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The fundamental theorem of calculus in the fifth edition of Stewarts Calculus text is stated as:

Suppose $f$ is continuous on $[a,b]$. Then:

1). If $g(x) = \int_a^x f(t)dt$, then $g'(x)=f(x)$

2). $\int_a^bf(x)dx=F(b)-F(a)$ where $F$ is any antiderivative of $f$.


So, when we interpret $g(x)$ as the function that tells us the "area so far" under the graph of $f$, I think $(1)$ is pretty straightforward... honestly, it seems like with all the brilliant minds that came before Newton/Leibniz that this is something that should have already been clearly understood.

So, the reason the FTC is so "amazing" is usually stated as "OMG, we can calculate the area under the integral by knowing the value of ANY of its antiderivatives at ONLY the two boundary points!"

However, I feel a bit cheated. We define the function $g(x)$ as a limit of Riemann sums that involve the entire curve. So yeah, it's going to give the entire area under the graph of $f$ from $a$ to $x$ even though we only plug $x$ into the function, but that's not to say the calculation of our function didn't involve the whole curve, you know?

After this, it follows quite directly from the mean value theorem that any two antiderivatives differ by a constant, and so we arrive at $(2)$.

Now, I hope I don't come off too sarcastic in this post, because I genuinely suspect that there is something that my feeble mind is not seeing that makes this more amazing, and that one of the more adequate thinkers that linger here can enlighten me.


edit:

antiderivatives and integrals are defined in very different ways. The fact that

$\int_a^x f(t)dt$

is an antiderivative of $f(x)$ is very cool.

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    I don't think you're interpreting the theorem correctly. – Randall Jul 20 '21 at 16:27
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    Differentiation is a process whereby we attempt to understand "what is the instantaneous rate of change of a function?" Alternatively: "What is the slope of the tangent line to the graph at a point?" Integration is a process whereby we attempt to answer "What is the area bounded by the graph of the function?" There is absolutely no reason to think that these two questions would be related in that one is a kind of inverse of the other. Why would finding an area be possible by finding a function whose tangent line slopes are given by the function whose area we are calculating? – Arturo Magidin Jul 20 '21 at 16:30
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    Indeed, versions of (1) were known to James Gregory and Isaac Barrow, predecessors to Newton. – Matthew Leingang Jul 20 '21 at 16:31
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    Also, (2) is a direct consequence of MVT only if $f$ is continuous. But the theorem is true for integrable $f$. That's the harder proof. – Matthew Leingang Jul 20 '21 at 16:32
  • if it's so obvious and trivial then proving it should be no problem. You don't need to look at the book for the proof, right? – David C. Ullrich Jul 20 '21 at 16:32
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    "We define the function () as a limit of Riemann sums that involve the entire curve. So yeah, it's going to give the entire area under the graph of from to even though we only plug into the function." It seems like you're saying that the FTC is close to tautological or obvious, but I don't think that's true. The surprising part to me is not the definition of $g(x)$ via Riemann sums, but the fact that we can in many cases calculate $g(x)$ exactly. – Jair Taylor Jul 20 '21 at 16:33
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    Compare this to, say, Archimedes' methods, where each different kind of shape had to be examined carefully in an ad-hoc way, calculus gives us the tools to compute area very easily. – Jair Taylor Jul 20 '21 at 16:33
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    I voted to close. Whether one is subjectively impressed or unimpressed by this or that result is entirely a matter of personal opinion. Take it to Reddit or something. – Rivers McForge Jul 20 '21 at 16:33
  • Also, it's fundamental to the subject (we can't do integral calculus without it) whether or not it subjectively blows some random person on the Internet away. – Rivers McForge Jul 20 '21 at 16:34
  • I did not vote to close. Whether opinion-based or not, it's an interesting and valuable discussion. – Jair Taylor Jul 20 '21 at 16:35
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    IMO the question is equivalent to “Why is the Fundamental Theorem of Calculus fundamental?" and I think that's a good question. – Matthew Leingang Jul 20 '21 at 16:35
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    @MatthewLeingang except that is not at all how the question is phrased. – Randall Jul 20 '21 at 16:36
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    I also voted to close. I can understand, to some extent, the urge to give the OP some benefit of the doubt regarding this question. But to me the sarcastic and opinionated tone wipes that benefit out by inviting sarcastic and opinionated answers (which I am barely holding back... eeeerrgggghhhh.....) – Lee Mosher Jul 20 '21 at 17:03
  • is amazing because it allows computing explicitly many integrals. 1) is amazing because it gives a convenient way to define many new functions (e.g. logarithms)
  • – Taladris Jul 21 '21 at 01:51
  • @MatthewLeingang: right. And I am surprised it is not a duplicate. – Taladris Jul 21 '21 at 01:51
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    The fundamental theorem of calculus (version 2) can be interpreted as saying "the total change is the sum of all the little changes". From that perspective, it seems almost obvious. But, it's undoubtedly useful to know how to get the total change from the instantaneous rate of change, or vice versa. That's a basic thing we need to do all the time. The theorem only became "obvious" once the correct definitions were stated clearly, and it takes much investigation and exploration to pinpoint the right definitions. – littleO Jul 21 '21 at 07:26
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    @littleO "the total change is the sum of all the little changes" - Excellent, very pithy description of FTC. – Jair Taylor Jul 23 '21 at 21:01