It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. I know that certain functions are not integrable. Geometrically integration is finding the area under the curve of the graph of the given function.
So, in another way why is it not possible to find the area under the curve in case of certain functions? Well, well, well, this question actually can go really deep into the elementary ideas of integration, or, in general, finding the area under a curve. In the ancient times, and especially during the period when mathematicians such as Eudoxus and Archimedes lived, the main way in which one could calculate the area between non-linear curves was the Method of Exhaustion.
We will examine, in short, how would this work on calculating the area of a circle. Then, we have the following situation:. Now, the next step is to enclose the circle in some "tighter" canonical polygons. Then one can move a step forward to create a deca-exagon 16 angles canonical polygon etc. It is clear the the red polygon is always underestimating the area of the circle and the blue polygon is always oversetimating it.
Now, it is relatively easy, using the know formulas for the area of canonical polygons, to prove that both the red and the blue area do converge to the area of the circle.
However, since back then calculus was not something "known" with the modern-day meaning numbers were always treted as geometrical objects and not as abstract notions they used the following form of Eudoxus' principle:. After that, "convergence" this is a modern term; it was not used back then was ensured.
What was missing was to show that these two areas blue and red were both "converging" to the area of the circle. To do so since calculus and, hence, limits, were unknown an argument - again tiring, in the general case - of double contradiction was used.
It is remarkable that, in the same way one could find the volume of the sphere and several other 3D shapes. It is called Riemann Integral , however, in order to find the very solid geometric foundation of the notions of integral and integration we should investigate the so-called Darboux Upper and Lower Sums. Note now that, in this way, Rieman integration is "just" a more generalised implementation of the Method of "Exhaustion" that ancient mathematicians used.
Moreover, what we seek, in order to be able to say that a function is Riemann integrable, is for the two limits of the sequences of the estimations given by the Darboux sums do converge to the same number, which was also the wanted in the Method of Exhaustion, in spite of not being proved with modern day calculus but with "old-fashioned" - however, interesting - geometrical arguments.
So, to conclude, a function is not Riemann integrable exactly when the Method of Exhaustion cannot be applied to the curve that represents it graph. However, this does not apply to Lebesgue integrability I don't feel this was the case the OP requested, however, I can extend to Lebesgue integration, if requested.
So, with respect to the historical frame, Archimedes was very close to the notion of infinite, non-countable summation that we give to the Riemann integral, nowadays. I think this is more or less due to Banach-Tarski paradox, which shows a ball can be decomposed into a finite number of point sets and reassembled into two balls identical to the original one.
This means we cannot define measurement a conception rather related with integration for every point set. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Thus the area chosen to represent a single slice in a Riemann sum will be either its width or 0 depending upon whether we pick a rational x or not at which to evaluate our integrand in that interval.
For this function no matter how small the intervals are, you can have a Riemann sum of 0 or of b - a. In this case it is possible to use a cleverer definition of the area to define it. You can argue, in essence, that there are so many more irrational points than rational ones, you can ignore the latter, and the integral will be 0. If we consider the area under the curve defined by in an interval between -a and b for positive a and b, the area has an infinite positive part between 0 and b and an infinite negative part between -a and 0.
It is possible to define the area here so that these cancel out and meaning can be given to the net area. If you leave out the interval between -d and d for any small d, the remaining area is finite, and can be computed.
You can then take the limit of this area as d goes to 0.
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