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Inrectangular subintegral expression

  1. Jul 28, 2011 #1
    Normally, integration means summing the infinitesimal rectangles of height f(t) and width dt. The differential dt appears as a multiplier behind the integration mark, [itex]\int[/itex], therefore. But, what do you think about summing non-standard infinitesimal pieces, like [itex]\int{1 \over 1/dt + 1}[/itex]?

    For every dt, it gives some very small value and the sum converges. I know that I can solve this one by the rearrangement [itex]\int{1 \over 1/dt + 1} = \int{dt \over 1 + dt} = \int{dt}[/itex], as dt -> 0. I was puzzled when I had first time to compute this thing. How do you call call it? Is it a usual method to solve this kind of thing?
  2. jcsd
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