Finding the Derivative of an Integral: Can You Solve It?

[Icebreaker]

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.

 

[saltydog]

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.

 

[Hurkyl]

IIRC, it's not very difficult to work it out directly from the limit definition of the derivative and the appropriate continuity hypotheses.

 

[saltydog]

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.

 

[Hurkyl]

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

<br /> h(u, v) = \int_0^u f(v, r) g(r) \, dr<br />

 

[HallsofIvy]

By the way- there is no need to write that as a partial derivative. It's a function of t only.

 

[saltydog]

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. 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*}<br /> \frac{\partial}{\partial t}\int_0^t (t^2 r)Sin(r)dr &amp;=<br /> t^3 Sin(t)+\int_0^t \frac{\partial}{\partial t}(t^2 r Sin(r))dr \\ &amp;= <br /> t^3 Sin(t)+\int_0^t 2t rSin(r)dr \\ &amp;=<br /> t^3Sin(t)+2t[Sin(t)-tCos(t)]<br /> \end{align}<br />

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 ?

 

[arildno]

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)

 

[Icebreaker]

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 = <br /> <br /> \int_0^t {2t rsin(r)dr} = <br /> <br /> 2t \int_0^t {rsin(r)dr} = <br /> <br /> 2t (sin(t) - cos(t) - ( sin(0) - cos(0)) =<br /> <br /> 2t ( sin(t) - cos(t) +1)<br /> <br />.

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

Castilla

 

[saltydog]

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

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

The asterisk prevents equation numbering, the \\ skips lines, the &= aligns on the equal signs. I think you missin' that extra t.

 

[Castilla]

Thank you.