Double integration equivalence

Time to clarify my assumptions about integration.

The double integral,

 is shorthand for:
             

We begin at the innermost integral.   Integrate with respect to x, holding  y constant.  Then integrate the result with respect to y (holding x constant).  Each step is an ordinary, single-variable integral.  We may also change the order of integration:
 

Just remember to keep x-bounds with x, and  y-bounds with y.

These are convenient results, and after hammering through partial differentiation, integrating in any order makes intuitive sense.  This is the next chapter of Advanced Calculus, but I think the result is self-evident, and I want to clarify that assumption so I can find any mistakes.

  I have heard equality (2)  referred to as Fubini's Theorem, but the statement before me is, the two integrals are equivalent.  The only Theorem I have any business with must be the Fundamental Theorem of the Calculus.  If I break up (2) into infinite sums, the sum on the right side of the equal sign must be the same as that on the left.  Right?

Variable bounds, paramaterization, partial summation with discrete and continuous quantities are a few pages away.  I  need a working, base example: a continuous function with constant bounds.  Let me build that now with the simplest case: a dobule integral of the form (1), with constant bounds.

A FUNCTION OF A SINGLE VARIABLE

Let F(x) be a function whose derivative, F'(x) = f(x)  is continuous over the closed interval [a, b], i.e.
              
Then the Fundamental Theorem of Calculus states


PROOF:  This approach is abstract, but concise.  By continuity, we can write the integral as the limit of a sum


Where δ is the largest subinterval and ξ_k lies in the subinterval [x_k - x_k-1].
Now, we choose for our  ξ_k the points in each subinterval where the line tangent to F(x)  has the same slope as the segment joining the endpoints, F(x_k ) and  F(x_k-1):

             (Rise/Run) (Run)             =   (Rise)

Then the right hand side of (4) becomes,


Rearranging the terms,  we have

    _____________________________________________________


And the theorem is proved.  Let me try to extend this method myself.

A FUNCTION OF TWO INDEPENDENT VARIABLES

Beginning with equation (1):
     
First, consider the inner integral:
     
What relationship does f(x,y)dx have to F(x)?  It is the partial derivative F₂₁(x):
     

The result of the first integration will then be  F₂(x).  This is my principal assumption. Specifically,
     
I'm not sure how I feel about that.  The assumption is sufficient to make the sums work out.  And to use the Fundamental Theorem, we must already know the relationship between f and F.  I don't see a way around equality (8).   But I already know that F₂₁(x, y,) and F₁₂(x, y) are identical for 'nice' functions, and the FTC is just going to give that information back to me.

That's why  I need to work out the concrete terms, determine if it holds, and if so, how.  Otherwise I'm thinking in a circle.  I will use the notation of partial derivatives, but I did not begin with this assumption.  I tried several ways of indexing the sums, and in the end found that "indexing" was the same thing as labeling partial derivatives.  The FTC makes this process clear.

I.  Integrate with respect to x
     

Choose the ξ_k so that
     
This is a directional derivative.  There is no slope in the y direction because I'm not varying y. Hence the equality.  This is the crux.  Not-varying one of two independent variables is exactly the same thing as taking a partial derivative. The x intervals sum and cancel to a discrete value, independently of the y intervals.  Multiplication and addition are commutative and associative, so the sums and products of these values can be arranged in any order.  Taking the integrals separately is also exactly the same thing as saying the integration can be done in any order. x and y never mix in the infinite sums.  That's not a theorem; that's what we're doing.  This is what I mean about Fubini's theorem, whatever the hell it is.  Integrating the independent variables separately is how we insure there is only one answer.  Onward!

Now we have


Rearranging the terms,

   __________________________________________________________________



II.  Integrate with respect to  y
I will integrate the general form:
     
For the purposes of integration, this is the same as
     
and can be applied to (7)'.   The process is now the same as above.
     
Choose the ζ_k  so that
     

That's still a derivative on the left! (5)'' can alternately read,
                 
Thus we have


And the last line can again be rearranged so that all but two terms cancel:

   ______________________________________________________________



III. Equivalence
Now compute the result:


... with the happy result that giving one order of integration entails the equivalence of the other.

HOW'S MY DRIVING?

   I have the feeling double integration is applicable in a variety of circumstances, leaving this approach insufficient for many cases.  Or maybe I made mistakes, and it's all wrong.  Somebody is letting a 12-year old drive a semi, and both of those guys are me.  Tell me how my driving is.  Leave me a comment.  Correct a mistake.  Prattle on about Space-Wogs.

Fuckin' space wogs.  Always framping the scrumpets.

{Thanks again to CodeCogs for free LaTeX equation editing and hosting the resulting, tiny bitches images.}