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Integrating wrt a function

  1. Mar 6, 2010 #1
    When we integrate a function [tex]f(t)[/tex] with respect to t, we are finding the area under the curve [tex]f[/tex]. Intuitively, this is very clear.

    What is the intuition behind integrating a function with respect to another function?
    \int f(t)dg
    where g is itself a function of t?
  2. jcsd
  3. Mar 6, 2010 #2


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    I'm not sure there is an easy intuitive answer, beyond interpreting it mathematically.
    Since g is a function of t, you can apply the chain rule to give

    [tex] \int f(t)d[g(t)]=\int f(t)\frac{d[g(t)]}{dt}dt=\int f(t)\.{g}(t) dt[/tex]

    So you are finding the area under another curve that equals the first function weighted by the slope of the second.
  4. Mar 6, 2010 #3
    What if g is not differentiable? In fact, the most interesting case is when g is not even continuous. Look up the Riemann–Stieltjes integral
  5. Mar 6, 2010 #4
    Thanks for your attempt Marcus.

    l'Hopital, I suppose if g is not differentiable then the equation can still be solved using stochastic calculus (ie. Ito integrals).
    But my question has more to do with trying to understand what the integral represents, rather than a way of solving it.
  6. Mar 6, 2010 #5
    it's a way to weight the domain.
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