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Quick integration question

  1. Jul 5, 2012 #1
    Can somebody explain to me what this is called

    I mean what's happening on the left side of this equation
    Why does the T turn into y(T) and 0 into y(0)?

  2. jcsd
  3. Jul 5, 2012 #2
    Could you give some context behind the equations?
  4. Jul 5, 2012 #3
  5. Jul 5, 2012 #4


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    hi nhmllr! :smile:
    the variable of integration is changing from t to y

    the original limit was 0 < t < T

    y is a function of t

    so that's the same as y(0) < y(t) < y(T) :wink:

    (you need the same limit, written in the new variable)
  6. Jul 5, 2012 #5
    Ohhh... I think I see, Thanks
  7. Jul 5, 2012 #6


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    It's basically a change of variable, or if you prefer to think about it this way, a reversal of the usual method of substitution.

    If you're working out [itex]\int f(y)dy[/itex], where y is dependent on t, i.e. [itex]y = g(t)[/itex], then you can state:

    [itex]\int f(y)dy = \int f(g(t))dy = \int f(g(t))\frac{dy}{dt} dt = \int f(y)\frac{dy}{dt}dt[/itex].

    What happened there is I made a change of variables from y to t. [itex]dy = \frac{dy}{dt}dt[/itex]. You should be able to recognise that as the basis for substitution.

    In the example, they're just going in reverse.

    The reason the bounds change is that the bounds must follow the variable of integration. So if the integration is wrt t, the bounds will be [0,T]. If the integration is wrt y, the bounds will be [itex]y_0, y_T[/itex] where the lower bound refers to the y-value at t = 0 and the upper bound refers to the y-value at t=T.
  8. Jul 7, 2012 #7
    The limit of integration is from 0 to T and you're integrating with respect to the varible t, as shown by dt.
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