The Unapologetic Mathematician

How to Use the FToC

I’ll get back to deconstructing comics another time. For now, I want to push on with some actual mathematics.

After much blood, toil, tears, and sweat, we’ve proven the Fundamental Theorem of Calculus. So what do we do with it? The answer’s in this diagram:

This is sort of schematic rather than something we can interpret literally.

On the left we have real-valued functions — we’re being vague about their domains — and collections of “signed” points. We also have a way of pairing a function with a collection of points: evaluate the function at each point, and then add up all the values or their negatives, depending on the sign of the point. On the right we also have real-valued functions, but now we consider intervals of the real line. We have another way of pairing a function with an interval: integration!

At the top of the diagram, we can take a function and differentiate it to get back another function. At the bottom, we can take an interval and get a collection of signed points by moving to the boundary. The interval $\left[a,b\right]$ has the boundary points $\{a^-,b^+\}$, where we consider $a$ to be “negatively signed”.

Now, what does the FToC tell us? If we start with a function $F$ in the upper left and an interval $\left[a,b\right]$ in the lower right, we have two ways of trying to pair them off. First, we could take the derivative of $F$ and then integrate it from $a$ to $b$ to get $\int_a^b F'(x)dx$. On the other hand, we could take the boundary of the interval and add up the function values along the boundary to get $F(b)-F(a)$. The FToC tells us that these two give us the same answer!

To write this in a diagram seems a little much, but keep the diagram in mind. We’ll come back to it later. For now, though, we can use it to understand how to use the FToC to handle integration.

Say we have a function $f$ and an interval $\left[a,b\right]$, and we need to find $\int_a^bf(x)dx$. We’ve got these big, messy Riemann sums (or Darboux sums), and there’s a lot of work to compute the integral by hand. But notice that the integral is living on the right side of the diagram. If we could move it over to the left, we’d just have to evaluate a function twice and add up the results.

Moving the interval to the left of the diagram is easy: we can just read off the boundary. Moving the function is harder. What we need is to find an antiderivative $F(x)$ so that $F'(x)=f(x)$. Then we move to the left of the diagram by switch attention from $f$ to $F$. Then we can evaluate $F(b)-F(a)$ and get exactly the same value as the integral we set out to calculate. So if we want to find integrals, we’d better get good at finding antiderivatives!

This has an immediate consequence. Our basic rules of antiderivatives carry over to give some basic rules for integration. In particular, we know that integrals play nicely with sums and scalar multiples:

$\displaystyle\int\limits_a^bf(x)+g(x)dx=\int\limits_a^bf(x)dx+\int\limits_a^bg(x)dx$
$\displaystyle\int\limits_a^bkf(x)dx=k\int\limits_a^bf(x)dx$

February 18, 2008 - Posted by | Analysis, Calculus

1. The FToC diagram is very cool! Could you please post a higher-definition version? Much appreciated.

Comment by rod. | February 18, 2008 | Reply

2. I’ll see what I can do about forcing TeXshop to make something bigger. But trust me: you haven’t seen anything yet in terms of cool diagrams.

Comment by John Armstrong | February 18, 2008 | Reply

3. I couldn’t make my own TeX installationdo the enlargement for me – however, I could use some devious ghostscript trickery to boost the size without losing the prettiness.

The result can be found at

Enjoy.

Comment by Mikael Vejdemo Johansson | February 18, 2008 | Reply

4. By the way – the way I solved it was by
1) compiling with LaTeX instead of my usual PDFLaTeX.
2) dvips -x fact -y fact -O -5cm,-5cm mathochist.dvi
3) convert -trim mathochist.ps mathochist-size.png

where the -5cm will need to be twiddled to coerce the diagram back onto the page, and where the factor in -x and -y should stay the same. The results in my pictures are for factors of 2074 and 2986.

Comment by Mikael Vejdemo Johansson | February 18, 2008 | Reply

5. If that’s nothing in the way of cool diagrams, I can’t wait to see what’s coming up…

Tangentially, does anyone here know why we us $\partial$ for both taking boundaries and partial derivatives? I’ve been mystified by this since I first saw the $\partial$-as-boundary used.

Comment by Geoff | February 20, 2008 | Reply

6. I’m not sure why, actually. I tend to think of it as a “nonstandard d”. We use it for boundaries because the boundary operator is dual to exterior differentiation. It’s about homology, where differentiation is about cohomology.

“What are homology and cohomology doing here?” All in good time…

Comment by John Armstrong | February 20, 2008 | Reply

7. Thanks. At least that’s a start.

Is the fact that the boundary operator is dual to exterior differentiation (part of) what makes the generalized Stokes theorem so pretty?

Comment by Geoff | February 20, 2008 | Reply

8. Geoff: the duality is the Stokes theorem. In fact

(a) The array of different versions of the Stokes theorem all look like the diagram above.

(b) All these diagrams fit together in a very nice way.

(c) We can make the whole picture rigorous, instead of just schematic like it is now.

Comment by John Armstrong | February 20, 2008 | Reply

9. John @ 6.:
YAAAAAY! My good old friends are returning!

Let me guess – you’ll be leading up to de Rham?

Comment by Mikael Vejdemo Johansson | February 20, 2008 | Reply

10. Eventually, yes

Comment by John Armstrong | February 20, 2008 | Reply

11. […] way to use the FToC Let’s look back at that diagram that encapsulated the FToC. I want to point out something that’s going to seem really silly […]

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12. […] Since we’ve established the connection between integration and antidifferentiation, we’ll be concerned mostly with […]

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13. […] by Parts Now we can use the FToC as a mirror to work out other methods of finding antiderivatives. The linear properties of differentiation were […]

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14. […] of Variables Just like we did for integration by parts we’re going to use the FToC as a mirror, but this time we’ll reflect the chain […]

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15. […] we can start evaluating these integrals. The innermost integral goes immediately, using the fundamental theorem of calculus. An antiderivative of with respect to is , and so we […]

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16. […] Change of Variables in Multiple Integrals I In the one-variable Riemann and Riemann-Stieltjes integrals, we had a “change of variables” formula. This let us replace our variable of integration by a function of a new variable, and we got the same answer. This was useful because the form of the resulting integrand might have been simpler to work with in terms of using the fundamental theorem of calculus. […]

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