Integration by Parts in Calculus: Understanding the Process and Its Applications

[bugatti79]

FOlks,

I am self studying a book and I have a question on

1)what the author means by the following comment "Integrating the second term in the last step to transfer differentiation from v to u"

2) Why does he perform integration by parts? I understand how but why? I can see that the last term has no derivatives of v in it.

See attached jpg.

THanks

 

[HallsofIvy]

"The second term in the last equation" is (\partial F/\partial u')v'dx and, since that is a product of a function and a derivative, a rather obvious thing to do is to integrate by parts (\int U dV= UV- \int V dU) with U= v' and dV= (\partial F/\partial u')dx.

 

[bugatti79]

I can see what he has done but why? Why integrate by parts?

Thanks

 

[HallsofIvy]

Because integration by parts, being the "inverse" of the product rule for derivatives, is a natural way to integrate a product. Especially here where the product is of a function and a derivative: \int u dv.

 

[andrien]

integration by part has been done to arrive at something similar to euler lagrange eqn.it just occurs many times in physics in different form may be covariant form

 

[bugatti79]

Because integration by parts, being the "inverse" of the product rule for derivatives, is a natural way to integrate a product. Especially here where the product is of a function and a derivative: \int u dv.

integration by part has been done to arrive at something similar to euler lagrange eqn.it just occurs many times in physics in different form may be covariant form

Ok, was just wondering what he meant by transfer differentiation from v to u?

Anyhow, thanks folks.

 

[bugatti79]

"The second term in the last equation" is (\partial F/\partial u')v'dx and, since that is a product of a function and a derivative, a rather obvious thing to do is to integrate by parts (\int U dV= UV- \int V dU) with U= v' and dV= (\partial F/\partial u')dx.

Actually, shouldn't it be the other way around U= (\partial F/\partial u')dx and dV=v'

Then V=v since \int dV=\int v'dx and \displaystyle \frac{dU}{dx}= d(\frac{\partial F}{\partial u'})=\frac{d}{dx} \frac{ \partial F}{\partial u'}...?