Derivative of an integral

  • Thread starter Icebreaker
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  • #1
Icebreaker

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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.
 

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  • #2
saltydog
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Icebreaker said:
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.
Use Leibnitz's rule and don't forget the part with the integral of the partial with respect to t.
 
  • #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.
 
  • #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.
 
  • #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]
 
  • #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.
 
  • #7
saltydog
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saltydog said:
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.
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.
 
  • #8
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Saltydog, which is the antiderivative of [tex]r Sin(r)[/tex] with respect to

[tex]r[/tex] ???
 
  • #9
arildno
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Castilla said:
Saltydog, which is the antiderivative of [tex]r Sin(r)[/tex] with respect to

[tex]r[/tex] ???
Use intergation by parts on this one.
(You should get Sin(r)-rCos(r)+C)
 
  • #10
Icebreaker
Wow, so many replies in such a short time. Thanks; I'll be going over all this info.
 
  • #11
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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
 
  • #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:
  • #13
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Thank you.
 

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