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That is okay, I hope you are feeling better. If not then get some rest and we can continue later or tomorrow.
Your answer is correct! We use this formula to get more.
n is the number of intervals or something like that I suppose?
How do you get this formula?
Hi;
Yes, n is the number of intervals. The formula is derived in a lot of books. I am not that big on memorizing proofs or derivations so I do not recall it offhand. What is important for numerical work is the error estimate and the fact that it works!
This is known as the Trapezium Rule. Proving it can be done as follows.
It can be shown that
where U(f,P) and L(f,P) are the lower Darboux sums of f with respect to the partition P, defined by
and M_i and m_i are given by their usual definitions
Then, it is easy to see that
.A similar technique is used to show that a monotonic function on [a,b] is Riemann integrable on [a,b].
Epsilon is just some (arbitrarily small) positive real number. We claim that, if f is Riemann integrable on [a,b], then we can find some partition, P, of [a,b], such that the difference between the upper and lower Darboux sums is arbitrarily small. A partition P of [a,b] is a finite sequence of real numbers
such that .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.
The function can be shown to not be integrable, which would show that it diverges via the integral test -- but on the other hand, you would have to show that f is monotone decreasing on [1,infinity) before you can use that test, and I'm not sure if it is...
Bob bundy's reasoning is probably the most intuitive way to look at it. For a more rigorous proof, it may help to know some more details about f -- for instance, is f continuous? Whilst it is true that continuity => Riemann integrability, the converse is not true in general -- but f being continuous does make the proof a tad easier, without having to resort to measure theory. A good starting point might be how to apply Riemann's Criterion for Integrability here, i.e. f is Riemann integrable if, and only if;
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
.