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The two meanings of the integral sign

  1. Jan 25, 2010 #1
    The integral sign [tex]\int[/tex] can have two meanings:

    (1) It indicates the Riemann sum, in which we don't actually integrate, but add the rectangles under the curve.

    (2) It indicates the actual integration, in which we integrate (i.e. antidifferentiate) the function, either just to find the antiderivative (without limits but with a constant), or to find the area under a curve (with limits but without a constant).

    But almost all of the times the integral sign indicates (2).

    Is the above correct?
  2. jcsd
  3. Jan 25, 2010 #2


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    That IS integration!
    This is anti-differentiation.
    Of course, since the fundamental theorem of calculus proves that the often ridiculously easier (2) gives the exact answer to (1)
  4. Jan 26, 2010 #3


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    Not exactly. In (1), the integral sign does not represent the Riemann sum- it represents the limit of the Riemann sum where you take the limit in a specified way. As for (2), the Fundamental Theorem of Calculus says that it gives the same result as (1) so they really have the same meaning.
  5. Jan 26, 2010 #4

    D H

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    Plus an arbitrary constant, of course. Essentially, (1) is the definite integral (but note Hall's "not exactly" caveat) and (2) is the indefinite integral, aka the anti-derivative.

    Also note that while an anti-derivative of any well-behaved function always exists, finding that anti-derivative can be a difficult, if not impossible, task. Ofttimes the anti-derivative of some function is expressed in terms of a definite integral. For example, consider the anti-derivative of [itex]f(x)=\exp(-x^2)[/tex]. This anti-derivative cannot be expressed in terms of elementary functions. The corresponding definite integral is a very important function in statistics and elsewhere:

    [tex]\text{erf}(x) = \frac 2 {\sqrt{\pi}} \int_0^x e^{-t^2} dt[/tex]

    Just because the anti-derivative does not exist (in terms of elementary functions, that is) does not mean the definite integral cannot be computed.
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