Any proof for the definition of the definite integral

[Hurkyl]

It is difficult; it took a while before mathematicians came up with limits to allow them to rigorously deal with these things... and limits do seem awfully obtuse at first. But, the more you use them, the more sense they make.

And mathematicians like things to behave nicely too, so we define special classes of things that do behave nicely. For instance, an infinite series is "absolutely convergent" iff it's commutative and assocative. e.g.

1 + -1/2 + 1/4 + -1/8 + 1/16 + -1/32 + ...

is an absolutely convergent series becuase no matter how you rearrange and group these terms, you still get a sum of 2/3. However,

1 + -1 + 1 + -1 + 1 + -1 + ...

is not absolutely convergent, because it fails to be commutative and associative.

And it turns out that there is a simple criterion for a sequence to be in this class of functions (and this criterion is used as the definition): a series is absolutely convergent iff the series converges when you replace each term with its absolute value.


To relate this to what you said earlier, it turns out a (convergent) sequence is not absolutely convergent iff the sum of the positive terms and the sum of the negative terms are both divergent.

 

[ydnef]

Hey, this thread kind of died. Why? Can nobody think of anymore questions. Or is everything clear now to everybody. Maybe it just confused too much.

Anyway, here is one thought I would like to share. I first encountered quantum mechanics in a book where one would derive the Schrodinger equation using the notion of particles as wavepackets, and then afterwards the author would cleverly discard the derivation and keep the equation, arguing that the notion of a wavepacket may be useful to develop some intuition, but that in the end the idea cannot be taken too serious, and that everything in fact was far more general and abstract. Was the author doing first-time students of quantum mechanics an incredible favour or was he doing some unrepearable damage to them?

ydnef

 

[StonedPanda]

I've read the whole thread. I took AP Calc BC, so I have a year of calc under my belt (I got an A, so I guess that means I understood it). However, this is a question I'm still asking. WHY does the antiderivitive of a function give the area under the curve? I think this question is different from the one originally proposed -- i.e my question is different than why an infinite Reimann converges to an anti-derivitive. Or maybe I missed something... It's kinda late at night and I'm tired -- can anyone help? =p

 

[matt grime]

You have missed something - it is the fundamental thorem of calculus in action and has been explained in this thread, i think i did it twice.

 

[ydnef]

For developing some intuitive feel, I surely do recommend my own reflections on the matter, but be careful because this punk-kid of an ydnef has been abusing a lot of terminology and notation, which even confuses ydnef when he is not careful.

 

[matt grime]

That would be your system where you add up a countable number of (non-real) infinitesimals and get 1? Yep, that's the way forward...

 

[ydnef]

Yes you told me already that professional mathematicians have already reserved the word infinitesimal for some other mathematical idea and so forth. Got that. So okay, let's not call my dx's infinitesimals anymore, and think of every dx that appears as actually truly meaning ''limit dx->0''. Think of it part of me being too lazy and part of me wanting to keep the notation as tidy as possible.

I happen to think that it is useful in some practical cases to think of integration as some sort of ''adding of infinite terms'', like in some cases I like to think of matter as of being made out of point particles, though it does not take a genius to see that actually particles cannot really be points, they are more complex, and in some sense, nobody really really knows the true nature of matter, yet, but if you want to explain to a person for instance how television works, thinking of electrons as point particles is good enough. I know there is a flaw in that, but as long as it is not fatal, then it is not too bad I guess.

 

[matt grime]

You could of course just learn what the definition of integration is, and understand it properly, which might, just might, be considered the best way of doing it, seeing as it is the limit of finite sums approximating the area of the curve...

 

[ydnef]

You know what I've been thinking. That guy who invented/discovered non-standard analysis STOLE the term 'infinitesimal', not me. I mean, the term has been in existence ever since the time of leibniz and Newton, and who ever applied calculus between the time of its birth and somewhere between 1960 must have already had some sort of understanding connected to that word 'infinitesimal'. Hence, perhaps, all my confusion.

 

[mathwonk]

I am trying to discuss the answer to the question in thread 38, why is the FTC true? i.e. why does the antiderivative (of the height) give the area under the curve?

An equivalent question is why is the derivative of the area equal to the height of the curve?

The easiest way for me to understand anything is in a simple example. So take a constant function y = f(x) = c, for all x between a and b.

Then the area function A(x) = the area under the "curve" y = C, between a and x, is height times base = C times x-a = C(x-a) = Cx - Ca. So the derivative of this area function is C = height! So it is true in this case.


Now the next simplest case is a piecewise constant function, say y = C for x between a and r, and y = D for x between r and b, with a< r < b. let f(r) = D say (it does not matter).

Then the area function A(x) = C(x-a) for x between a and r, and equals

A(x) = C(r-a) + D(x-r), for x between r and b.

Thus the derivative of A exists except at r, and the derivative of A for x between a and r is C = the height, and the derivative of A for x between r and b is D = height.

And A is continuous. So here we are allowing as an "antiderivative" a function which is continuous everywhere, and differentiable where the original function is continuous, and has derivative equal to the height when the height is continuous. At those points where the original function is not continuous, we take the antiderivative to be whatever makes it continuous.

Note however that we have: total area = C(r-a) + D(b-r) = A(b)-A(a)
= C(r-a) + D(b-r) = the difference of the values of A at the endpoints a and b.


Now this continues to be true for all piecewise constant functions.

Moreover this property is preserved under uniform limits, so since every continuous function is a uniform limit of piecewise constant functions, and the antiderivatives also converge uniformly, it is still true for continuous functions.

I.e. if f is any continuolus function, and if A is an antiderivative of f, then the area under f equals A(b) - A(a). or equivalently, if A(x) is the area function from a to x, then A'(x) = f(x).

does this help? I realize it ain't full, but sometimes partial explanations help more than full ones.

 

[Ethereal]

I found the following interactive proof for the FTC:

http://archives.math.utk.edu/visual.calculus/4/ftc.9/

But for some reason, after looking through the whole proof, I still don't know why the anti-derivative of an integrable function for a certain range gives you the area under the graph. They said "Let $$A(x) = \int_a^x f(t) dt$$ where A(x) is the area from x=a to x=t" But why is it possible for the area under the graph to be expressed as the anti-derivative in the 1st place?

 

[HallsofIvy]

You are missing the point.

$$A(x) = \int_a^b f(t) dt$$ is defined as "the area bounded by y= f(x) above (assumed to be positive), y= 0 below, x= a on the left, and x= b on the right", not as an "anti-derivative".

The proof you give then uses the basic properties of area (if A and B are disjoint sets, then the area of A U B is area(A)+ Area(B)) to show that lim (A(x+h)-A(x))/h IS f(x) and so A itself IS an anti-derivative.

 

[mathwonk]

mr ethereal, have you read my post #45 above? I have tried to lay out the proof of FTC as simply as possible. I.e. it is true for piecewise constant functions, and hence also for their uniform limits, hence for all continuous functions. What do you think of that?

 

[Ethereal]

Actually, I don't follow your proof at all. I haven't studied maths formally yet, so I have difficulty understanding what a "piecewise constant function" is etc. Apparently a quick visit to mathworld.wolfram.com didn't help at all. Sorry.

EDIT: I think I just found out what those terms meant. I also happened to find this on the internet which seems similar to yours:

http://www.ma.hw.ac.uk/~robertw/F11UB3/slides1.pdf

 

[PaulScheduler]

Proofs For Volume

The method I know of is based on the principle of revolving a function around an axis. This is accomplished in one of two ways by using integration. The first is based on the area of a circle Vs the area of a cylinder. Often both can be used when set up appropriately.

I always pick the area of the circle method, whenever possible. This is because I see the function best that way.

Sphere:

Using the equation for a circle Y^2 + X^2 = R^2

Solve for Y


Replace the function f(x) or Y equal into the eqn pi * F(x)^2 , the area of a circle

This revolves any function around the X axis, summing up all areas, resullting in a volume

answer should be 4/3 pi R^3 Volume


2) Do the same for a line segment, except revolve it to get the volume of a cone

3) Volume of a cylinder is Y= a constant the simplest


Then take the definiete integral from 0 to R over dx

 

[PaulScheduler]

Earlier someone wanted to prove the volumes of different shapes, sphere, cone,etc. The above is for that #15 or #16

 

[Hurkyl]

This thread was four years old!