du_x/dx1 = Ax1^2 x3 + Bx2x3

x1, x2 , x3 are functions and A and B are constants.

how to find u_x ?

Your derivative term is with respect to x1 only. So your integral will only be with respect to x1. x2 and x3 would be treated as constants even though they are functions.

u_x = 1/3(Ax1^3 x3) + Bx1x2x3 + C

but in the example my professor did in class, when he integrated he a function of different varible instead of a constant.... for example he had f(y) (or f(x2) in place of the constant C.....



This post was edited by FamilyGuyViewer on Oct 2 2017 11:29pm

You'll have to get feanur in here. My Cal C is rusty.

But I see what you mean. Since the y in the example you posted is a "constant" with respect to the integral operation, the arbitrary constant isn't just a constant but instead an unknown function of y which includes its own arbitrary constant.



u_x = 1/3(Ax1^3 x3) + Dx3 + Bx1x2x3 + Ex2x3 + C

You would evaluate the integral as two separate integrals due to the addition. You would end up with 2 arbitrary constants. But they can be combined into a single arbitrary constant. Likewise x2 and x3 in the second term would end up each having their own arbitrary coefficient, but these can also be combined into E.

Of course the latter term with the E may be more complicated as f(x2)f(x3). But I'm too lazy to think about all that at the moment.

This post was edited by timmayX on Oct 2 2017 11:50pm

so what im confused about is when do i add a constant versus adding a function of a different variable after integrating

You would add a function for any term in which there is a non integrated function. Since you are only integrating over x1, the arbitrary term for the A-term in the original problem after integration must include an x3 function. In the second term it must include both an x2 and x3 function. What I am unclear on is the distribution of the second term's functions versus the arbitrary constant.


Feanur would do a much better job explaining this, I'm sure.

du_x/dx1 = Ax1^2 x3 + Bx2x3

u_x(x₁ , x₂ , x₃) = A.(x₁³ / 3).x₃ + B.x₁.x₂.x₃ + C(x₂ , x₃)
where C is a function that does not depend on x₁.

When derivating C with respect to x₁, only 0 remains (since, precisely, it doesn't depend on x₁).

Without any other information, it's impossible to know anything about C.
It could be as simple as : C(x₂ , x₃) = x₂ + x₃
or as complicated as : C(x₂ , x₃) = log(x₂² + x₃⁶ + 1) / (sin x₂ + cos x₃ + exp(x₂ - x₃) )



What you must add is a constant - with respect to the variable you're working on.
In other word, you add a function that depends on any other variables (see examples above).

Thank you for the lesson. As generic and arbitrary as possible.





This post was edited by timmayX on Oct 3 2017 01:29pm

Integrate it

This post was edited by BERRISWEET on Oct 3 2017 01:57pm