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I'm aware this question has been asked before but I believe I have a unique specific question. I have read, and believe I understand the proof of:

$$ \int_{a}^{b}\ f'(x){\mathrm{d} x} = f(b)-f(a) .$$

A very good explanation and proof is given by littleO at Why does the fundamental theorem of calculus work? I'm sure there are other good proofs. He points out that "the total change is the sum of all the little changes".
I understand. I want to step back a bit and ask, why on earth would you expect that knowing something about the end points (the anti-derivatives at a and b), would tell you all you need to know about a property of the entire thing (the area under the curve). It doesn't matter if the function is $f(x) = 1$; $f(x) = 2x$ or $f(x) = 100 * x^{99}$. The area under the curve will be the same from 0 to 1, and we can figure that out just from knowledge of just the end points. What is the intuition?

Put another way, I (believe) I understand the proof, but I'm wondering why one should expect that knowledge of the end points tell us all we need to know about a property that has to deal with the entire interval (the area under the curve). I'm hoping insight might prepare me for ideas in physics about the surface of a ball telling us what we need to know about every point inside the ball. Thanks

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    $\begingroup$ In my opinion, one wouldn’t. That’s why the FTC is such a mind-boggling result (IMHO): it tells us that two very different questions (what is the slope of the tangent to the graph at a given point? What is the area under graph of the function?) are actually intimately connected and are, in a sense, “inverses” of each other. To me, it is a surprising result, not an intuitive one. $\endgroup$ Commented May 2, 2021 at 22:19
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    $\begingroup$ Think of it as a continuity/flow equation: Think of it in terms of sequences. suppose you have a sequence $f_1,...$ and the differences $d_1 = f2-f1,...$ then you have $f_n -f_1 = d_{n-1}+d_{n-2}+...+d_1$. Think of the derivative as the difference. $\endgroup$
    – copper.hat
    Commented May 2, 2021 at 22:21
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    $\begingroup$ Take a look at this Pavel Grinfeld's video. I like that explanation a lot. $\endgroup$ Commented May 2, 2021 at 22:38
  • $\begingroup$ Maybe it helps to view the fundamental theorem "from the perspective" of an antiderivative $F$ instead of from the viewpoint of $f = F'$: We're saying the net change in $F$ over $[a, b]$ is equal to the "sum" (integral) of the infinitesimal increments $f(x)\, dx = F'(x)\, dx$ as $x$ runs from $a$ to $b$. If you'll forgive a rhetorical question, why would one not believe that to happen regardless of $F$ (perhaps under mild technical hypotheses, such as $F'$ exists and is continuous)? :) $\endgroup$ Commented May 2, 2021 at 23:25
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    $\begingroup$ "why on earth would you expect that knowing something about the end points [...] would tell you all you need to know about a property of the entire thing" ... That aspect of FTC is one of those unexpected things that make math fascinating. One day, you'll learn that Green's Theorem shows how "path integrals" say something about what happens over the interior a region by studying what happens over its boundary; likewise for "surface integrals" and beyond. FTC is just the beginning of a whole boundary-info-encodes-interior-info thing. $\endgroup$
    – Blue
    Commented May 3, 2021 at 11:34

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One way of thinking about it is: $$I=\int_a^bf'(x)\,dx=\int_a^b\frac{df}{dx}dx$$ now let $u=f\Rightarrow \frac{du}{dx}=\frac{df}{dx}\Rightarrow dx=du\frac{dx}{df}$ now change the limits and sub in: $$I=\int_{f(a)}^{f(b)}\frac{df}{dx}\frac{dx}{df}du=\int_{f(a)}^{f(b)}du=u|_{f(a)}^{f(b)}=f(b)-f(a)$$


EDIT

In terms of why the antiderivative would tell us the area of the whole thing, It might make some sense to think of the antiderivative as the area from zero up to a given point, in other words: $$\int_0^xf(X)dX=F(x)-F(0)$$ and now from one point to another would just be: $$\int_a^bf(x)dx=\int_0^bf(x)dx-\int_0^af(x)dx=[F(b)-F(0)]-[F(a)-F(0)]=F(b)-F(a)$$ We can thinking of it the opposite way too, $F'(x)=f(x)$, or the function is just the derivative of its antiderivative, meaning the function gives us the rate of change of its area and when we integrate it (sum it over a continuous interval) we are getting the total change of area

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Suppose you have a bunch of numbers $y_1,y_2,\dots,y_n$, and you want to add them all up, but you don't know how. I tell you to do the following. Start with some arbitrary number $s_0$, then add $y_1,y_2,\dots$ to it one by one: $$\begin{align} s_1 &= s_0 + y_1, \\ s_2 &= s_1 + y_2, \\ &\vdots \\ s_n &= s_{n-1} + y_n. \end{align}$$ Then, I claim, the sum you want is simply $s_n - s_0$.

Now meditate upon your original questions:

I want to step back a bit and ask, why on earth would you expect that knowing something about the end points ([the values $s_0$ and $s_n$]), would tell you all you need to know about a property of the entire thing ([the sum of all the numbers]). It doesn't matter if the [numbers are $y_i=1$; $y_i=2i$ or $y_i=100i^{99}$]. The [sum] will be the same from [$1$ to $n$], and we can figure that out just from knowledge of just the end points. What is the intuition?

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  • $\begingroup$ Thank you for the answer Rahul, and I did mediate on my original question :). I believe you are trying to say than s(n) is already the addition of all the previous numbers. Just like the anti-derivative (the integral) at b is the addition of all the previous little areas. However, this doesn't explain the extraordinary fact that the anti-derivative appears (to me anyhow) to be something completely different as well. It is the function whose derivative at b is f(b). Perhaps I'm trying to ask "Why is a function whose derivative is f at b, also the total area under the curve?". Thanks! $\endgroup$
    – Dave
    Commented May 4, 2021 at 13:44
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I've been thinking about it for a bit and think I've come up with a way to understand it.

The function we want to find the area under is the derivative of the antiderivative, which is what we plug numbers into to evaluate definite integrals.

The function is the rate of change of the antiderivative, so the antiderivative accumulates the change of the original function over time.

So when you plug in a value into the antiderivative, you're finding the accumulated change of the original function from x=0 to that point. If you want from 2 to 8, simply subtract the accumulated change from 0 to 2 from 0 to 8.

I believe this is how it works. Not sure if this is a correct way of thinking about it. Not sure if I explained it well. Just a Calc II student.

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