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

In summary, the conversation discusses a mathematical equation involving integrals and the use of the divergence theorem. The statement in question is shown to be incorrect and a counterexample is provided. The context is in solid deformation and finite element analysis, with the problem being in 1D. There is a discussion on whether the equation depends on context and a clarification is sought on the use of the divergence theorem. It is concluded that the equation is not a general mathematical relationship and its truth or falsehood depends on the specific context and meaning of the symbols involved.
  • #1
hoomanya
90
0
Hi,

I was wondering if the following statement is correct:

[itex]\int[/itex]([itex]\frac{du}{dx}[/itex])[itex]^{2}[/itex]dV=[itex]\int[/itex][itex]\frac{du}{dx}[/itex]dV[itex]\int[/itex][itex]\frac{du}{dx}[/itex]dV

Please help!

Thanks!
 
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  • #2
hoomanya said:
Hi,

I was wondering if the following statement is correct:

[itex]\int[/itex]([itex]\frac{du}{dx}[/itex])[itex]^{2}[/itex]dV=[itex]\int[/itex][itex]\frac{du}{dx}[/itex]dV[itex]\int[/itex][itex]\frac{du}{dx}[/itex]dV

Please help!

Thanks!
No because the left hand side has just [itex]dV[/itex], whereas the right hand side has [itex]dV^2[/itex] because you have two of them multiplying together.
 
  • #3
hoomanya said:
Hi,

I was wondering if the following statement is correct:

[itex]\int[/itex]([itex]\frac{du}{dx}[/itex])[itex]^{2}[/itex]dV=[itex]\int[/itex][itex]\frac{du}{dx}[/itex]dV[itex]\int[/itex][itex]\frac{du}{dx}[/itex]dV

Please help!

Thanks!
Counterexample: let u(x) = x, v = x.
 
  • #4
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:

[itex]\int[/itex]([itex]\frac{du}{dx}[/itex])[itex]^{2}[/itex]dV=[itex]\frac{1}{V}[/itex][itex]\int[/itex]u.n dS[itex][/itex][itex]\int[/itex]u.n dS

Cheers!
 
  • #5
hoomanya said:
Thank you both. pmsrw3 I am not bothered with solving the integral.
?

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.

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

[itex]\int[/itex]([itex]\frac{du}{dx}[/itex])[itex]^{2}[/itex]dV=[itex]\frac{1}{V}[/itex][itex]\int[/itex]u.n dS[itex][/itex][itex]\int[/itex]u.n dS

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

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?

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
hoomanya said:
Sorry, I don't get the counter example.
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.



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.
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

[tex]\int_a^b \frac{du}{dx}dx = u(b) - u(a)[/tex]

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. [itex]\int(\frac{du}{dx})^2 dV[/itex] is no more special than [itex]\int u dV[/itex].

By the way, does it depend on the context or is it a mathematical relationship (if it's correct)?
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 [itex]x^2 \geq 0[/itex] is true or false unless I tell you what kind of values are allowed for x.
 
  • #8
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
hoomanya said:
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?
Maybe you should explain where this question comes from and what you're trying to do.
 
  • #10
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
I think I know where the first equation I wrote is coming from... To do with finite elements ... :D
 

Related to Integration of a term that is squared, cubed, etc.

1. How do you integrate a term that is squared?

To integrate a term that is squared, you can use the power rule for integration. This rule states that when integrating a term with a variable raised to a power, you can add 1 to the power and then divide by the new power. For example, if you have the term x^2, you would add 1 to the power, making it x^3, and then divide by the new power, giving you (x^3)/3 as the integrated term.

2. Can you integrate a term that is cubed?

Yes, you can integrate a term that is cubed using the same power rule as for a squared term. You would add 1 to the power, making it x^4, and then divide by the new power, giving you (x^4)/4 as the integrated term.

3. Is there a different rule for integrating a term that is raised to a higher power?

No, the power rule can be used for any term that is raised to a power, regardless of how high the power is. You would simply add 1 to the power and then divide by the new power to integrate the term.

4. Can you integrate a term that is squared and has a coefficient?

Yes, when integrating a term with a coefficient, you can simply bring the coefficient outside of the integral and then integrate the remaining term using the power rule. For example, if you have the term 2x^2, you would bring the 2 outside of the integral, giving you 2∫x^2dx, and then integrate x^2 using the power rule to get 2(x^3)/3 as the final integrated term.

5. How do you handle integration of a term that is raised to a negative power?

When integrating a term with a negative power, you can use the power rule in reverse. Instead of adding 1 to the power, you would subtract 1, and instead of dividing by the new power, you would multiply by the reciprocal of the new power. For example, if you have the term x^-2, you would subtract 1 from the power, making it x^-3, and then multiply by the reciprocal of the new power, giving you -1/(x^3) as the integrated term.

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