Getting a bit philosophical on us, here...

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Could you go into a little more detail on how the assumption that the value exists, and how it is that by making this assumption, we can later prove that it, in fact, does exist.

Take a bucket outside and fill it with an arbitrary amount of water. How many liters is it? Even though you don't know exactly, certainly this figure *exists*, right? Since it exists, give it a name, A. Now take a measuring cup and pour the water into the cup, exactly one cup at a time, counting how many cups it holds. This simple measuring system gives you 1 cup, 2 cups, 3 cups, etc., and with each "iteration" as a mathematician would call it, the number gets closer and closer to A. Voila, when the iteration is complete, we have calculated A exactly.

This is not so different than what we are doing here. We look at a curve and we know that there is area under it, and that the amount of area is an unquestionably unique number. We're coming up with an iterative system of approximating this area which after a sufficient number of iterations, gives us the answer to any level of accuracy. So no, we are not assuming A exists and using this assumption to prove it does indeed exist. We know it exists, and we are attempting to calculate it to an arbitrary level of accuracy.

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Also, how would you ever be able to choose an $\displaystyle \epsilon$ if you don't choose the correct value of A?

Here you are correct. In general, given a $\displaystyle \delta$ it is not possible to determine how many rectangles are necessary to find A with an accuracy within $\displaystyle \pm \epsilon$. This changes for different functions.

A "tangible" example, whatever that may mean

$\displaystyle

\int_a^b f(x) dx=\lim_{\Delta\rightarrow 0}\sum f(x_i)\Delta

$

Let $\displaystyle f(x)=2x$, $\displaystyle a=0$, $\displaystyle b=1$, and $\displaystyle x_i=i*\Delta$. (Keep in mind the sum goes from $\displaystyle 0$ to $\displaystyle n$, where $\displaystyle n\Delta=b-a=1$.)

$\displaystyle

\int_0^1 2x dx=\lim_{\Delta\rightarrow 0}\sum_{i=0}^{n} 2(i*\Delta)\Delta

$ = $\displaystyle \lim_{\Delta\rightarrow 0}2\Delta^2\sum_{i=0}^{n} i$

Now recall the formula we all learned in discrete math, the story that little Gauss discovered a shortcut to adding all the numbers from 1 to 100 with this formula: $\displaystyle \sum_{i=0}^{n} i = \frac{1}{2}n(n+1)$

So, $\displaystyle \lim_{\Delta\rightarrow 0}2\Delta^2\sum_{i=0}^{n} i$ = $\displaystyle \lim_{\Delta\rightarrow 0}2\Delta^2\frac{1}{2}n(n+1)$ = $\displaystyle \lim_{\Delta\rightarrow 0}\Delta^2(n^2+n)$

Now, remember $\displaystyle n\Delta=1$, so...

$\displaystyle \lim_{\Delta\rightarrow 0}\Delta^2(n^2+n)$ = $\displaystyle \lim_{\Delta\rightarrow 0}\Delta^2(\frac{1}{\Delta^2}+\frac{1}{\Delta})$= $\displaystyle \lim_{\Delta\rightarrow 0}(1+\Delta)$ = **1**

Ergo, A=1 QED.

*Note: the invention of integrals and calculus was intended as a cohesive system of legal shortcuts to allow us $\displaystyle not$ to have to do these kinds of calculations.

**Notice the second to last step, that $\displaystyle A=\lim_{\Delta\rightarrow 0}(1+\Delta)$ . So if we want to know A within an accuracy of $\displaystyle \pm\epsilon$, well, $\displaystyle |1+\Delta-1|=|\Delta|<\epsilon$ . So in this case, for this function, $\displaystyle \delta=\epsilon$.

They say a picture is worth a thousand words. Well this post is getting a little convoluted even for me. (Giggle)

Generalizing your example

Backing up through the proof, remember the last few steps, where $\displaystyle n\Delta=b-a=1$? Well, on the interval $\displaystyle [0,x],~~ n\Delta=x$. Observe:

$\displaystyle

\lim_{\Delta\rightarrow 0}\Delta^2(n^2+n)

$ = $\displaystyle

\lim_{\Delta\rightarrow 0}\Delta^2(\frac{x^2}{\Delta^2}+\frac{x}{\Delta})

$ = $\displaystyle

\lim_{\Delta\rightarrow 0}(x^2+\Delta x)

$ = $\displaystyle x^2$

So $\displaystyle F(x)=\int_0^x 2tdt=x^2$

This slightly more technical definition is shorthanded by $\displaystyle F(x)=\int 2xdx=x^2$. Again, the proofs are more technical than you think.