Proving the fundamental theorem of calculus using limits

["Don't panic!"]

Would it be a legitimate (valid) proof to use an $\epsilon$-$\delta$ limit approach to prove the fundamental theorem of calculus?
i.e. as the FTC states that if $f$ is a continuous function on $[a,b]$, then we can define a function $F: [a,b]\rightarrow\mathbb{R}$ such that $$F(x)=\int_{a}^{x}f(t)dt$$
Then $F$ is differentiable on $(a,b)$, and $\forall\;\; x\in (a,b)$,
$$F'(x)=f(x)$$ Is it possible to prove that this is true by showing that for any given $\varepsilon >0$ we can find a $\delta >0$ such that $$\Bigg\vert\frac{F(x+h)-F(x)}{h}-f(x)\Bigg\vert <\varepsilon$$ whenever $0<\delta < h$?

(I've found these set of notes https://math.berkeley.edu/~ogus/Math_1A/lectures/fundamental.pdf that uses this approach and it seems legitimate and easy to follow, but I'm unsure whether it's valid or not)?!

(Also, is it valid to say that $F'(x)=f(x)\Rightarrow F'=f$, i.e. the derivative of the function $F$ is equal to the original function $f$, such that $F(x)=F(a)+\int_{a}^{x}\frac{dF}{dt}dt$?)

 

[lavinia]

You can definitely use the Newton quotient method to prove the fundamental theorem of calculus.
I would try approximating the integral from x to x+h by hf(x) and thus the Newton quotient by f(x).
I would then use the continuity of f to show that this approximation differs from the actual integral by epsilon as h goes to zero.

 

[PeroK]

Then $F$ is differentiable on $(a,b)$, and $\forall\;\; x\in (a,b)$,
$$F'(x)=f(x)$$ Is it possible to prove that this is true by showing that for any given $\varepsilon >0$ we can find a $\delta >0$ such that $$\Bigg\vert\frac{F(x+h)-F(x)}{h}-f(x)\Bigg\vert <\varepsilon$$ whenever $0<\delta < h$?

Why wouldn't that be valid? (Although, you've got $h$ and $\delta$ the wrong way round in the final inequality.)

 

["Don't panic!"]

You can definitely use the Newton quotient method to prove the fundamental theorem of calculus.
I would try approximating the integral from x to x+h by hf(x) and thus the Newton quotient by f(x).
I would then use the continuity of f to show that this approximation differ from the actual integral by epsilon as h goes to zero.

So the proof given in the link I gave would be a valid one then?! (Sorry for going on a bit, just wanted to double check)

Why wouldn't that be valid? (Although, you've got hh and δ\delta the wrong way round in the final inequality.)

Whoops, yes thanks for pointing out the mistake. Yes, I couldn't see anything wrong with it myself, but I'm not particularly confident in myself, so wanted to clarify it.

 

[lavinia]

So the proof given in the link I gave would be a valid one then?! (Sorry for going on a bit, just wanted to double check)

Try doing the proof as I sketched it for yourself.

 

["Don't panic!"]

Would the following be correct?

Let $f$ be a continuous function on $[a,b]$ and define a function $F: [a,b]\rightarrow\mathbb{R}$ such that $$F(x)=\int_{a}^{x}f(t)dt$$
Now, for small $h\in\mathbb{R}$ we can approximate the difference between the values of $F$ at $x$ and at $x+h$ by $hf(x)$, i.e. $$F(x+h)-F(x)\approx hf(x)$$
As $f$ is continuous on $[a,b]$ we know that, for a given number $t\in [x,x+h]$ the following limit exists $$\lim_{h\rightarrow 0}f(t)=f(x)$$
In other words, there exists a $\delta >0$ such that for any $\varepsilon >0$, $$0<h<\delta\;\;\Rightarrow\;\; \bigg\vert f(t)-f(x)\bigg\vert <\varepsilon$$ This implies that $$f(x)-\varepsilon <f(t)< f(x)+\varepsilon \\ \Rightarrow\;\;\int_{x}^{x+h}\left(f(x)-\varepsilon\right)dt < \int_{x}^{x+h}f(t)dt < \int_{x}^{x+h}\left(f(x)+\varepsilon\right)dt \\ \Rightarrow\;\; \left(f(x)-\varepsilon\right)h < \int_{x}^{x+h}f(t)dt < \left(f(x)+\varepsilon\right)h \\ \Rightarrow\;\; f(x)-\varepsilon < \frac{1}{h}\int_{x}^{x+h}f(t)dt < f(x)+\varepsilon$$ which further implies that $$\bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)\bigg\vert < \varepsilon$$ and as such the limit $$\lim_{h\rightarrow 0} \frac{1}{h}\int_{x}^{x+h}f(t)dt = f(x)$$ exists.
Finally we note that $$F(x+h)-F(x)=\int_{a}^{x+h}f(t)dt - \int_{a}^{x}f(t)dt \\ \qquad\qquad\qquad\quad\;= \left(\int_{a}^{x}f(t)dt +\int_{x}^{x+h}f(t)dt\right) - \int_{a}^{x}f(t)dt \\ \qquad\qquad\qquad\quad\;= \int_{x}^{x+h}f(t)dt$$
and so $$\lim_{h\rightarrow 0} \frac{1}{h}\int_{x}^{x+h}f(t)dt= \lim_{h\rightarrow 0} \frac{F(x+h)-F(x)}{h} = F'(x) = f(x)$$

 

[PeroK]

Some comments:

Would the following be correct?

Let $f$ be a continuous function on $[a,b]$ and define a function $F: [a,b]\rightarrow\mathbb{R}$ such that $$F(x)=\int_{a}^{x}f(t)dt$$
Now, for small $h\in\mathbb{R}$ we can approximate the difference between the values of $F$ at $x$ and at $x+h$ by $hf(x)$, i.e. $$F(x+h)-F(x)\approx hf(x)$$

I'm not sure why you included this, as you don't use it. That's the "informal" proof, just to use that approximation.

As $f$ is continuous on $[a,b]$ we know that, for a given number $t\in [x,x+h]$ the following limit exists $$\lim_{h\rightarrow 0}f(t)=f(x)$$
In other words, there exists a $\delta >0$ such that for any $\varepsilon >0$, $$0<h<\delta\;\;\Rightarrow\;\; \bigg\vert f(t)-f(x)\bigg\vert <\varepsilon$$

You need to sort out the order of $\epsilon-\delta$ here. And also, you need to think about how $t$ relates to this.

Hint: it should be $\epsilon$ then $\delta$ then $h$ then $t$

I think the rest is fine, although you might consider why you can take $h > 0$. This seems to be a flaw in the link as well.

This implies that $$f(x)-\varepsilon <f(t)< f(x)+\varepsilon \\ \Rightarrow\;\;\int_{x}^{x+h}\left(f(x)-\varepsilon\right)dt < \int_{x}^{x+h}f(t)dt < \int_{x}^{x+h}\left(f(x)+\varepsilon\right)dt \\ \Rightarrow\;\; \left(f(x)-\varepsilon\right)h < \int_{x}^{x+h}f(t)dt < \left(f(x)+\varepsilon\right)h \\ \Rightarrow\;\; f(x)-\varepsilon < \frac{1}{h}\int_{x}^{x+h}f(t)dt < f(x)+\varepsilon$$ which further implies that $$\bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)\bigg\vert < \varepsilon$$ and as such the limit $$\lim_{h\rightarrow 0} \frac{1}{h}\int_{x}^{x+h}f(t)dt = f(x)$$ exists.
Finally we note that $$F(x+h)-F(x)=\int_{a}^{x+h}f(t)dt - \int_{a}^{x}f(t)dt \\ \qquad\qquad\qquad\quad\;= \left(\int_{a}^{x}f(t)dt +\int_{x}^{x+h}f(t)dt\right) - \int_{a}^{x}f(t)dt \\ \qquad\qquad\qquad\quad\;= \int_{x}^{x+h}f(t)dt$$
and so $$\lim_{h\rightarrow 0} \frac{1}{h}\int_{x}^{x+h}f(t)dt= \lim_{h\rightarrow 0} \frac{F(x+h)-F(x)}{h} = F'(x) = f(x)$$

 

["Don't panic!"]

Thanks for your feedback.

You need to sort out the order of ϵ−δ\epsilon-\delta here. And also, you need to think about how tt relates to this.

Hint: it should be ϵ\epsilon then δ\delta then hh then tt

So would it be the following?

As $f$ is continuous at $x\in [a,b]$, we have that for any $\varepsilon >0$ there exists a $\delta >0$ such that $0<h< \delta$ (where $h>0$ is some small positive number). Then, for all $t\in [x,x+h]$ that satisfy $0<(t+h)-t=h< \delta$, we have that the following inequality holds $\bigg\vert f(t)-f(x)\bigg\vert <\varepsilon$, and as such the following limit exists $$\lim_{h\rightarrow 0}f(t)=f(x)$$ where $f(x)\in\mathbb{R}$

 

[PeroK]

That's essentially correct but if you break it down a bit more I think it's more logical. For example:

Let $\epsilon > 0$

As f is continuous $\exists \delta > 0$ such that $|t - x| < \delta \Rightarrow |f(t) - f(x)| < \epsilon$

Let $0 < h < \delta$ ...

 

["Don't panic!"]

As f is continuous ∃δ>0\exists \delta > 0 such that |t−x|<δ⇒|f(t)−f(x)|<ϵ|t - x| < \delta \Rightarrow |f(t) - f(x)| < \epsilon

Let 0<h<δ0 < h < \delta ...

Is what you've written the case because of the following:

$f$ is continuous $\forall\;\; t\in [a,b]$ and in particular at $t=x$ and so $\exists\delta >0$ such that $\vert t-x\vert <\delta\Rightarrow\big\vert f(t)-f(x)\big\vert <\varepsilon$, i.e. the following limit exists: $$\lim_{t\rightarrow x}f(t)=f(x)$$ Given this we can consider the interval $[x,x+h]$ (where $h>0$ is some small number) such that $$x\leq t\leq x+h \;\;\Rightarrow\;\; 0\leq t-x \leq h<\delta$$ which motivates us to assume the same $\delta >0$ such that we have $0<h<\delta$. We can then use that, from the above criteria, when $t\in [x,x+h]$ we have that $\vert t-x\vert <\delta$ and so we know that the following inequality holds, $f(x)-\varepsilon <f(t)< f(x)+\varepsilon$. From this, we can use that $$f(x)-\varepsilon <f(t)< f(x)+\varepsilon\;\;\Rightarrow\;\;\int_{x}^{x+h}\left[f(x)-\varepsilon \right]dt < \int_{x}^{x+h}f(t)dt < \int_{x}^{x+h}\left[f(x)+\varepsilon \right]dt \\ \Rightarrow \left[f(x)-\varepsilon \right]h <\int_{x}^{x+h}f(t)dt < \left[f(x)+\varepsilon \right]h \;\;\Rightarrow\;\; f(x)-\varepsilon <\frac{1}{h}\int_{x}^{x+h}f(t)dt < f(x)+\varepsilon$$ and this implies that $$\Bigg\vert \frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)\Bigg\vert <\varepsilon$$ whenever $0<h<\delta$ and the rest follows...?

 

[PeroK]

You are over-complicating things. What I wrote is based on the clear and simple definition of the limits we are considering. To extend this a bit further:

Let $x \in (a, b)$

Let $\epsilon > 0$

As $f$ is continuous at $x$ $\exists \delta > 0$ such that $|t - x| < \delta \Rightarrow |f(t) - f(x)| < \epsilon$

Let $0 < h < \delta$

$t \in [x, x+h] \Rightarrow |t-x| < \delta \Rightarrow |f(t)-f(x)| < \epsilon$

$\Rightarrow \int_{x}^{x+h}|f(t)-f(x)|dt < h\epsilon \Rightarrow |\int_{x}^{x+h}f(t)-f(x)dt| < h\epsilon \Rightarrow |\int_{x}^{x+h}f(t)dt -hf(x)| < h\epsilon$

$\Rightarrow |\frac{1}{h}\int_{x}^{x+h}f(t)dt -f(x)| < \epsilon$

$\therefore \lim_{h \rightarrow 0^{+}}\frac{1}{h}\int_{x}^{x+h}f(t)dt = f(x)$

Which, hopefully, is clear and logical.

 

["Don't panic!"]

Thanks for your input, I admit I do have a tendency to over-complicate things at times, apologies for that!

t∈[x,x+h]⇒|t−x|<δ

Does this follow because if $t\in [x,x+h]$ then $x\leq x\leq x+h$ which implies that $0\leq t-x \leq h < \delta$ (in the limit definition doesn't it have to be $0<\vert t-x\vert<\delta$ though?)

Also, can we get away with just considering $h>0$ by noting that $f$ is continuous in this interval and hence the right-hand limit must equal the left-hand limit and so the derivative $F'(x)$ exists?