Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

missyfredtom

  • 4 years ago

i really need someone to help me

  • This Question is Closed
  1. Safari321
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    WITH WHAT?

  2. missyfredtom
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    integrals and reimann sums

  3. Safari321
    • 4 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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].

  4. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy