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Leibnitz Rule For an Integral.

  1. Oct 14, 2012 #1
    I found a particular integral in my stat book.

    [itex] \frac{d}{ d\theta}\int^{b(\theta)}_{a(\theta)}f(\theta,t)dt =

    \int^{b(\theta)}_{a( \theta)}\frac{ \partial}{ \partial \theta}f( \theta ,t)dt +

    f( \theta, b( \theta)) \frac {\partial b(\theta)}{ \partial \theta} -

    f(\theta, a(\theta))\frac{ \partial a(\theta)}{\partial \theta} [/itex]

    Why is this the case? Why is it not...

    [itex] \int^{b(\theta)}_{a(\theta)}f(\theta,t)dt =

    \int^{b(\theta)}_{a( \theta)}\frac{ \partial}{ \partial \theta}f( \theta ,t)dt +

    \frac{d}{ d\theta} [F( \theta, b( \theta)) - F(\theta, a(\theta))]

    [/itex]



    EDITED: Fixing LaTeX, as per usual. Sorry Folks.
     
    Last edited: Oct 14, 2012
  2. jcsd
  3. Oct 16, 2012 #2

    arildno

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    1. Set up the limit definition of the derivative.
    2. Now, you have to add zero in a creative way within the limit expression.
    3. For example:
    [tex]\int_{a(\theta+h)}^{b(\theta+h)}f(\theta+h,t)dt=\int_{a(\theta+h)}^{b(\theta+h)}f(\theta+h,t)dt-\int_{a(\theta+h)}^{b(\theta)}f(\theta+h,t)dt+\int_{a(\theta+h)}^{b(\theta)}f(\theta+h,t)dt=\int_{b(\theta)}^{b(\theta+h)}f(\theta+h,t)dt+\int_{a(\theta+h)}^{b(\theta)}f(\theta+h,t)dt[/tex]
    The first integral is readily seen to converge (when divided by h and letting h go to zero) to the second term in your first line's RHS.
    With the two remainding integrals within your limiting expression, add another creative zero to get the other two terms in your first line's RHS
     
    Last edited: Oct 16, 2012
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