# Derivative of an integral

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
Homework Helper
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.

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

saltydog
Homework Helper
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
Staff Emeritus
Gold Member
I suppose you could always use the chain rule... consider

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

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

saltydog
Homework Helper
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.

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

$$r$$ ???

arildno
Homework Helper
Gold Member
Dearly Missed
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)

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

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

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

\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}

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

Last edited:
Thank you.