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What exactly does integration signify?

  1. Jun 15, 2013 #1
    I thought about posting this in the math forums, but I'm curious about what integrals actually represent as applied to physics so I'm posting it here. I feel that others will conceptually benefit from this conversation as well.

    Differentiation has a clear and easily understandable meaning and application since there are rates everywhere in the physical world.

    We also all know that integration is simply the "opposite" of differentiation and it can be thought of as the area under a curve by summing the infinitely many rectangles of ever decreasing width.

    ... So that all makes sense, but can someone talk about how integration fits into the physical world? Derivatives are to rates as integrals are to... what?
  2. jcsd
  3. Jun 16, 2013 #2


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    As far as physics goes I just view them as sums of infinitesimal elements. Mathematically this isn't really correct since summation is a specific type of integration but its fine for physics. For example if I have charge distributed over a region I would start with an infinitesimal element of the region on which there is some charge and add up all such elements throughout the region to get the total charge. This ends up just being the integral of the charge density over that region. Geometrically it's a way of dealing with quantities associated with continuous distributions as opposed to discrete distributions. However as I said this is my handwavy way of looking at integrals in physics because mathematically summation is a special case of integration.
  4. Jun 16, 2013 #3
    An indefinite integral, as you probably know, is merely the antiderivative of a function. A definite integral is the change in the antiderivative's value between the two points equal to the limits. For example, if you have velocity as a function of time and you take the integral from t=0 to t=3 then the result will be the total displacement between time 0 and time 3. An integral represents the change in the antiderivative between two values of the independent variable. If you want me to explain how that has anything to do with the area under the curve, I will.
  5. Jun 16, 2013 #4
    The two fundamental theorems of integration. I think i studied somewhere in highschool. So for me the definite and the indefinite integrals are two different things.

    The first fundamental theorm states that
    if df/dx = g, then
    {integral}g*dx = f + C

    This means that the indefinite integration is inverse function of diffrenciation.

    The second fundamental theorem of integration talks about the definite integrals. It states that

    {integral_a to b} f(x)*dx = Area under the curve from a to b.

    From the second fundamental theorem the concept of sum of infinitesimal elements is derived.

    As you said that difrentiation has a clear meaning so is indefinite integration.

    Even the definite integral has a clear meaning in physics, it is the are under the curve or the sum of infinitesimal elimens(both are same things).
  6. Jun 16, 2013 #5


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    I think this is the best way of seeing integrals. Both in physics and in mathematics. Using the usual epsilon-delta theory of analysis, this isn't a rigorous approach. However, you might be interested that we can make this rigorous in nonstandard analysis. There, an integral really is adding up infinitesimal elements.
  7. Jun 16, 2013 #6


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    You could try "derivatives are to the rate of change of a quantity as integrals are to the quantity itself"....

    Although to really make it click, you might try using integration to derive ##s=\frac{1}{2}at^2+x_0## from ##v=at##; and to derive the formula for the volume of a sphere.

    (That's what worked for me many many years ago)
  8. Jun 17, 2013 #7
    Totally agree, the same worked for me. And, to take the derivative of 1/2 at^2 + x_0 to realize that it is just at, so that if you forget either but remember one of them you can still easily find the other.
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