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Question about integration shouldn't be too difficult

  1. Mar 4, 2009 #1
    Really, I should know the answer to this, but...

    Suppose I'm trying to perform an integration with respect to [itex]t[/itex]:

    [tex]
    \int_0^T f(\phi(t)) dt
    [/tex]

    So my function [itex]f[/itex] is explicitly a function of [itex]\phi[/itex], and [itex]\phi[/itex] depends on time [itex]t[/itex]. But then suppose I end up being able to write the integral as

    [tex]
    \int_0^T g(\phi(t)) \frac{d \phi}{dt} dt.
    [/itex]

    Can I just cancel the [itex]dt[/itex] and perform an integral with respect to [itex]\phi[/itex]? If so, I need to change the limits of integration, right?
     
  2. jcsd
  3. Mar 4, 2009 #2
    Yes, this is my understanding.

    Example: Find the mass of a rod of length L and uniform mass density D.

    M = integral of dm

    D = dm/dx

    so D dx/dm = 1 and

    M = integral of (1 dm) = integral of (D dx/dm dm) = integral of(D dx)

    Since you're now working in distance-space, you just switch the limit to the distance-space limit, namely, 0 -> L.
     
  4. Mar 4, 2009 #3
    Also known as http://en.wikipedia.org/wiki/Integration_by_substitution" [Broken].
     
    Last edited by a moderator: May 4, 2017
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