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.

  • $\begingroup$ What mean "infinitesimal"? $\endgroup$ Sep 23, 2016 at 20:26
  • $\begingroup$ infinitely small en.wikipedia.org/wiki/Infinitesimal $\endgroup$ Sep 23, 2016 at 20:27
  • $\begingroup$ Read the argument I give below, it should clear this up for you. $\endgroup$ Sep 23, 2016 at 20:28
  • $\begingroup$ @JoaquinBrandan: OK, what mean "infinitely small" in your context. Do you have a rigorous definition? If no, your "proofs" cannot be correct. $\endgroup$ Sep 23, 2016 at 20:30
  • $\begingroup$ As i said in my second line, "I'm not trying to get a rigorous proof here", however i did define it as δy $\endgroup$ Sep 23, 2016 at 20:31

1 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).


$\textbf{Edit}$ (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!

  • $\begingroup$ Hi faraad, thanks for your answer. I'm aware of this, it's just that i'm trying to get a more intuitive (at least more intuitive for me) approach. I know that in my case $f$ in an antiderivative of $f'$, and i'm aware that my result is just FTC, what i'm trying to understand is if my way of thinking of the FTC is also valid because it makes more sense to me, I know that my result is correct (as stated by FTC), but I dont know if the logic i'm using to justify it is really valid or if i'm just fooling myself. $\endgroup$ Sep 23, 2016 at 20:36
  • $\begingroup$ I made an edit for the intuitive approach which I think is really clear. $\endgroup$ Sep 23, 2016 at 20:50
  • $\begingroup$ Thanks, i cant quite see the reasoning behind the last step where the sum is equal to the change in $f$, where did that come from?. does this means that the logic i developed is flawed somewhere?. $\endgroup$ Sep 23, 2016 at 20:54
  • $\begingroup$ I have a really hard time reading what others wrote. I try to get the big picture of what they are saying and then go form there. Really, if you draw a picture you'll see what I mean. Think of how much $\{[x_i^*, x_i^* + \Delta x]\}_{i=1}^n$ can cover if $\Delta x$ is very small and the partition is really big. However you see the problem now with a non-rigorous argument, it's just too hand-wavy. $\endgroup$ Sep 23, 2016 at 21:00
  • $\begingroup$ Sorry, I just understood that, i missed the fact that the last sum wasnt for the derivative function but for the antiderivative. Then what i dont understand is why the last two sums are similar? $\endgroup$ Sep 23, 2016 at 21:01

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