Ok so i have been really digging into calculus recently and i'm trying to really figure out why $f(b)- f(a) = \int_{a}^bf'(x)$
I'm not trying to get a rigorous proof here, but i also dont want to base my understanding with flawed logic.
Ok here i go:
Let's say I have
$f(x)$ and $f'(x)$
Let's say that I'm interested in the values of $f(x)$ between some $x$ values $[a,b]$, I call the total change in the function $\Delta y$.
$\Delta y$ = $f(b)-f(a)$
I can think of the total change in the function ($\Delta y$) as the sum of infinitesimally small changes so let's say $\Delta y$ is the sum of infinitesimally small changes $\delta y$.
I guess it could be defined as:
$δy$ = $\lim_{n \to \infty} \frac {f(b)-f(a)}{n}$
So the total change in the function is:
$\Delta y$ = $\sum δy$ = $\lim_{n \to \infty} \sum _{i=1}^n \frac {(f(b)-f(a)}{n}$
I can assume that there is very little $x$ distance between one $\delta y$ and the next, so the distance will tend to cero.
$\delta x \rightarrow 0$
I will take a leap here and say that I could get the slope of each point by calculating $\frac {\delta y}{ \delta x}$, therefore I can compute the values between $f'(a)$ and $f'(b)$
given that, i could calculate the area under $f'(x)$ between $(a,b)$ by multiplying each point of the function with a very small width, i know that the distance between each $\delta y$ and the next is $\delta x$ so:
Area of an infinitesimal rectangle = $\frac {\delta y}{ \delta x} * \delta x$
Total area = $\sum \frac {\delta y}{ \delta x} * \delta x$
Therefore i cancel the $\delta x$
Area of infinitesimal rectangle = $\delta y$
Total area = $\sum \delta y$
And we earlier defined $\Delta y$ as $\sum \delta y$, that means that
Total area = $\Delta y$
Therefore the area under the curve between $f'(a)$ and $f'(b)$ is equal or close to $f(b)-f(a)$
Sorry for the long read, I want to know if this reasoning is correct, infinite thanks to those who read the whole thing and want to help me.