# 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

Staff Emeritus
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 $(a,b)$?!