Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof of Definite Integral?

  1. Feb 8, 2012 #1
    Proof of Definite Integral??

    Riemann Sum helps us to find the are under a curve.But I couldn't understand connection between Riemann Sum and Definite Integral.I mean;

    [itex]\int^{b}_{a}f(x).dx=F(b)-F(a)[/itex] How can I prove F(b)-F(a) in definite integral??

    Last edited by a moderator: Feb 8, 2012
  2. jcsd
  3. Feb 13, 2012 #2
    Re: Proof of Definite Integral??

    Well, the connection between the two is that the definite integral is essentially using an infinite amount of Riemann sums to calculate the area under a curve. Integration allows for the area to be calculated without having to add up the area of each Riemann sum manually.

    Now, [itex]\int^{b}_{a}f(x)dx[/itex] is equal to [itex]F(b)-F(a)[/itex] because [itex]F(b)[/itex] calculates the area from 0 to b, [itex]F(a)[/itex] calculates the are from 0 to a. Subtracting these two values will cut the area [itex]F(a)[/itex] calculated from [itex]F(b)[/itex] leaving just the area from a to b.

    I hope I was able to provide the information that you needed for you to finish your proof.
  4. Feb 13, 2012 #3


    User Avatar
    Science Advisor

    Re: Proof of Definite Integral??

    Hey Calculuser and welcome to the forums.

    You actually can't do this for any f(x): it needs to what is called "Riemann-Integrable" for this to happen.

    There are strict conditions that can be representing through formal mathematical statements but the basic idea is that your anti-derivative must be analytically differentiable and continuous everywhere across the domain (i.e. from a to b).

    If this is not the case, then you have to either separate your intervals into ones where this happens or use a different integration technique (like for when you are modelling things like Brownian motion).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook