Proving the fundamental theorem of calculus.

Question

I have tried proving the fundamental theorem of calculus:

Assuming: f,g,F are continuous,differentiable in the appropriate range(s)

By definition, $$ \int _{x=a}^{x=b}g(x)dx=\lim_{n \to \infty} \sum_{r=0}^{n}\dfrac{b-a}{n}g(a+ r\dfrac{b-a}{n})\tag{E1}$$ Which can be better expressed as: $$\int _{v=v_{1}}^{v=v_{2}}g(v)dv=\lim_{n \to \infty} \sum_{r=0}^{n}\dfrac{v_{2}-v_{1}}{n}g(v_{1}+ r\dfrac{v_{2}-v_{1}}{n})\tag{E2}$$

where $v$ is some quantity.

Now consider the integral:

$I= \int _{a}^{b}f(t)dt$.

Suppose there exists a function $F(t)$ such that $F'(t)=f(t)$.Thus,

$I= \int _{a}^{b}f(t)dt= \int _{a}^{b}\dfrac{dF(t)}{dt}dt=\int _{t=a}^{t=b}dF(t)$, which is equivalent to: $$ \int _{F(t)=F(a)}^{F(t)=F(b)} dF(t)$$

I believe that saying $F(t)=F(a)$ is just an equivalent statement to $t=a$ think Using $E2$,this becomes,(since $v=F(t)$, and $g(v)=1$) :

$$\lim_{n \to \infty} \sum_{r=0}^{n}\dfrac{F(b)-F(a)}{n}$$

Which is just $F(b)-F(a).$

(This is of course intuitively obvious: adding up small changes $dF$ will simply be the overall change in $F$, viz: $F(b)-F(a)$, however I wished to avoid this line of reasoning since this will entail discussions on "rigorously" treating/defining differentials)

If we consider a function $I(x)=\int _{a}^{x}f(t)dt$, then by the same reasoning, we will have:

$I(x)=F(x)-F(a)$.

Which implies, $I'(x)=F'(x)=f(x)$

According to me, This completes the proof of both parts: part 1 and the evaluation theorem also.

However, this, in my view is different from the proof given in Thomas'-calculus (or just any standard textbook), since it does not make use of the Mean value theorem anywhere.

Also, this proof seems to be significantly shorter.

All of this makes me believe there is something wrong, or "non-rigorous" about my proof. My guess is the cancellation of $dt$ with $dt$. If anything else, I will be glad if someone points it out.

Answer 1

You have not stated precisely the parts of FTC you want to prove so I'll need to guess. In Wikipedia's page on the FTC, there are two parts. I'll refer to them as the first and second part:

"First part" Let $f$ be a continuous real-valued function defined on a closed interval $[a, b] .$ Let $F$ be the function defined, for all $x$ in $[a, b],$ by $$ F(x)=\int_{a}^{x} f(t) d t $$ Then $F$ is uniformly continuous on $[a, b]$ and differentiable on the open interval $(a, b),$ and $$ F^{\prime}(x)=f(x) $$ for all $x$ in $(a, b)$.

"Second part" Let $f$ be a real-valued function on a closed interval $[a, b]$ and $F$ an antiderivative of $f$ in $[a, b]$ : $$ F^{\prime}(x)=f(x) $$ If $f$ is Riemann integrable on $[a, b]$ then $$ \int_{a}^{b} f(x) d x=F(b)-F(a) $$

In the first half of your Question, you want to end up with the result $\int_a^b f(x) dx = F(b)-F(a)$. So this means you are trying to prove something like the second part.

The "cancellation of $dt$" can actually be interpreted (while staying in the realm of e.g. Riemann integration, or for nice functions, the definition of an integral that you gave) as a theorem called change of variables ($F$ is the new variable), and usually comes after the proof of the FTC and so its use here would be circular. Therefore you should avoid this, but now you are stuck with $\int_a^b \frac{dF}{dt}(t) dt$, and when you apply (E1), its not clear to me how to obtain $\lim_{n \to \infty} \sum_{r=0}^{n}\dfrac{F(b)-F(a)}{n}$.

(In a chatroom, it was confirmed that OP was thinking of $dF(t)$ not as a Stieltjes type object, but rather as $dF$ where $F$ is a new variable; hence the alleged applicability of E1.)

For the second half of your Question, it seems you have also assumed the existence of an antiderivative $F$. In fact, the existence of an antiderivative is part of the FTC. Actually, the first part of the FTC shows that $I$ is differentiable, and is an antiderivative, i.e. $I'=f$. These two things are proven at the same time in Wikipedia.

So I believe what you have shown is: assuming

• $(A)$ the second part of the FTC (this is the 'by the same reasoning' bit),
• $(B)$ that $I$ is differentiable, and
• $(C)$ that there is an antiderivative $F$,

then in fact $I(b)=F(b)-F(a)$ (and hence $I'=F'=f$). This is true:

$$I(b) \overset{\text{defn of }I}= \int_a^b f \overset{(C)}= \int_a^b F' \overset{(A)}= F(b)-F(a) \overset{(B)}\implies I'=F'=f.$$ But it is much less than the full FTC, which does not assume $(A)$, $(B)$ and $(C)$.

(See also the Corollary in wikipedia, which also has a nice short proof, but it requires the first part of the FTC; under your regularity assumptions, its the same as the second part of the FTC, so you can skip that proof.)