Fundamental theorem of calculus - question & proof verifying

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1. Mar 4, 2015

"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 $$\lim_{h\rightarrow 0}\frac{F(x+h)-F(x)}{h}=F'(x)=f(x)$$

Last edited: Mar 4, 2015
2. Mar 4, 2015

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.

Last edited: Mar 4, 2015
3. Mar 4, 2015

"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?

4. Mar 4, 2015

nuuskur

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 $F(a)\neq 0$

5. Mar 4, 2015

"Don't panic!"

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 $$F(x)=F(a)+\int_{a}^{x}F'(t)dt$$

6. Mar 4, 2015

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.

7. Mar 4, 2015

"Don't panic!"

Apologies, hadn't meant to cause confusion or come across agressive. 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?

8. Mar 4, 2015

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.

Last edited: Mar 4, 2015
9. Mar 4, 2015

"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 $(a,b)$?!