- #1

Icebreaker

[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.

- Thread starter Icebreaker
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- #1

Icebreaker

[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

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.Icebreaker said:

[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.

- #3

Hurkyl

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- #4

saltydog

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[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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[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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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:saltydog said:

[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.

[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] ???

[tex]r[/tex] ???

- #9

arildno

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Use intergation by parts on this one.Castilla said:Saltydog, which is the antiderivative of [tex]r Sin(r)[/tex] with respect to

[tex]r[/tex] ???

(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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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.

[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:

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Thank you.

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