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Integral and a derivative definition

  1. Mar 22, 2013 #1
    Hello,

    I just whant to know what mathematical rule alows me to do this? I mean i think it is u-substitution, but i am not sure how it is done here? It is weird to me as it seems that ##dt## just cancel out and limits are changed...

    $$
    \int\limits_{0}^{t} \frac{dv}{dt} \cdot mv \gamma(v)\, dt = \int\limits_{0}^{v} mv \gamma(v)\, dv
    $$
     
  2. jcsd
  3. Mar 22, 2013 #2

    pwsnafu

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    Science Advisor

    Integration by Substitution:
    Suppose ##I## is an interval. Let ##g : [a,b] \to I## have continuous derivative. Let ##f : I \to \mathbb{R}## be continuous. Then
    ##\int_{g(a)}^{g(b)} f(x) \, dx = \int_{a}^{b} (f \circ g)(t) \, g'(t) \, dt##
     
  4. Mar 22, 2013 #3
    This is a basic formula yes i know, but how does this solve my case?
     
  5. Mar 22, 2013 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    "Integration by substitution" is, essentially, the "chain rule" reversed. It is valid because the chain rule for differentiation is valid.
     
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