Riemann integration

The last chapter (Chapter 11) of my second-year module G12MAN Mathematical Analysis consists of a brief (one-lecture) introduction to Riemann integration. This material is motivated in terms of questions of antidifferentiation and area. The proofs of the lemmas and theorems are not included here (see books for details), but the main definitions are given in full, along with illustrative examples and diagrams, and the statements of the main theorems. Material discussed includes: partitions of intervals; Riemann lower and upper sums (approximation using rectangles); the Riemann lower and upper integrals; Riemann integrability of functions, and the Riemann integral. Examples are given of functions which are/are not Riemann integrable: in particular, continuous real-valued functions on closed intervals are Riemann integrable. The lecture concludes with the statements of the (first) Fundamental Theorem of Calculus and the Mean Value Theorem of Integral Calculus.


5 responses to “Riemann integration

  1. I dont get this post or its point.

    I dont see why a function must be continuous for it to be Riemann Integrable. It stands to reason that any function – continuous or discontinuous – has a finite Riemann Sum if it is DEFINED all all points in an interval.

    As long as it is defined everywhere in an interval (ie, the function is defined everywhere on a continuous domain, even if its range is discontinuous), then I see no reason why f(x) would fail to return a value. As long as f(x) produces a real output value for all x in [a,b], then it is, by definition, Riemann Integrable.

    Is it not? If not then why not?


    • Hi,
      There are several issues raised here.
      – You are certainly correct, a function does not have to be continuous in order for it to be Riemann integrable. However, the one-way implication is true: every continuous real-valued function on a closed interval [a,b] is Riemann integrable there. The converse to this is false. However, I did not (I hope!) claim that Riemann integrable functions must be continuous.
      – Whenever f is a bounded real-valued function on [a,b], then you can define the Riemann upper sums and the Riemann lower sums for f. This does not work if the function f is unbounded, though. For example, if the function f is, say, 1/x for x not equal to 0, but f(0)=0, then you can not define Riemann lower/upper sums for f on [-1,1].
      – Even when f is a bounded, real-valued function on [a,b], you can have problems. For example, if the function f(x) is defined to be 1 when x is rational but 0 when x is irrational, and you try to find the Riemann integral on the interval [0,2], say, then all of the Riemann upper sums come out to be 2, while the Riemann lower sums are all 0. As a result, the Riemann upper integral is 2, and the Riemann lower integral is 0. Since these are different, the Riemann integral does not exist here, and f is not Riemann integrable on [0,2].
      Joel 18/12/09


      • No, youre fine. Maybe its the fact that I havent slept in over a day, I dont know. Its just that when you say an if then statement the way you did, it just implied to me a required condition. You said it the right way, sure, but seemingly implying that it was a condition and a necessity. I mean, if it wasnt a necessity then why even say it? Perhaps implication is necessity is just an artifact of language. And yes, the converse is obviously false.

        Again, though, your bit on bounded Riemann sums seems off/wrong to me. Finite sums, youre right and I see it. But infinite sums? If the “rectangles” are infinitesimally wide then your bit on bounded intervals and piecewise functions seems irrelevant.

        Of course, if you mean that the sum cannot diverge infinitely, as it would in 1/x over [-1,1] then I agree – for infinite sums. But for finite sums, I dont see it as a problem… why? Because a loss of accuracy is implied in any finite sum, isnt it? Again, the question isnt what the area IS, the question is whether or not the Riemann Sum is defined and evaluable for what it is… with no regard for what its intended to represent.

        Perhaps Im just yapping ignorantly.


  2. how do you use riemann sums to evaluate functions of the form f(x)={-x, 0<(less than equal to)x1} taking the integral of f(x)dx from -1 to 1?


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