# Proving the fundamental theorem of calculus using limits

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1. Feb 24, 2015

### "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 $F(x)=F(a)+\int_{a}^{x}\frac{dF}{dt}dt$?)

Last edited: Feb 24, 2015
2. Feb 24, 2015

### 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.

Last edited: Feb 24, 2015
3. Feb 24, 2015

### PeroK

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

4. Feb 24, 2015

### "Don't panic!"

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)

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.

5. Feb 24, 2015

### lavinia

Try doing the proof as I sketched it for yourself.

6. Feb 24, 2015

### "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 $$\lim_{h\rightarrow 0} \frac{1}{h}\int_{x}^{x+h}f(t)dt= \lim_{h\rightarrow 0} \frac{F(x+h)-F(x)}{h} = F'(x) = f(x)$$

7. Feb 24, 2015

### PeroK

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

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.

8. Feb 24, 2015

### "Don't panic!"

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 $f(x)\in\mathbb{R}$

Last edited: Feb 24, 2015
9. Feb 24, 2015

### 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 $0 < h < \delta$ ...

10. Feb 25, 2015

### "Don't panic!"

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...?

11. Feb 25, 2015

### 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.

Last edited: Feb 25, 2015
12. Feb 25, 2015

### "Don't panic!"

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

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 continous in this interval and hence the right-hand limit must equal the left-hand limit and so the derivative $F'(x)$ exists?