Finding the gradient of an integral

[tomelwood]

Homework Statement

I am getting quite confused as to the concepts behind this task. I have a function given as a double integral, and am asked to find the gradient of it.
However, I have no notes on how to do this, so it is either a very simple task, or the lecturer has once again missed things out.

Calculate the gradient of F(x,y) = \int^{1}_{0}\int^{1}_{0}sin(sxy)e^{t}dsdt
and also F(x,y) = \int^{x}_{0}yte^{t}dt + \int^{1}_{0}sin(xyt)dt

Homework Equations

The Attempt at a Solution

I don't see why we can't just differentiate under the integral sign, essentially leaving the same expressions but without the integral signs. Then that leaves the question of what to do with the limits of integration.
I'm not even sure what topic this might be. Line integrals?
Any help at all would be greatly appreciated.

 

[LCKurtz]

Homework Statement

I am getting quite confused as to the concepts behind this task. I have a function given as a double integral, and am asked to find the gradient of it.
However, I have no notes on how to do this, so it is either a very simple task, or the lecturer has once again missed things out.

Calculate the gradient of F(x,y) = \int^{1}_{0}\int^{1}_{0}sin(sxy)e^{t}dsdt
and also F(x,y) = \int^{x}_{0}yte^{t}dt + \int^{1}_{0}sin(xyt)dt

Homework Equations

The Attempt at a Solution

I don't see why we can't just differentiate under the integral sign, essentially leaving the same expressions but without the integral signs. Then that leaves the question of what to do with the limits of integration.
I'm not even sure what topic this might be. Line integrals?
Any help at all would be greatly appreciated.

You can differentiate under the integral sign. But you aren't differentiating with respect to the variable of integration so the integrals don't disappear. For example if you want to differentiate

\int^{1}_{0}\int^{1}_{0}sin(sxy)e^{t}dsdt

with respect to x, you get

F_x(x,y) =\int^{1}_{0}\int^{1}_{0}\frac{\partial \sin(sxy)e^{t}}{\partial x}dsdt = \int^{1}_{0}\int^{1}_{0}sy\cos(sxy)e^{t}dsdt

which you can integrate and simplify. Also I don't see anything to stop you from doing the integral first before you calculate the gradient.