ stays between $f(x)\pm\varepsilon$ if $h$ is close enough to $0$. Hence the expression on the right side in $(1)$ is between $$ \frac 1 h \int_x^{x+h} (f(x)\pm\varepsilon)\,dt = \frac 1 h (f(x)\pm\varepsilon)h = f(x)\pm\varepsilon. $$ So we can make $\dfrac{A(x+h)-A(x)}{h}$ as close as desired to $f(x)$ by making $h$ close enough to $0$.

In other words $\lim\limits_{h\to0}\dfrac{A(x+h)-A(x)}{h}=f(x)$. Thus $A'(x)$ exists and is equal to $f(x)$.


Maybe it was in the early books.

This geometric argumentation is reminding of early work on geometric curves from around 1650-1700, those scientists (Gregory, Barrow, Leibniz, Newton) discovered and knew that relationship between the area under a curve and that curve.

To quote from this nice talk:

The modern statement of the FTIC is the result of centuries of refinement of the original understanding and requires considerable unpacking if students are to understand and appreciate it.

So there are reasons why it is not done this way in modern texts.

In the end even the idea of assigning meaningful "area" to all subsets of the plane has been given up, after recognizing some tricky cases. See Maßproblem

Here are some historic links if you want to see that early methods.


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