alt.hn

4/23/2026 at 1:09:56 AM

Fundamental Theorem of Calculus

 https://david.alvarezrosa.com/posts/fundamental-theorem-of-calculus/

by dalvrosa

4/23/2026 at 1:19:05 PM

> Let f ... be Riemann integrable and F ... differentiable.

What many people don't notice the first time they read this in the fundamental theorem of Calculus is that this is a double criteria. That f needs to be integrable seems like an extraneous point when F is differentiable. This holds also for the Lebesgue integral. The understanding is usually that if F is differentiable then its derivative is integrable, that is, people understand the integral as an anti-derivative but the Riemann/Lebesgue integral version of the fundamental theorem of calculus only proves that if the function you want the anti-derivative of is integrable, so you have this separate requirement to prove that f is integrable having already proven F to be differentiable (to f).

However, this theoretical (because if you aren't a mathematician you won't be bothered by this sticking point, you'll just insist that the integral is the anti-derivative when an anti-derivative exists) defect is ameliorated by the Henstock–Kurzweil integral which is (I feel) a lot easier to define and understand than the Lebesgue integral. It is practically the Riemann integral with just a minor tweak: the delta in the delta-epsilon proof is allowed to vary by location (essentially, as you approach non-integrable singularities, you tend the delta towards zero).

For the Henstock-Kurzweil integral, if F is differentiable then f is (Henstock-Kurzweil) integrable. This happens because not every derivative is Riemann or Lebesgue integrable, you need a stronger integral.

by BlackFly

4/23/2026 at 1:54:03 PM

Henstock-Kurzweil is a neat teaching trick. Often also because it shows that definition of riemann integration is not the only possible one. It leads a good motivation for lebesque later but also to of importance of spaces.

by ghighi7878

4/23/2026 at 3:27:55 PM

Does it usually get taught in the undergrad maths curriculum?

by viscousviolin

4/23/2026 at 4:37:34 PM

I wasn’t taught it, but that was forty years ago

by asplake

4/23/2026 at 11:49:23 AM

> This post introduces the Riemann integral

Sweet! I'm keen to learn about the basic fundamentals of calculus!

> For each subinterval ...(bunch of cool maths rendering I can't copy and paste because it's all comes out newline delimited on my clipboard) ... and let m<sub>k</sub> and M<sub>k</sub> denote the infimum and supremum of f on that subinterval...

Okay, guess it wasn't the kind of introduction I had assumed/hoped.

Very cool maths rendering though.

As someone who never passed high school or got a degree thanks to untreated ADHD, if anyone knows of an introduction to the basic fundamentals of calculus that a motivated but under educated maths gronk can grok, I would gratefully appreciate a link or ten.

by EdwardDiego

4/23/2026 at 12:33:27 PM

If you dive into Analysis (the underlying theory behind calculus) this book - "How to Think About Analysis" by Lara Alcock is the book I wish I had when I studied it. Calculus by Spivak is the book I learnt from but it is probably not the easiest, it is very thorough though.

by chadcmulligan

4/23/2026 at 1:58:59 PM

Table of contents look really helpful to understanding.

by globalnode

4/23/2026 at 11:56:32 AM

Yeah, judging by the terseness, this is clearly aimed at undergrads. Then again, this is covered in literally every calculus class, so I'm not sure who this is supposed to be for.

by mr_mitm

4/23/2026 at 6:20:19 PM

Math Academy will diagnose exactly where your gaps are and build you up in a series of bite-sized lessons where you actively solve problems with immediate feedback until you've demonstrated mastery.

It can take you from knowing your multiplication tables all the way to all the math you need for an engineering and/or CS degree.

by sn9

4/23/2026 at 1:15:08 PM

ChatGPT.

You can ask for a syllabus first, then go through it.

It's interactive, and it covers in detail everything you don't get. You can ask for an unlimited amount practice material, exercises, flashcards, or anything you want.

by bmacho

4/23/2026 at 12:11:38 PM

Fair, sorry about that

by dalvrosa

4/23/2026 at 1:20:52 PM

> if anyone knows of an introduction to the basic fundamentals of calculus that a motivated but under educated maths gronk can grok, I would gratefully appreciate a link or ten.

"The basic fundamentals of calculus" usually go under the name "real analysis".

You have many options for studying it.

MIT OpenCourseWare: https://ocw.mit.edu/courses/18-100a-real-analysis-fall-2020/

Free calculus-through-nonstandard-analysis textbook: https://people.math.wisc.edu/~hkeisler/calc.html

Lean4 game implementing Alex Kontorovich's undergrad course: https://adam.math.hhu.de/#/g/alexkontorovich/realanalysisgam... (also includes videos of the course lectures)

I like the idea of the lean4 game, because if you do your work in lean you'll know whether you've made a mistake.

("Standard analysis" uses limiting behavior to ask what would happen if we were working with infinitely large or infinitesimally small values, even though of course we aren't really. "Nonstandard analysis" doesn't bother pretending and really uses infinitely large and infinitesimally small values. Other than the notational difference, they are the same, and a proof in one approach can be easily and mechanistically converted into the same proof in the other approach.)

Note that the ordinary course of study involves learning to do calculus problems first (in a "calculus" class), and studying the fundamentals second (in an "analysis" class). The textbook I linked is a "calculus" textbook, but there is a bit more focus on the theoretical backing because you can't rely on the student to learn about nonstandard analysis somewhere else.

by thaumasiotes

4/23/2026 at 12:22:50 PM

Here's my understanding: 1: In the 'olden days' the area A(x) under the graph f(x) used to be approximated as a Riemann sum. 2: Using limits, as the delta x in the Riemann sum->0, we'd call that an integral and set it to be the exact area A(x). 3: If we then look at some small change in A(x), we might notice f(x) = A'(x)... mind blown. 4: since we can now say A is an anti-derivative of f, we have A(x)=F(x)+C (we have to add the C because the derivative of a constant is 0). 5: Using logic and geometry we have C=-F(a) which leads to... 6: The area under the graph f between [a,b] is A = F(b)-F(a). 7: We don't have to cry anymore about pages of Riemann sum calculations.

by globalnode

4/23/2026 at 12:09:16 PM

I recommend Math Academy + Mathematica + YouTube + ChatGPT, Gemini, or Claude Opus and a LOT of motivation.

by UltraSane

4/23/2026 at 11:50:52 AM

Khan academy

by moi2388

4/23/2026 at 11:05:03 AM

I've studied the proofs before but there's still something mystical and unintuitive for me about the area under an entire curve being related to the derivative at only two points, especially for wobbly non monotonic functions.

I feel similar about the trace of a matrix being equal to the sum of eigenvalues.

Probably this means I should sit with it more until it is obvious, but I also kind of like this feeling.

by mchinen

4/23/2026 at 12:12:47 PM

> there's still something mystical and unintuitive for me about the area under an entire curve being related to the derivative

the discrete version is much clearer to me. Suppose you have a function f(n) defined at integer positions n. Its "derivative" is just the difference of consecutive values

     f'(n) = f(n+1) - f(n)
Then the fundamental theorem is just a telescopic sum:

     f(b) - f(a) = \sum_a^b f'(n)

by enriquto

4/23/2026 at 12:44:17 PM

Very. Very nice

by dalvrosa

4/23/2026 at 11:16:47 AM

It is not determined by the derivative, it's the antiderivative, as someone else mentioned. The derivative is the rate of change of a function. The "area under a curve" of the graph of a function measures how much the function is "accumulating", which is intuitively a sum of rates of change (taken to an infinitesimal limit).

by ironSkillet

4/23/2026 at 12:12:43 PM

Thanks for bringing some intuition!

by dalvrosa

4/23/2026 at 1:58:32 PM

If you think of it as being an accumulator function it can feel a bit more natural - the _definition_ of this accumulator is that, F(x) is the area from 0 to x

The fact that the derivative of this accumulator function is equal to the original function, this is the fundamental theorem of calculus, and I violently agree with you that this part is shockingly, unexpectedly beautiful

by nimonian

4/23/2026 at 3:52:39 PM

Thank you for wording it better than I could. On the computational side, the integral as a function that measures how the another function accumulates intuitively lets you measure the area under the curve between two points only requires two variables.

Where it gets interesting is that if I were to naively construct this accumulator in a non-computational, infinitesimal manner, I would probably start using the idea that the infinitesimal accumulation is related to the instantaneous slope and using that somehow with the previous accumulation. But this is going the opposite direction of derivatives that the fundamental theorem uses.

I understand there are a few other types of integrals that slice things different ways that might be more intuitive. I vaguely recall some (non fundamental theorem) proof that made Riemann integrals click for a while.

by mchinen

4/23/2026 at 12:35:36 PM

If I tell you I have function f with f(a) = 10 and on it's path from a to b, the graph first increaes by 5 units then by another 10, and then later on drops by 25 units, you can immediately deduce that f(b) = f(a) + (+5 +10 -25) = 0. The fundamental theorem of calculus uses the same concept:

To see why \int_a^b f(x) dx = F(b) - F(a) with F'(x) = f(x),

we replace f with f' (and hence F with f) and get

\int_a^b f'(x) dx = f(b) - f(a).

Re-arranging terms, we get

f(b) = f(a) + \int_a^b f'(x) dx.

The last line just says: The value of function f at point b is is the value at point a plus the sum of all the infinitely many changes the function goes through on its path from a to b.

by aquafox

4/23/2026 at 4:24:30 PM

how does this work for these two different paths which arrive at the same endpoint:

+5 +10 +0 -25 = -15

+5 +12 -2 -25 = -15

They have different graphs if you consider the values above sampling points, this is what parent is asking.

by dist-epoch

4/23/2026 at 8:11:58 PM

the derivative becomes a telescoping sum. it doesn't matter how many points you insert or what the values are because x-x=0

let x_i be in (a,b) with any i drawn from [0,N] and x_0=a and x_N=b

then

int_a^b d/dx F = (F(x_1) - F(x_0))+ (F(x_i+1) - F(x_i))+ (F(x_n) - F(x_n-1)) = F(x_N) - F(x_0)

the derivative produces a ratio of differences and the integral is a weighted sum of those same differences

every middle term is added and subtracted. and only the endpoints remain with opposite sign

by nh23423fefe

4/23/2026 at 11:58:46 AM

The antiderivative at x is defined as the area under the curve from 0 to x, which the Riemann sum gives a nice intuition for how you can get from the derivative.

So to get the area under the curve between a and b, you calculate the area under the curve from 0 to b (antiderivative at b) and subtract the area under the curve from 0 to a (antiderivative at a).

At least that's my sleep deprived take.

by magicalhippo

4/23/2026 at 12:14:16 PM

I took calculus in high school and college, and I don't think any of my instructors explained the intuition as well. So sleep-deprived or not, it's a great one!

by 1980phipsi

4/23/2026 at 11:09:55 AM

You meant antiderivative?

by sambapa

4/23/2026 at 12:07:53 PM

There is some geometric intuition in wikipedia page for this theorem you may like :)

by dalvrosa

4/23/2026 at 1:41:40 PM

I have internalised that in mathematics nice things come in bouquets. If there is a thing defined with properties A, B, C, and there is an other thing defined with properties D, E, F, then chances are that those 2 things are the same thing, because there are only so few nice concepts.

There are many types of examples, and many different reasons why I don't find a particular connection or connection type surprising. So I can concentrate on memorising them, and building intuition.

For the Fundamental Theorem of Calculus:

  - int f' = f: the sum of the change is the thing itself. E.g. pour water in the bathtub, if you sum the rate you pour, that's the total water in the bathtub
  - int f' = f(t2) - f(t1) : same but water differences between 2 times.
  - (int f)' = f: the rate of the sum is the function itself. If you go and integrate your function f, the integrate function's change rate at x is f(x) 
  - and so on.
Also someone mentioned discrete functions, partial sums and difference series are indeed easier. Say, F is your gross money and f is your monthly salary, or F is gross amount of rain and f is daily rain. Summing a series or taking differences between 2 consecutive data points are each other's inverses.

> the area under an entire curve being related to the derivative at only two points

This is a very wrong sentence. The area under f on [a,b] is not related to the derivative of f at a and b. The area under f on [0,x] is a real function F(x) by definition, and there is nothing surprising that the area of f on [a,b] is F(b)-F(a). Simple interval arithmetic. Granted F, the integral function, is related to f: F' = f.

tl;dr : in the "fundamental theorem of calculus" there are 2 main observations:

  - the operators 'sum of' and 'change rate' are each other's inverse and commute: 
  
       (int f)' = int (f') = f
                              
       F' = f  <==>  int f = F
  
  - from interval arithmetics:
  
       S(a,b) = S(0,b) - S(0,a)

by bmacho

4/23/2026 at 11:03:27 AM

> f is Riemann integrable iff it is bounded and continuous almost everywhere.

FWIW, I think this is the same as saying "iff it is bounded and has finite discontinuities". I like that characterization b/c it seems more precise than "almost everywhere", but I've heard both.

I mention that because when I read the first footnote, I thought this was a mistake:

> boundedness alone ensures the subinterval infima and suprema are finite.

But it wasn't. It does, in fact, insure that infima and suprema are finite. It just does NOT ensure that it is Riemann integrable (which, of course the last paragraph in the first section mentions).

Thanks for posting. This was a fun diversion down memory lane whilst having my morning coffee.

If anyone wants a rabbit hole to go down:

Think about why the Dirichlet function [1], which is bounded -- and therefore has upper and lower sums -- is not Riemann integrable (hint: its upper and lower sums don't converge. why?)

Then, if you want to keep going down the rabbit hole, learn how you _can_ integrate it (ie: how you _can_ assign a number to the area it bounds) [2]

[1] One of my favorite functions. It seems its purpose in life is to serve as a counter example. https://en.wikipedia.org/wiki/Dirichlet_function

[2] https://en.wikipedia.org/wiki/Lebesgue_integral

by emacdona

4/23/2026 at 11:08:53 AM

> FWIW, I think this is the same as saying "iff it is bounded and has finite discontinuities".

It is not: for example, the piece-wise constant function f: [0,1] -> [0,1] which starts at f(0) = 0, stays constant until suddenly f(1/2) = 1, until f(3/4) = 0, until f(7/8) = 1, etc. is Riemann integrable.

"Continuous almost everywhere" means that the set of its discontinuities has Lebesgue measure 0. Many infinite sets have Lebesgue measure 0, including all countable sets.

by mjdv

4/23/2026 at 11:14:36 AM

Ah, thanks for the clarification! Would it have been accurate then to have said:

"iff it is bounded and has countable discontinuities"?

Or, are there some uncountable sets which also have Lebesgue measure 0?

by emacdona

4/23/2026 at 11:25:07 AM

No, it's really sets of measure zero. The Cantor set is an example of an uncountable set of measure 0: https://en.wikipedia.org/wiki/Cantor_set

The indicator function of the Cantor set is Riemann integrable. Like you said, though, the Dirichlet function (which is the indicator function of the rationals) is not Riemann integrable.

The reason is because the Dirchlet function is discontinuous everywhere on [0,1], so the set of discontinuities has measure 1. The Cantor function is discontinuous only on the Cantor set.

Likewise, the indicator function of a "fat Cantor set" (a way of constructing a Cantor-like set w/ positive measure) is not Riemann integrable: https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%9...

by jfarmer

4/23/2026 at 11:19:41 AM

No that's not true either. A quick Google will reveal many examples, in particular the "Cantor set".

by ironSkillet

4/23/2026 at 11:19:21 AM

The Cantor set is uncountable and has Lebesgue measure 0.

by thaumasiotes

4/23/2026 at 12:10:48 PM

Great example

by dalvrosa

4/23/2026 at 11:09:29 AM

"Almost everywhere" means "everywhere except on a set of measure 0", in the Lebesgue measure sense.

Here's an example of a Riemann integrable function w/ infinitely many discontinuities: https://en.wikipedia.org/wiki/Thomae%27s_function

Anyone interested in this should check out the Prologue to Lebesgue's 1901 paper: http://scratchpost.dreamhosters.com/math/Lebesgue_Integral.p...

It gives several reasons why we "knew" the Riemann integral wasn't capturing the full notion of integral / antiderivative

by jfarmer

4/23/2026 at 11:12:31 AM

"Almost everywhere" is precisely defined, and it is broader than that. E.g. the real numbers are almost everywhere normal, but there are uncountably many non-normal numbers between any two normal reals.

by bandrami

4/23/2026 at 11:13:45 AM

"almost everywhere" can mean the curve has countably infinite number of discontinuities

by sambapa

4/23/2026 at 11:10:02 AM

“Almost everywhere” is a mathematical term and can mean two things (I think):

- except finitely many, or

- except a set of measure zero.

by Jaxan

4/23/2026 at 12:10:23 PM

Here is used in the Lebesgue measure theory sense

by dalvrosa

4/23/2026 at 10:13:47 AM

What is the font used on the site?

by bikrampanda

4/23/2026 at 10:34:58 AM

Today I learned there's a CSS property for styling the first letter of a paragraph, neat. (https://css-tricks.com/almanac/properties/i/initial-letter/)

--edit: The font used for those initials is called Goudy Initialen: https://www.dafont.com/goudy-initialen.font

by viscousviolin

4/23/2026 at 12:42:29 PM

I love this -- I'll have to do something like that for my site. I always liked the big initials on the start of a paragraph. Though it feels a bit more prose-applicable than for non-fiction writing.

by quibono

4/23/2026 at 12:13:12 PM

Technically "it depends on the browser settings," but the body font Alegreya is served directly by the site, so I think it would be the one used in almost all cases.

The math fonts used in the formulas are just the ones provided by KaTeX, which I think are just TeX's default math fonts.

by WCSTombs

4/23/2026 at 10:30:34 AM

That font, and how it's integrated with the math looks amazing. Katex for the math?

by crispyambulance

4/23/2026 at 10:36:21 AM

Seems like Katex from the scripts getting loaded. I love the design too, kinda medieval-chic.

by viscousviolin

4/23/2026 at 10:59:26 AM

Looks more early modern to me. :)

by eru

4/23/2026 at 10:19:19 AM

Alegreya

by genezeta

4/23/2026 at 12:09:28 PM

Already replied :)

The source code of the website is open if you wanna check it out!

by dalvrosa

4/23/2026 at 8:06:31 PM

Thanks. Your website looks really nice!

by bikrampanda

4/24/2026 at 7:17:02 AM

Glad that you like it :)

by dalvrosa

4/23/2026 at 11:45:53 AM

Good job, David. Have a lollipop. Now learn & write up the proof that the Henstock-Kurzweil integral integrates _every_ derivative. This is what we had in my calculus class on top of the outdated Riemann integral.

by shmoil

4/23/2026 at 4:01:25 PM

That how you got the taste for lollipops?

by random3