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I was trying to understand, how does the proof of the FTC1 ties to FTC2. Can someone please correct me, if my reasoning is wrong?

The first part tells us, that if function $f$ is continous over the interval $[a,b]$, then:

$f(x) = F'(x)$, where $F(x) = \int_a^xf(t)dt$, for every $x \in [a,b]$

Now would it be correct (?) to write this as:

$F'(x) = f(x)$ // $\int$

antiderivative of both sides, to get rid of the derivative on the left, and we get:

$F(x) = \int f(x)$

And then just plug it in:

$\int_a^bf(t)dt = \int_a^bf(t)dt - \int_a^af(t)dt = \int f(b) - \int f(a) $

So we are left with the notion that a definite integral of function $f$ from $a$ to $b$ is the antiderivative of $f$ evaluated at $b - f$ evaluated at $a$: which is how we calculate integrals

Is my reasoning good? ( for a beginner of course, this is not meant to be a professional proof :))

Bruno
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    Hey Bruno. These arguments are fine- but I think you would like to fix your last line into $F(b)-F(a) $, instead of these not well defined integrals. – Ron Abramovich Jan 10 '21 at 20:55
  • @RonAbramovich thanks for help, I agree with your comment $F(b) - F(a)$ is how it is written in the textbooks. But to be clear, this still means, that to calculate the definitie integral, we actually substract an undefined integral at two points, right? – Bruno Jan 10 '21 at 21:03
  • The power of FTC part 2 (which deals with evaluation of an integral as difference between values of anti-derivative) can be understood only when we know the definition of an integral. An integral is usually defined as a complicated limit of a complicated sum and it is quite a surprise that it can be evaluated so easily via anti-derivatives. – Paramanand Singh Jan 16 '21 at 13:33
  • See more details in this answer : https://math.stackexchange.com/a/1768053/72031 – Paramanand Singh Jan 16 '21 at 13:37

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