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Notice that the left end of any interval after the first is the right end of another.

Thus Ln and Rn differ only by the contribution from the first and last intervals:

which approaches 0 as n increases if f(b) and f(a) are finite.

Thus Ln approaches Rn.

Notice also that Tn is the maximum possible sum and Bn the minimum possible sum obtainable here; the true area must lie between them.

Finally notice that if f(x) is an increasing function of x, (when x < y then f(x) f(y)) then Rn = Tn and Ln = Bn. Thus, for any bounded increasing function f, Bn approaches Tn as n increases, and all possible sums approach the true area as limit.

We have just shown that the definite integral or area is well defined for bounded increasing functions. By symmetry, it is so for bounded decreasing functions as well, and hence for those bounded functions for which b-a can be broken up into a finite number of pieces on each of which f is increasing or decreasing.

Which integrals are not covered by this result?

1. Those that are unbounded between a and b, like if a is negative and b positive.

2. Those for which b - a is infinite. (The will exist only if the contribution from sufficiently far off values are negligeable.)

3.Those who wobble to much; two examples are:

with a negative and b positive.

The function that is 1 on rational numbers and 0 elsewhere.

Some functions that are unbounded have bounded area under  them, some integrals over unbounded regions make sense and other do not. Integrals involving these are called improper integrals, we shall discuss them much later. Wobbling is not as bad as it seems here for defining area. Be careful to make sure that integrals exist before you use them!