# Proving/deriving first fundamental theorem of calc

1. May 6, 2013

### 1MileCrash

I know this is going to be atrociously bad, but I like to try to prove things.

$\frac{f(x+\Delta x) - f(x)}{ \Delta x} = \frac{\Delta y}{\Delta x}$

$=> f(x+\Delta x) - f(x) = \frac{\Delta y}{\Delta x} \Delta x$

$=> f(x+\Delta x) = \frac{\Delta y}{\Delta x} \Delta x + f(x)$

Now, express x as (x2 + delta x) and define f(x) in the same way we just defined f(x+deltax)

$=> f(x+\Delta x) = \frac{\Delta y}{\Delta x} \Delta x + \frac{\Delta y}{\Delta x} \Delta x + f(x2)$

We could now express x2 as (x3 + delta x) and define f(x2) in a similar way again.

$=> f(x+\Delta x) = \frac{\Delta y}{\Delta x} \Delta x + \frac{\Delta y}{\Delta x} \Delta x + \frac{\Delta y}{\Delta x} \Delta x + f(x3)$

(I should have subscripts or something similar, because these delta y/delta x's are not all the same thing, but bear with me :)

It is clear that repeating this process over and over becomes a sum of all delta x/ delta y

$=> f(x+\Delta x) = \sum \frac{\Delta y}{\Delta x} \Delta x$

If we take the limit of both sides as delta x approaches 0, the left side becomes f(x) and the right side becomes the definition of the integral.

$f(x) = \int f'(x) dx$

I know a lot of things are messed up about it, I think I need to name all my different x's and y's for starters, and I'm sure I said a lot of things that didn't make sense as well, but this is a 10 minute attempt.

Any input is appreciated.

Last edited by a moderator: May 8, 2013
2. May 6, 2013

### mathman

You have the general idea.

3. May 6, 2013

### 1MileCrash

That's quite surprising.

4. May 6, 2013

### Stephen Tashi

Doesn't that demonstration use $\triangle y$ to stand for more than one quantity? Is $\triangle y$ going to be $f(x_2) - f(x_1)$ or is it going to be $f(x_3) - f(x_2)$ etc. ? I think you need subscripts to distinguish the $\triangle y$'s.

I think your idea amounts to proving the fundamental theorem of calculus from an agrument using the Calculus of Finite Differences.

5. May 7, 2013

### 1MileCrash

Yes, as noted those delta y's are all different. I definitely need subscripts, I'll try to rewrite it sometime today.

I don't know what the Calculus of Finite differences is, but I'll check it out. My direction was just based off "the derivative is the quotient of a difference, and the integral is the sum of a product." Knowing that that was inverse-like, I figured that solving for f(x+h) in the derivative definition could give me something integral-like.

The only hurdle is that the derivative is "of a point" while the integral isn't, which I'm guessing is the reason why the sum of all of the delta y/delta x makes an appearance, as opposed to two terms as in the derivative.

Last edited: May 7, 2013
6. May 7, 2013

### Stephen Tashi

A basic idea in the Calculus Of Finite Differences is the relation between summation and differencing.
Let $h$ be a given constant. For a function $G(x)$ , define the function $F$ whose domain will be the integers by $F(k) = G(kn).$.

Define the function $f(k) = \triangle F(k) = F(k+1) - F(k)$

If we face the problem of doing a summation $S = \sum_{i=1}^n f(n)$ then this amounts to the easy "telescoping" sum:

$S = f(1) + f(2) + .. .+ f(n) = (F(2)-F(1)) + (F(3)-F(2)) + ... +(F(N+1)-F(N))$
$= F(N+1) - F(1)$

One can approach doing finite a summation of a function $f$ as a process of seeking an "anti-difference" function for $f$.

7. May 7, 2013

### Stephen Tashi

I meant $F(k) = G(kh)$ of course.