# Integration of a term that is squared, cubed, etc.

1. Jul 9, 2011

### hoomanya

Hi,

I was wondering if the following statement is correct:

$\int$($\frac{du}{dx}$)$^{2}$dV=$\int$$\frac{du}{dx}$dV$\int$$\frac{du}{dx}$dV

Thanks!

2. Jul 9, 2011

### BrianMath

No because the left hand side has just $dV$, whereas the right hand side has $dV^2$ because you have two of them multiplying together.

3. Jul 9, 2011

### pmsrw3

Counterexample: let u(x) = x, v = x.

4. Jul 9, 2011

### hoomanya

Thank you both. pmsrw3 I am not bothered with solving the integral.

Can I write it as something like this then, using the divergence theorem:

$\int$($\frac{du}{dx}$)$^{2}$dV=$\frac{1}{V}$$\int$u.n dS$\int$u.n dS

Cheers!

5. Jul 9, 2011

### pmsrw3

???

But the example I gives shows that your speculated relationship doesn't hold. It doesn't matter whether you want an explicit solution or not.

Wait a minute, I think this needs some clarification. Is dV a volume element? In how many dimensions, and what is x? What's the context?

6. Jul 9, 2011

### hoomanya

Sorry, I don't get the counter example. Thank you for your quick responses.

V is a volume integral, x is a dimension and u is displacement. The problem is 1D and the context is solid deformation and finite element analysis.

By the way, does it depend on the context or is it a mathematical relationship (if it's correct)?

Thanks again.

7. Jul 9, 2011

### pmsrw3

If u(x) = x, du/dx = 1. If v=x, dv=dx. The left-hand-side integral is therefore x+C1. The RHS integrals comes to (x+C2)(x+C3). Obviously the LHS and RHS aren't equal.

OK, then I completely don't get the question. If you're in one dimension and it's a volume integral, doesn't that mean V = x? In one dimension, the surface of a region is just two points, the left and right ends (assuming the region is contiguous), and the normal vector n is -1 at one end, +1 at the other end. In other words, the divergence theorem in one dimension is nothing but the familiar

$$\int_a^b \frac{du}{dx}dx = u(b) - u(a)$$

I think you have the idea that the fact the the integrand is the square of the derivative of something makes it special. But it doesn't, you know: ANY (well-behaved) function can be expressed as the square of the derivative of something. $\int(\frac{du}{dx})^2 dV$ is no more special than $\int u dV$.

Both could be true. The truth or falsehood of an equation will depend on what the symbols mean. For instance you can't tell me whether the inequality $x^2 \geq 0$ is true or false unless I tell you what kind of values are allowed for x.

8. Jul 9, 2011

### hoomanya

Thanks a lot.If the problem was not 1D and you had to write the left hand side of the equation in terms of a product the two terms I have on the RHS(after using the divergence theorem), would there be a way? Or is there a context or a situation this or something similar would hold?

9. Jul 9, 2011

### pmsrw3

Maybe you should explain where this question comes from and what you're trying to do.

10. Jul 9, 2011

### hoomanya

Thanks for trying to help. It would be too complicated to explain the entire context. I posted the question hoping this is a general mathematical relations or something close. But you helped understanding that it probably isn't. So thanks for that!

11. Jul 10, 2011

### hoomanya

I think I know where the first equation I wrote is coming from.... To do with finite elements ... :D