Essentially, you choose how you want to split up the interval. For instance, P = {0, 1/2, 1} and Q = {0, 1/2, 3/4, 1} are both partitions of [0,1]. (In fact, for this particular example, we say Q is a refinement of P.)
A perfectly good question is: Why is this useful? It turns out that many functions which aren't seemingly 'nice' can actually be integrated -- in particular, functions whose set of points of discontinuity is everywhere dense in their domain can be Riemann integrable, which can be quite a non-intuitive result.
]]>where P is a partition of [a,b], and U(f,P) and L(f,P) are the lower Darboux sums of f with respect to the partition P, defined by
where
and follow their usual definitions:It is probably doable with a careful choice of P. The fact that f is never positive might be of some use here!
This is also part of a more general property called the order property of integrals, i.e. if f,g are Riemann integrable on [a,b], then
.]]>I believe it is the direct result of the definition of 'integral'.
An integral is a summation process. In the area example in my diagram proceed as follows:
Divide the space into vertical strips with width Δx. The height of each strip is f(x) which will be negative in your case.
Calculate the area of one strip = f(x).Δx
Add up all the areas for strips starting at 'a' and ending at 'b' = ∑f(x).Δx
Now let the widths tend to zero and find the limit as Δx -> zero of ∑f(x).Δx
This is denoted by ∫ f(x).dx
As all the f(x) are negative, the sums will be too, and so the limit will be too.
It is possible to extend this idea to summate things other then areas. But it is also possible to 'convert' such problems into area problems, so the result will generalise.
Bob
]]>I've tried a number of equations for the above statement, and from the answers I get it seems like it's true?
If it is true, is there any theory that I can use to explain the statement?