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Derivative of an integral

  1. Sep 14, 2005 #1
    Can anyone help me solve this?


    [tex]\frac{\partial}{\partial t} \int_{0}^{t} f(r,t)g(r)dr[/tex]

    That's the generalized form of an equation I'm trying to solve.
     
  2. jcsd
  3. Sep 14, 2005 #2

    saltydog

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    Use Leibnitz's rule and don't forget the part with the integral of the partial with respect to t.
     
  4. Sep 14, 2005 #3

    Hurkyl

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    IIRC, it's not very difficult to work it out directly from the limit definition of the derivative and the appropriate continuity hypotheses.
     
  5. Sep 14, 2005 #4

    saltydog

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    Oh yea Icebreaker, try this one both ways ok:

    [tex]\frac{\partial}{\partial t}\int_0^t (t^2 r)Sin(r)dr[/tex]

    Do the integration first then take the derivative is one way. Then for the second way use Leibnitz's rule or do what Hurkyl said.
     
  6. Sep 14, 2005 #5

    Hurkyl

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    I suppose you could always use the chain rule... consider

    [tex]
    h(u, v) = \int_0^u f(v, r) g(r) \, dr
    [/tex]
     
  7. Sep 14, 2005 #6

    HallsofIvy

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    By the way- there is no need to write that as a partial derivative. It's a function of t only.
     
  8. Sep 15, 2005 #7

    saltydog

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    Details, details guys. Icebreaker I suspect, and this is only a hunch and I could be wrong, but that partial, the integral, function of two variables is well maybe a little intimidating. Hey, they're always posting stuf in here I find intimidating. :yuck: Here's Leibnitz' rule applied to this problem:

    [tex]\frac{d}{dt}\int_0^t G(r,t)dr=G(t,t)+\int_0^t \frac{\partial G}{\partial t} dr[/tex]

    So:

    [tex]\begin{align*}
    \frac{\partial}{\partial t}\int_0^t (t^2 r)Sin(r)dr &=
    t^3 Sin(t)+\int_0^t \frac{\partial}{\partial t}(t^2 r Sin(r))dr \\ &=
    t^3 Sin(t)+\int_0^t 2t rSin(r)dr \\ &=
    t^3Sin(t)+2t[Sin(t)-tCos(t)]
    \end{align}
    [/tex]

    But check it to make sure I didn't make any errors.
     
  9. Sep 15, 2005 #8
    Saltydog, which is the antiderivative of [tex]r Sin(r)[/tex] with respect to

    [tex]r[/tex] ???
     
  10. Sep 15, 2005 #9

    arildno

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    Use intergation by parts on this one.
    (You should get Sin(r)-rCos(r)+C)
     
  11. Sep 15, 2005 #10
    Wow, so many replies in such a short time. Thanks; I'll be going over all this info.
     
  12. Sep 26, 2005 #11
    Saltydog, I get another result for [tex] \int_0^t \frac{\partial}{\partial t} {t^2 rsin(r)dr [/tex].

    I got this:

    [tex] \int_0^t \frac{\partial}{\partial t} {t^2 rsin(r)dr =

    \int_0^t {2t rsin(r)dr} =

    2t \int_0^t {rsin(r)dr} =

    2t (sin(t) - cos(t) - ( sin(0) - cos(0)) =

    2t ( sin(t) - cos(t) +1)

    [/tex].

    By the way, how can I cut these long phrases?

    Castilla
     
  13. Sep 26, 2005 #12

    saltydog

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    Check out the align code here (do a "quote" to check out the commands):

    [tex]
    \begin{align*}
    \int_0^t \frac{\partial}{\partial t} t^2 rsin(r)dr &= \int_0^t 2trSin(r)dr \\
    &= 2t\int_0^t rSin(r)dr \\
    &= 2t\left[(Sin(r)-rCos(r))\right ]_0^t \\
    &= 2t(Sin(t)-tCos(t))
    \end{align}
    [/tex]

    The asterisk prevents equation numbering, the \\ skips lines, the &= aligns on the equal signs. I think you missin' that extra t.
     
    Last edited: Sep 26, 2005
  14. Sep 26, 2005 #13
    Thank you.
     
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