This is more of an explanation/sketch proof than anything and I don't know if it helps at all.

(I'm taking $f(x)$ to be a perfectly integrable function.)

The area under a curve can be split up into evenly-spaced rectangles as you said, where each rectangle has a height $f(t_i)$ where $t_i$ is its ' $x$-coordinate' (i.e. $t_1$, $t_2$, ...) and length (say) $\delta t$. This is just an approximation of course. If we let $A(x)$ represent the area under the graph from (some fixed constant) $a$ to (the variable) $x$, then $A(x)$ is approximately $\sum _a^{x}f(t_i)\delta t$ where the sum is taken over all the $t_i$ which lie between $a$ and $x$.

Now, if we let $\delta t$ go to 0 (i.e. we increase the number of rectangles) the approximation for $A(x)$ becomes better and better. Now, as $\delta t$ goes to 0 the sum expression will tend to some limit and we can write this limit (so that it looks like the sum) as ${\int }_a^{x}f(t)\mathrm{dt}$ and we call it (as you know) the integral of $f(t)$ wrt $t$ between $a$ and $x$. (So far this has nothing whatsoever to do with differentiation.)

Hence, by the previous argument, we have $A(x)={\int }_a^{x}f(t)\mathrm{dt}$. (This is so far basically what you said, but to be honest I got a bit confused by your argument so I thought I'd just say it again!)

Now, let's show the link between integration and differentiation.

Try also to visualise what's going on. We can intuitively convince ourselves that the rate of change of the area is the function itself (consider two very thin strips of the area next to each other - like our rectangles before.)

Okay, after that bad explanation let's go at it more convincingly (and mathematically! You can probably find a better intuitive way of thinking about it.):

Before we go on make sure you know and (at least, sort of) understand $g\text{'}(x)=\underset{h\to 0}{lim}[g(x+h)-g(x)]/h$.

Suppose we have a curve $f(x)$ like before. Now, the area between two variable points $x$ and $x+h$ is $A(x+h)-A(x)$ (as it's the area from $a$ to $x+h$ minus the area from $a$ to $x$).

Also, within our range between $x$ and $x+h$ there will be two values of $x$ which will give the smallest value of $f(x)$ and the largest value. Call these $x_{min}$ and $x_{max}$ respectively.

Now, we can enclose the whole area between $x$ and $x+h$ in a rectangle with height $f(x_{max})$ and length $(x+h)-(x)=h$, i.e. with area $f(x_{max})h$ and this rectangle has area greater than the actual area under the graph.

Likewise, we can form a rectangle with area $f(x_{min})h$ and this will have area less than the actual area under the curve.

(If you don't get this try drawing a diagram of an arbitrary function and marking everything on.)

Thus we get

$f(x_{min})h\le A(x+h)-A(x)\le f(x_{max})h$

Dividing by $h$ gives

$f(x_{min})\le [A(x+h)-A(x)]/h\le f(x_{max})$

Now let's see what happens to this inequality as $h$ goes to 0.

The left term $f(x_{min})$ goes to $f(x)$ as $x_{min}$ will go to $x$ (it's like $x_{min}$ is being crunched into a closer and closer distance from $x$ when $h$ becomes smaller and smaller).

Likewise, $f(x_{max})$ also goes to $f(x)$ for the same reason.

The middle term is $\underset{h\to 0}{lim}[A(x+h)-A(x)]/h$ which is $A\text{'}(x)$.

Thus after taking the limit $h\to 0$, our inequality is now

$f(x)\le A\text{'}(x)\le f(x)$

from which we conclude that $A\text{'}(x)=f(x)$ (and this result is so important that it's called the fundamental theorem of calculus).

Now, if you're still following (and I haven't bored you to death!), this shows that the result of integration and anti-differentiation differs only by a constant term.

So, to get the area under a curve we can use the anti-derivative function of the curve.

Let the anti-derivative of $f(x)$ be $F(x)$. Thus by the previous result $A(x)=F(x)+C$ where $C$ is some constant.

Now, $A(x)$ is the area under the curve from $a$ to $x$ and so $A(a)=0$. So, putting $x=a$ in our expression $A(x)=F(x)+C$ we get $0=F(a)+C$, or $C=-F(a)$.

Thus, $A(x)=F(x)-F(a)$.

This is the result that we use all the time when we first do definite integrals and that was where it comes from. From it by putting in a constant, say $b$, we get that the area under the curve from $a$ to $b$ is $F(b)-F(a)$. (i.e. the difference of the anti-derivatives of our curve at $a$ and $b$.)

Well, that's about it...

Please ask if anything is unclear (if you read it all),

Marcos

I don't actually understand what you're trying to say for either of those points (particularly the second one) and if this doesn't answer your question can you please explain a bit more what you meant?

For the first point,

You said:

"the area of each of those strips is $f(x)$ or $y$-length...''

Not quite. [Note: I'll use $t$ as the ' $x$-coordinate' of each strip as it represents many values (as opposed to $x$) so we don't confuse it with the $x$ in, say, $A(x)$ or $f(x)$ which represents a single value.]

Each strip hs area $f(t)\delta t$.

I'll try to make it a bit clearer (particularly what I mean by a limit):

If we split up the area under $f(x)$ between $x=0$ and $x=1$ into 100 strips, then $\delta t=1/100$ and each strip will have area $f(t)/100$. Thus an approximation for the area can be found by adding the area of these 100 strips.

If we now increase this number of strips to 10000, then each strip will have area $f(t)/10000$. Thus a better approximation for the area can be found using these 10000 strips. We can get better and better approximations by increasing the number of strips (hence decreasing the value of $\delta t$). We can't actually talk about having an infinite number of strips with $\delta t$ as this will then lead to a meaningless statement: We would have $\infty \times f(t)\times 0$ and now, what on earth does $\infty \times 0$ mean? (Is it 0, 1 or some other number you care to mention?)

So we must restrict ourselves to just considering what happens as $\delta t$ gets smaller (but doesn't actually ever become 0) and the number of strips increases (but is always a finite number).In this case we find the 'limit' of this sum of the areas of rectangles:

The limit is a value which our expression becomes closer and closer to as $\delta t$ gets smaller and the number of strips gets larger (we can get as close as we care to name by making these quantities small enough and large enough respectively), but which our expression doesn't ever actually become necessarily (although it may).

As an example to see if you're following: what is the limit of $1/x$ as $x$ becomes larger and larger? It's 0 because as $x$ becomes larger $1/x$ gets closer and closer to 0 ( $1/x$ is never equal to 0 though, but if you told any number, however close to 0, I could find a value of $x$ that wouldgive $1/x$ closer to 0 than the valueyou gave me - this is what the `limit' idea means).

For the second part,

Here's a diagram of a possible situation:


The two boxes I had in mind were the (overlapping) green boxes. The three arrows (two on the right and one at the bottom) show where they are more clearly.

The yellow area is $A(x+h)-A(x)$.

Quote:

"for the largest rectangle, the area as delta x tends to zero, again becomes basically a thin strip f(x) or y-length and the same for the smallest rectangle...

you've shown the above in your inequality statements....''

Again, not quite. This isn't what I've shown.

The inequality was originally $f(x_{min})h\le A(x+h)-A(x)\le f(x_{max})h$.

This is the inequality involving areas, (after dividing by $h$ we are no longer dealing with the area of strips). If we look at what happens in this inequality (the original) as $h$ gets smaller we get that $f(x_{min})h$ will get closer and closer to 0, and so will $f(x_{max})h$. Also $A(x+h)$ will get closer and closer to $A(x)$ when $h$ becomes smaller, and so we get the correct but useless inequality, $0\le A(x)-A(x)\le 0$. So instead we choose to divide by $h$ to begin with. From now on, each of the terms of the inequality isn't an area of anything anymore. By the definition of differentiation (ask if you need more on this), $A\text{'}(x)=\underset{h\to 0}{lim}[A(x+h)-A(x)]/h$, so whenever we see the RHS we can replace it by $A\text{'}(x)$. That's what the middle term of the inequality in my previous post becomes upon taking the limit of $h$ getting smaller...

Thus our inequality becomes $f(x)\le A\text{'}(x)\le f(x)$ and we thus conclude that $A\text{'}(x)=f(x)$ (purely from the inequality).

Maybe it helps you consider it like this (this is definitely not a proof - just an illustration you may find more intuitive):

Take the case of some graph $f(x)$, and the area under it $A$. Now, if we consider a very thin strip of the area, $\delta A$ (i.e. a small part of the area) we can approximate it as a rectangle and it will have height $f(x)$ and a length $\delta x$ (i.e. a small increase in $x$).

So, $\delta A$ is approximately $f(x)\delta x$.

Dividing by $\delta x$,

$\delta A/\delta x$ is approximately $f(x)$.

Now, this approximation gets better as $\delta x$ gets smaller and smaller. But, as $\delta x$ gets indefinitely small $\delta A/\delta x$ becomes $\mathrm{dA}/\mathrm{dx}$, and hence $\mathrm{dA}/\mathrm{dx}=f(x)$.

I hope that was better. Please ask if there's anything you don't understand - and sorry for the very long post! I got carried away...

Marcos

it's quite interesting...i didn't know this...thanks...

is $\delta y/\delta x$ pronounced as delta y/delta x?

with regards to the difference between $\delta y/\delta x$ and $\mathrm{dy}/\mathrm{dx}$ [(is this pronounced (dee y/dee x?)...what does the 'd' stand for...?]

so the gradient between any two points (no matter how far apart) is $\delta y/\delta x$ ...but in terms of using the limit as $x$ or $h$ or whatever tends to zero we say $\mathrm{dy}/\mathrm{dx}$....the $\mathrm{dx}/\mathrm{dy}$ indicates a very small change in $y$ and $x$...

thus the reason why we say or write ' $\mathrm{dx}$' in $\int f(x)\mathrm{dx}$ and not $\delta x$ is b/c we are using an infinitesmall $x$...?

thanks again...

All your pronunciations are correct I think...

It's convention that " $\delta x$'' represents "a small change in $x$''. So, yes, $\delta y/\delta x$ could be the gradient between two points (although we would drop the term "small'' here!).

The " $\mathrm{dx}$'' can be looked at (informally) as "an infinitesimal change in $x$'' but in differentiation the $\mathrm{dy}/\mathrm{dx}$ is an entity on its own (it's not a fraction, or whatever). The "d'' notation is convenient though because things like $\mathrm{dy}/\mathrm{dx}\times \mathrm{dx}/\mathrm{dy}=1$ do indeed hold (but we would need to prove this first), and so do many other fraction-looking results.

Again, with integration we can view $\int f(x)\mathrm{dx}$ as meaning (informally) "the infinite sum of $f(x)\mathrm{dx}$'s'' - i.e. of rectangles with infinitesimal width, but again we can't assume things like $\int f(x)\times \mathrm{du}/\mathrm{dx}\times \mathrm{dx}=\int f(x)\mathrm{du}$ purely from this informal definition, although once we prove it the notation automatically suggests the result.

So, it's basically a useful notation because things like $\mathrm{dy}$ and $\mathrm{dx}$ do indeed behave (in most cases) how we would (intuitively) expect really small quantities to behave...

Marcos

P.S. I think the "d'' notation is Leibniz's whereas the $f\text{'}(x)$ notation is Newton's (both useful in different cases).