Fundamental Theorem of Calculus

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 infinite many practice material, exercises, flashcards, or anything you want.

"The No Bullshit Guide to Math and Physics"

1910 book, but actually does the job well

"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.

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

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)

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

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.

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.

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. 

> 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.

Now F, the sum, is related to f: F' = f.

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

  - summing and change rate are inverses: 
  
       (int f)' = int (f') = f
                               
       F' = f  <==>  int f = F
  
  - interval arithmetic

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.

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

- except finitely many, or

- except a set of measure zero.

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

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

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