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Math Help - Integration confusion

  1. #1
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    Integration confusion

    I'm having a look at a "proof" or motivating illustration of the Integrating Factor method for solving a first-order linear ODE, but can't quite understand the precise way in which one goes from step (1) below to step (2) here. I understand that one is to integrate both sides, but I think I'm a bit unclear as to how the integration is being done from t_0 to t when the integrand itself is to involve t, and how the term [TEX]I(t_0)w(t_0)[TEX] arrives in the LHS. And presumably, on the RHS we are free to just change to a dummy variable in the integrand while we integrate over t?

    (1) \frac{d}{dt}I(t)w(t) = I(t)r(t)

    Integrating both sides gives

    (2) I(t)w(t) - I(t_0)w(t_0) = \int_{t_0}^t{I(\sigma)r(\sigma)d \sigma}

    Could anyone please elaborate?
    Last edited by icarus92; July 29th 2014 at 05:00 PM.
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  2. #2
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    Re: Integration confusion

    In general

    $f(t)-f(t_0)=\displaystyle{\int_{t_0}^t} \left(\dfrac{d}{d\sigma} f(\sigma)\right) ~d\sigma$

    This is just the fundamental theorem of calculus.

    Just apply this directly to your formula.

    writing the first line as

    $\dfrac{d}{d\sigma}I(\sigma)w(\sigma)=I(\sigma) r(\sigma)$

    we substitute into the integral formula above to get

    $I(t)w(t)-I(t_0)w(t_0)=\displaystyle{\int_{t_0}^t} \left(\dfrac{d}{d\sigma}I(\sigma)w(\sigma)\right) ~d\sigma$

    $I(t)w(t)-I(t_0)w(t_0)=\displaystyle{\int_{t_0}^t}I(\sigma) r(\sigma) ~d\sigma$
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  3. #3
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    Re: Integration confusion

    Quote Originally Posted by icarus92 View Post
    I'm having a look at a "proof" or motivating illustration of the Integrating Factor method for solving a first-order linear ODE, but can't quite understand the precise way in which one goes from step (1) below to step (2) here. I understand that one is to integrate both sides, but I think I'm a bit unclear as to how the integration is being done from t_0 to t when the integrand itself is to involve t, and how the term [TEX]I(t_0)w(t_0)[TEX] arrives in the LHS. And presumably, on the RHS we are free to just change to a dummy variable in the integrand while we integrate over t?
    We can change to a dummy variable on both sides!

    (1) \frac{d}{dt}I(t)w(t) = I(t)r(t)
    \int_{t_0}^t \frac{d}{ds}I(s)w(s) ds= \int_{t_0}^t I(\sigma)r(\sigma) d\sigma

    Integrating both sides gives

    (2) I(t)w(t) - I(t_0)w(t_0) = \int_{t_0}^t{I(\sigma)r(\sigma)d \sigma}

    Could anyone please elaborate?
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