Trying an alternate Proof of the Fundamental Theorem

[Valour549]

The proofs of the Fundamental Theorem of Calculus in the textbook I'm reading and those that I have found online, basically show us:
1) That when we apply the definition of the derivative to the integral of f (say F) below, we get f back.
$$F(x) = \int_a^x f(t) dt$$
2) That any definite integral of f can be found using any of its anti-derivatives F and taking the difference when evaluated at the upper and lower limits.

However, there is no proof that actually starts from the derivative of f (say G), applies the definition of the definite integral to it, and arrives back at f. In any case this is my attempt at it (starting below the red line). Normally when evaluating a definite integral using Riemann sum, the rules of the sigma notation are invoked allowing the breaking down into smaller sums. But with the absence of an actual function in this case I'm stuck and need some help.

 

[andrewkirk]

There is no reason to consider the derivative of the function $f$. Indeed, $f$ does not need to be differentiable for the fundamental theorem to hold. In fact $f$ does not even need to be continuous. All that is necessary is that it be the derivative of $F$.

 

[chiro]

Hey Valour549.

I'm not sure what your question is exactly? Are you trying to dispute the proof or perhaps find a different way of proving the result?

It's a little bit unclear what you are trying to get at.

 

[Valour549]

There is no reason to consider the derivative of the function $f$. Indeed, $f$ does not need to be differentiable for the fundamental theorem to hold. In fact $f$ does not even need to be continuous. All that is necessary is that it be the derivative of $F$.

What you say is true if you're starting with $F$, an antiderivative of $f$, undergoing differentiation and becoming $f$.

But it should also work the other way round; namely we start with $G$, the derivative of $f$, undergoing integration and becoming $f$. Because the FTC tells us that "A

Now I understand that there is no guarantee that $G$, the derivative of $f$, is going to be integrable. Because there are functions like $ln(x)$ that is differentiable on (0,∞), and yet its derivative $1/x$ is not integrable on [1,∞).

However, there are certainly functions whose derivatives are integrable, and I want to attempt to prove that in those cases, applying integration to those derivative gives back the original function.

Hey Valour549.

I'm not sure what your question is exactly? Are you trying to dispute the proof or perhaps find a different way of proving the result?

It's a little bit unclear what you are trying to get at.

Please read the above if you can. But to sum it up, the FTC says (quoting Calculus 7E by Stuart):

1) "If $f$ is integrated and then the result is differentiated, we arrive back at the original function."

I have no problem with this, and I've included the proof given in the book in my opening post (above the red line).

2) "If we take a function $f$, first differentiate it, and then integrate the
result, we arrive back at the original function, but in the form $f(b)-f(a)$."

I am trying to prove this, starting with the definition of the derivative of $f$ (which I call $G$) then integrating it from $a$ to $b$. My attempt of the proof is below the red line in my opening post, but I am stuck at the third step.

 

[andrewkirk]

@Valour549 OK that is clearer now. Given that explanation, I think what you are trying to prove is what is sometimes called the 'Second Fundamental Theorem of Calculus'. The first theorem is about the derivative of an integral. The second is about the integral of a derivative. The two are discussed and contrasted in this wikipedia article and there is a proof of the Second theorem here. For consistency with the first theorem it uses $F$ for the original function and $f$ for $F'$, which is different from your notation. Note that the proof assumes that the derivative $f$ is Riemann-integrable. A key part of the proof is the use of the Mean Value Theorem.

 

[Valour549]

Spoiler

I've read that Wikipedia article and the proofs on it, and yes I understand the importance of that proof because it allows us to easily calculate definite integrals of $f$ simply by making use of anyone of its antiderivatives (as stated in point 2 of my opening post).

For consistency with the first theorem it uses $F$ for the original function and $f$ for $F'$, which is different from your notation.

That is exactly the problem! I don't want to start a proof with the derivative of $F$ from the the first theorem (where $F$ is defined by an integral of $f$), integrating it, then acting surprised that the result returned in the form of $F$.

What I want is to start with:

$G$, the derivative of $f$, using the definition of a derivative (NB: see how I'm purposely avoiding $F$?)
Assuming $G$ is Riemann-integrable, I want to show that the integral of $G$ from $a$ to $b$ gives the result $f(b)-f(a)$. In order to do this, I begin my proof by applying the definition of a definite integral to $G$.

I really don't know how to put this any clearer... I mean I even wrote "want to show..." clearly in blue on the piece of paper.

 

[Stephen Tashi]

I'd think the general idea is to use a "telescoping sum" like $\sum_{i=1}^n { h ( \frac{ f(a+(i+1)h) - f(a + i h)}{h})} = f (a + (n+1)h) - f( a + h) \approx f(a + nh) - f(a ) \approx f(b) - f(a)$.

Is that where your proof is headed ?

 

[Svein]

However, there is no proof that actually starts from the derivative of f (say G), applies the definition of the definite integral to it, and arrives back at f.

Well, there is. I will not do the proof (it is fairly long), but only state the theorem.

Theorem (H. Lebesgue). The derivative φ(x) of an absolutely continuous function F(x) defined on the closed interval [a, b] is summable, and for every x $$\int_{a}^{x}\phi (x) dx = F(x)-F(a)$$.

 

[andrewkirk]

That is exactly the problem! I don't want to start a proof with the derivative of F from the the first theorem (where F is defined by an integral of f), i

I can only suggest you read the linked wiki section more carefully, in particular the statement of the second theorem. It does not define F as the integral of f. In the second theorem F is defined to be simply an arbitrary real-valued function that is differentiable on $[a,b]$. The definition of F in the first theorem is inapplicable. The fact that F turns out to be is equal to an integral based on its derivative f is what the second theorem sets out to prove.

I think the problem is that you have never formally stated what it is that you are trying to prove. I encourage you to do that before trying to prove anything. Chances are you will then see that what you are trying to prove is - subject to renaming a couple of functions - the same as the second theorem.

 

[Valour549]

The fact that F turns out to be is equal to an integral based on its derivative f is what the second theorem sets out to prove. Chances are you will then see that what you are trying to prove is - subject to renaming a couple of functions - the same as the second theorem.

While the proof on Wikipedia and the proof I am working on both ultimately aim to prove the statement of FTC2, namely $\int_{a}^{b}f(x) dx = F(b)-F(a)$, I think we go about it in different ways.

The problem is I see no reason why my way shouldn't work. It proves the above statement from left to right, and is more close in form to how FTC1 was proved.

The proof of FTC1 applied the definition of a derivative to a definite integral ; my proof of FTC2 applies the definition of a definite integral to a derivative.

Anyways with help from @Stephen Tashi I was able to complete it. I see some similarities between it and the one on Wikipedia (which seems to go from right to left). Although I think it is intuitive from the diagram that $h = \Delta x = (b-a)/n$, I would still appreciate help so I can make that part more rigorous.

Spoiler

[PLAIN]https://lh4.googleusercontent.com/DJOy7J49-ZKRWTqso3FVHOSoLgIyprqqZBcy3hci8ox1dlojyG2mS5DjOA8AE3i1or5Q9VCV2aRb2Cg=w1680-h962[/SPOILER]

 

[pwsnafu]

You're definition of Riemann sum (for the Riemann integral) is wrong. If you are choosing intervals using $(b-a)/n$ then $x_i^* \in [x_i, x_{i+1}]$ must be arbitrary. Or do vice versa: allow unrestricted partitions with left point for choice of $x_i^*$.

 

[Stephen Tashi]

Your proof deals with an integrall defined as a limit of what can be called a "regular Riemann sums". Pwnsafu wants proof for more general Riemann sums.

You have an infomal proof for regular Riemann sums, there are still some issues when it comes to a rigorous proof for such sums. One issue is the validity of the reasoning:

If $\lim_{h\rightarrow 0} p(h) = L$ and $lim_{n \rightarrow \infty} q(n) = 0$ then $lim_{n\rightarrow \infty} p(q(n)) = L$.

When you establish a definite relation between "n" and "h", you are implicitly saying that you will get the same answer for a limit involving $n\rightarrow\infty, h\rightarrow 0$ no matter what the functional relation is between "n" and "h" as long as $lim_{n \rightarrow \infty} h(n) = 0$. For a rigorous proof, you need to justify that assumption.

Another possible issue is the distinction among "double limits" and various "iterated limits" The definition of a "double limit" such as $\lim_{n \rightarrow \infty, h \rightarrow 0} g(n,h)$ makes it technically different than either of the two "iterated limits" $\lim_{n\rightarrow\infty} ( lim_{h \rightarrow 0} g(n,h))$ and $\lim_{h\rightarrow 0}( lim_{n\rightarrow\infty} g(n,h) )$. There are examples where the two iterated limits exist, but are unequal. I'd have to think carefully about your steps to see if they invovle any assumptions about iterated limits. Anyway, it's worth keeping the dangers of iterated limits in mind for future work.