# Derivative of an integral

Can anyone help me solve this?

$$\frac{\partial}{\partial t} \int_{0}^{t} f(r,t)g(r)dr$$

That's the generalized form of an equation I'm trying to solve.

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Icebreaker said:
Can anyone help me solve this?

$$\frac{\partial}{\partial t} \int_{0}^{t} f(r,t)g(r)dr$$

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.

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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.

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Oh yea Icebreaker, try this one both ways ok:

$$\frac{\partial}{\partial t}\int_0^t (t^2 r)Sin(r)dr$$

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.

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

$$h(u, v) = \int_0^u f(v, r) g(r) \, dr$$

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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.

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saltydog said:
Oh yea Icebreaker, try this one both ways ok:

$$\frac{\partial}{\partial t}\int_0^t (t^2 r)Sin(r)dr$$

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:

$$\frac{d}{dt}\int_0^t G(r,t)dr=G(t,t)+\int_0^t \frac{\partial G}{\partial t} dr$$

So:

\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}

But check it to make sure I didn't make any errors.

Castilla
Saltydog, which is the antiderivative of $$r Sin(r)$$ with respect to

$$r$$ ?

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Castilla said:
Saltydog, which is the antiderivative of $$r Sin(r)$$ with respect to

$$r$$ ?
Use intergation by parts on this one.
(You should get Sin(r)-rCos(r)+C)

Wow, so many replies in such a short time. Thanks; I'll be going over all this info.

Castilla
Saltydog, I get another result for $$\int_0^t \frac{\partial}{\partial t} {t^2 rsin(r)dr$$.

I got this:

$$\int_0^t \frac{\partial}{\partial t} {t^2 rsin(r)dr = \int_0^t {2t rsin(r)dr} = 2t \int_0^t {rsin(r)dr} = 2t (sin(t) - cos(t) - ( sin(0) - cos(0)) = 2t ( sin(t) - cos(t) +1)$$.

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

Castilla

\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}