Can someone explain why the fundamental theorem of calculus works?

>>8543772
>>>/g/
brainlet get out

>>8543772
The derivative is the rate of change, the integral is the accumulative effect. If you have an equation for the speed at which an object is moving, taking the integral will give you the total distance covered.

Why does calculus do this? Because that's what is was designed to do.

>>8543772
Sit tight, I need latex to prove it and explain intuitively why each step works

>>8543827
Thank you!

>>8543772
[math]
\int_{a}^{b} f(x)dx = lim_{n\to \infty} \sum_{i = 1}^{n} f(x_i)\Delta x_i
[/math]
Here we have the definition of the definite integral. Given that the mesh size of our partition of the interval (a, b) goes to 0, we can rewrite this as
[math]
lim_{n\to \infty}\sum_{i = 1}^{n} \frac{dF(x_i*)}{dx_i*}\Delta x_i
[/math]
where f(x) = F'(x)
By the definition of the derivative, we can see that
[math]
\lim_{n \to \infty}\sum_{i = 1}^{n} \frac{dF(x_i*)}{dx}\Delta(x_i) = lim_{n \to \infty} \sum_{i = 1 }^{n} \frac{ F(x_{i+1}*) - F(x_i)}{x_{i+1}* - x_i*} \Delta x_i
[/math]
Further, it can be argued that in the limit,
[math]
x_{i+1}* - x_{i}* = x_{i+1}-x_{i} = \Delta{x_i}
[/math]
Expanding this, we ultimately see our result
[math]
\int_{a}^{b} fdx = \int_{a}^{b} \frac{dF}{dx}dx = F(b)-F(a)
[/math]

obvious corollary of the stokes theorem geez anon do you even differential geometry

seriously now, read something on the integral as a function of the upper limit. take a function f(t) and fix a point A, now consider a new function F(x) defined as the area under f(t) between A and x, in another words this is the integral from A to x of f(t). so what happens when u differentiate F(x) ? this derivative is the rate of change of F(x), so what is the "instantaneous" increment of the area between A and x? it's exactly the function value f(x). draw a picture and you'll see it. for me, this is the most intuitive fact about the relationship between derivative, antiderivative and integration.

>>8543855
I should mention that
[math]
x_i^{*} \in (x_i, x_{i+1})
[/math]
and that since
[math]
\Delta{x_i} = (b-a)/n \implies \Delta{x_i} \to 0
[/math]
then
[math]
\forall \epsilon > 0, \exists N \text{ s.t. } n >N \implies x_{i}^* - x_{i} < \epsilon
[/math]
which is the reason
[math]
x_{i+1}^* - x{i}^* = \Delta x_i
[/math]
in the limit

>>8543855
>Given that the mesh size of our partition of the interval (a, b) goes to 0, we can rewrite this as
Stopped reading here.
Explain this rigorously.

>>8543876
Well, I guess this probably requires a little more argument than even my second post explained. The mesh size is rigorously defined as the maximum difference between [math]
x_{i} \text{ and } x_{i+1}
[/math], then for any meshing (not just the nice one I chose in my second post for simplicity)
then given
[math]
lim_{n \to \infty} |mesh(I)| = 0
[/math]
with mesh(I) being the partition of [a,b]
then the 3rd latex line of my second post holds.

>>8543885
Err, I forgot to explain why the third line holds.
Given that for any
[math] i \in [1, n]
[/math]
and a meshing of I = [a, b], say mesh(I),
then
[math]
x_{i+1} - x_{i} < |mesh(I)|
[/math]
so if
[math]
lim_{n \to \infty} |mesh(I)| = 0 \implies \forall \epsilon > 0 \exists N \text{ s. t. } n > N \implies max({x_{i+1} - x_{i} \text{ s.t. } i \in [1, n]},\text{ } (x_{i}, x_{i+1}) \in [a, b]) < \epsilon
[/math]
then for any other pair, this must also hold true (since for all
[math]
(x_i, x_{i+1}) \text{, }x_{i+1}-x_{i} \leq max(x_{i+1} - x_{i})
[/math])

>>8543908
Anyway, I'm sure there's still several holes in the argument that keep it from being fully rigorous. I'm sorry that I'm not quite educated enough to see them, but to the OP: from this, just see that given the constraints in the statement of the fundamental theorem of calculus, it follows pretty succinctly from the definitions you learned in calc 1. This is much less important, however, than the generalizations into higher dimensions that you learn in Multivariable calculus.

let F(x) be the integral from a to x of a continuous function f

it is easy to prove that F' =f
now let's take another function whose derivative is f, call it G.
then (F-G)'=0
this G= F+ c where c is a constant.
thus F(x) = G(x)-G(a)