relation between derivative and integral regarding area and distance traveled

Question

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.

Answer 1

There are two things you have to do here which are Fundamental Theorem of Calculus and another result which involves functions of one variable which have the same derivative. I will just prove the latter since FTC is online in numerous places. Let $f$ be continuous on $[a,b]$ and define;

$$g(x):= \int_{a}^x f(t) \ dt$$

Then $g$ is continuous and by FTC $g'(x) = f(x)$. Now suppose you have another function $F(x)$ such that $F'(x) = f(x)$. This is the second result now in which I stated I would prove. Define a new function $h(x) = g(x) - F(x)$ then $h'(x) = 0$ which implies $h(x) = c$ for some $c \in \mathbb{R}$. Hence, $g(x) = F(x)+c$ with $g,F$ having their same definitions as above we get;

$$g(a) = F(a) + c = 0 \Rightarrow c = -F(a) \Rightarrow g(b) = \int_{a}^b f(t) \ dt = F(b)+c= F(b) - F(a) $$

In your case $f$ is the anti-derivative of $f$ and in this case $F$ is the anti-derivative of $f$, so the result is really just FTC $II$ (given above).

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$\textbf{Edit}$ (non-rigorous): If you want some kind of intuitive result you note that for a very,very small $\Delta x$ and large $N$;

$$f'(x) \approx \frac{f(x+\Delta x) - f(x)}{\Delta x} \Rightarrow \int_{a}^b f'(x) \approx \sum_{i=1}^N f'(x_i^*) \ \Delta x \approx \sum_i^N f(x_i^* + \Delta x) \approx f(b) - f(a)$$

for the last approximation, look at the graph of a continuous function and think about what that sum is doing if you think of taking a really big partition!