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Precision of integrals

  1. Aug 10, 2014 #1


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    I have read that even the simplest integrals (like y=x2) might need some correction if we want to reach an extreme precision. Is that really so?
    Can you explain why or give me some useful links?
  2. jcsd
  3. Aug 10, 2014 #2


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    Define "extreme precision".

    Integrals which can be done analytically (such as the integral of [itex]x^2[/itex]) cannot be more precise.

    If you can only do an integral numerically then you are constrained by the fact that computers can only ever do a finite number of calculations on a finite subset of the rational numbers, so there's always going to be some error; the question is whether the error can be made small enough so that you can ignore it in the context of whatever your actual problem is.
  4. Aug 10, 2014 #3
    No. ## \int_a^b x^2 \,dx = 2(b - a) ## there is no "correction" involved.

    Can you explain why you think this or give us a useful link?
  5. Aug 10, 2014 #4


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    Don't you mean,

    [tex] \int_a^bx^2 dx = \frac{1}{3}\left( b^3 - a^3 \right)? [/tex]
  6. Aug 10, 2014 #5


    Staff: Mentor

    If you use a numerical technique, the results are generally imprecise. As already mentioned in this thread, if you are lucky enough to know the antiderivative of the function you're integrating, the results will be exact.
    There are lots of numerical methods for integration - trapezoid rule, Simpson's rule, many others. See http://en.wikipedia.org/wiki/Numerical_integration for more information.
  7. Aug 10, 2014 #6

    Oops yes of course, thank you - now how did that happen? :blush:
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