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Quick question about integrals

  1. Jan 17, 2010 #1
    An integral can be defined as the limit (as n approaches infinity) of the sum of f(x)times Delta x. It has been said that f(x) can be taken at any point of the Delta x interval (at the right side, at the left side, at the center, etc.). My question is: can I pick f(x) at one point of the Delta x interval (let's say, the left) in one rectangle, and then in the next rectangle pick it in another point (let's say, the center), and so on, or once I make my choice of where I will choose the point, I must be consistent?
     
  2. jcsd
  3. Jan 17, 2010 #2
    Yes, you can take it at any point inside the rectangle side. It doesn't change anything since dx apreaches to zero, so every where on the rectangle side aproaches to the same f(.) value. It will not change the value of the integral. But in applied mathmatics, if you use a computer software or a microcontroller (DSP chip, etc) to calculate the inegral, it will change the result very very very little for sufficiently enough dx values assuming that function is not changing too fast (i.e.; it doesn't have discontinuities). The error will be neglegible.
     
  4. Jan 17, 2010 #3
    Yes, I understand I can take it at any point inside the rectangle side, but can I pick it at different points in each rectangle? For example, I take it at the center in the first, at the left in the second, at the right in the third, etc. I'm inclined to think this is true, because as you said "it doesn't change anything since dx apreaches to zero, so every where on the rectangle side aproaches to the same f(.) value."
     
  5. Jan 17, 2010 #4
    Yes you can take it at different points for each rectangle.
     
  6. Jan 17, 2010 #5
    OK, thanks for your help!
     
  7. Jan 18, 2010 #6

    HallsofIvy

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    Science Advisor

    One of the requirements for a function to be "Riemann integrable" is that any such choice must give the same result. One consequence of that is, knowing the integral does not depend on the choice of point inside each interval, you can choose points that make the sum as easy as possible.

    An example where that does NOT work is the function defined by f(x)= 0 if x is rational, 1 if x is irrational. Divide the interval from 0 to 1 into whatever intervals you like. There will be both rational and irrational numbers in each interval. If you choose always irrational points in each interval, the value of the function will always be 1 and the limit will be 1. But if you choose always rational points, the value of the function will always be 0 and the limit will be 0.

    However, precisely because the limit depends on the choice of points and those two limits are NOT the same, that function is NOT integrable.
     
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