Fundamental theorem of calculus - question & proof verifying

["Don't panic!"]

I understand that the fundamental theorem of calculus is essentially the statement that the derivative of the anti-derivative F evaluated at x\in (a,b) is equal to the value of the primitive function (integrand) f evaluated at x\in (a,b), i.e. F'(x)=f(x). However, can one imply directly from this that F'=f\;\;\forall x\in (a,b), such that F(b)-F(a)=\int_{a}^{b}F'(x)dx\;\;?

Also, I have attempted to prove the FTC and am interested to know whether my attempt is valid?!

Let f be a continuous function on [a,b] and let F: \mathbb{R}\rightarrow\mathbb{R} be a function that is continuous on [a,b] and differentiable on (a,b), defined such that F(x)=\int_{a}^{x}f(t)dt where x\in (a,b).
As f is continuous on [a,b] we therefore know that for any \varepsilon >0\;\;\exists\delta >0 such that 0<\vert t-x\vert <\delta\;\;\Rightarrow\;\;\big\vert f(t)-f(x)\big\vert <\varepsilon\quad\forall x \in [a,b]
Hence, consider the interval (x,x+h)\subseteq (a,b) and let 0<h<\delta. Now clearly x<t<x+h\;\;\Rightarrow\;\;0<t-x<h<\delta and so \big\vert f(t)-f(x)\big\vert <\varepsilon\quad\forall t \in (x,x+h) which implies that in this interval f(x)-\varepsilon<f(t)<f(x)+\varepsilon Using that f(x)\leq g(x)\;\;\Rightarrow\;\;\int_{a}^{b}f(x)dx\leq \int_{a}^{b}g(x)dx it follows that \int_{x}^{x+h}\left[f(x)-\varepsilon\right]dt<\int_{x}^{x+h}f(t)dt<\int_{x}^{x+h}\left[f(x)+\varepsilon\right]dt \\ \Rightarrow \left[f(x)-\varepsilon\right]h<\int_{x}^{x+h}f(t)dt<\left[f(x)+\varepsilon\right]h \\ \Rightarrow f(x)-\varepsilon<\frac{1}{h}\int_{x}^{x+h}f(t)dt<f(x)+\varepsilon \\ \Rightarrow\Bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt - f(x)\Bigg\vert <\varepsilon and as such the limit \lim_{h\rightarrow 0^{+}}\frac{1}{h}\int_{x}^{x+h}f(t)dt=f(x) exists.
Next we consider the interval (x-h,x)\subseteq (a,b) and again let 0<h<\delta. We see that x-h<t<x\;\;\Rightarrow\;\;-h<t-x<0\;\;\Rightarrow\;\;0<-(t-x)<h and so, following the same procedure as before \big\vert f(t)-f(x)\big\vert <\varepsilon\quad\forall t \in (x-h,x) \\ \Rightarrow\int_{x-h}^{x}\left[f(x)-\varepsilon\right]dt<\int_{x-h}^{x}f(t)dt<\int_{x-h}^{x}\left[f(x)+\varepsilon\right]dt \\ \Rightarrow \left[f(x)-\varepsilon\right]h<\int_{x-h}^{x}f(t)dt<\left[f(x)+\varepsilon\right]h\\ \Rightarrow f(x)-\varepsilon<\frac{1}{h}\int_{x-h}^{x}f(t)dt<f(x)+\varepsilon \\ \Rightarrow\Bigg\vert \frac{1}{h}\int_{x-h}^{x}f(t)dt - f(x)\Bigg\vert <\varepsilon and as such the limit \lim_{h\rightarrow 0^{-}}\frac{1}{h}\int_{x-h}^{x}f(t)dt=f(x) exists.
Therefore, as the right-handed and left-handed limits exist and are equal, we have that for any \varepsilon >0\;\;\exists\delta >0 such that \left(0<\vert t-x\vert <h<\delta\;\;\Rightarrow\;\; 0<h<\delta\right)\;\;\Rightarrow\;\;\bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)\bigg\vert <\varepsilon i.e. the limit \lim_{h\rightarrow 0}\frac{1}{h}\int_{x}^{x+h}f(t)dt=f(x) exists. As \frac{1}{h}\int_{x}^{x+h}f(t)dt= F(x+h)-F(x) this implies that \lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}=F'(x)=f(x)

 

[nuuskur]

The same holds if F\colon\mathbb{R}\to\mathbb{R}. The demands apply only to the closed interval a,b such that F is differentiable for every x in the open interval a,b. It implies nothing on how F behaves in its entire domain. If you define F from [a,b] to R, then it somewhat implies that the Fundamental theorem is only applicable to functions that are continuous in their entire domain.

 

["Don't panic!"]

You're right, I think I have edited my original post correctly?
Is the proof (that I gave above) itself ok though?

 

[nuuskur]

Let ff be a continuous function on [a,b][a,b] and let F:R→RF: \mathbb{R}\rightarrow\mathbb{R} be a function that is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), defined such that

F(x)=∫xaf(t)dt​

You define F(x) = \int_a^x f(t)dt The way I understand it is that F(a) = 0. So does the fundamental theorem only work if (for this case) F(a) = 0? The constant plays no role there once you dive into the limit definition, but it works for every constant : F(a) does not have to be 0 or have I mis-understood that part?

The limit part is handled nicely and concisely. The only thing I would be curious of would be if F(a)\neq 0

 

["Don't panic!"]

The way I understand it is that F(a) = 0. So does the fundamental theorem only work if (for this case) F(a) = 0? The constant plays no role there once you dive into the limit definition, but it works for every constant : F(a) does not have to be 0 or have I mis-understood that part?

I'm not sure I understand why F(a)=0? As far as I understand a is taken to be arbitrary, and similarly, there is no reason a priori for F(a)=0. The definition of F is just that, a way of defining a function that we wish to consider in our proof. We assume this, and then show that if this is the case then F'(x)=f(x), however, this does not imply that F(a)=0 as any two anti-derivatives that differ by a constant give the same result, so if we had F(a)\neq 0, then \left(F(x)-F(a)\right)'=F'(x)=f(x). The proof does not mean that we can immediately conclude that F(x)=\int_{a}^{x}F'(t)dt however, if I've understood it correctly it does mean that we can conclude that F(x)=\int_{a}^{x}F'(t)dt +c where c is some constant. Using the other part of the FTC (that I didn't state in my OP) we can then conclude that F(x)=\int_{a}^{x}F'(t)dt +c=F(x)-F(a)+c\;\;\Rightarrow\;\; c=F(a) and therefore F(x)=F(a)+\int_{a}^{x}F'(t)dt

 

[HallsofIvy]

You said, originally, that you were defining F(x) by F(x)= \int_a^x F'(t)dt. From that it follows immediately that F(a)= \int_a^a F'(t)dt= 0 no matter what F' was. That was what nuuskar was complaining about. Now you have changed to F(x)= F(a)+ \int_a^x F'(t)dt which fixes that problem. Of course you would have to have another equation defining F(a) separately.

 

["Don't panic!"]

Apologies, hadn't meant to cause confusion or come across aggressive. It was two separate questions in my first post, the first was that can one imply from F'(x)=f(x) that F(b)-F(a)=\int_{a}^{b}F'(x)dx? And then whether my proof is correct or not?

 

[nuuskur]

That implication is a bit tough for me too. I tend to think that \int_a^b F'(x)dx = F(b) - F(a) implies Riemann integrability on [a,b] specifically which does not come from F'(x) = f(x) , not directly
With the defined F that I growled at, the argument is conclusive for one specific case and not for every case. You would need to redefine your F such that it accounts for every case possible.

 

["Don't panic!"]

Yeah, it has been bothering me, as I would've thought that from F'(x)=f(x) that one can imply that F'=f\;\;\forall x\in (a,b) as x was chosen arbitrarily from the interval (a,b)?!