|Oct30-03, 10:39 PM||#1|
Can someone explain the notation of the parts formula pls.,.. it's very very confusing.
int [u dv] = uv - int [v du]
very confuzing, ... made me think it was the product formula lol. And I still don't understand what does dv and dx and all these d stuff standfor in integrals.
they have those dv = 1dx.. i can understand that,... dv/dx = 1, they just moved dx over , but then they made 1/dx = dv... i got confuzed...
|Oct30-03, 10:50 PM||#2|
i think you should look at that equation as so...
∫f(x)g'(x)dx = f(x)g(x)-∫g(x)f'(x)dx
for example a simple integral:
Find ∫x sinx dx.
∫x sinxdx= f(x)g(x)-∫g(x)f'(x)dx
= -x cosx + ∫cosx dx
= -x cosx + sinx + C (where C is a Constant m8)
Hope it helped. i can do some more examples if you like. [;)]
|Oct31-03, 07:44 AM||#3|
Actually, the "integration by parts" formula should make you think of "product rule" (hopefully without laughing too hard).
Integration by parts is the opposite of using the product rule to differentiate. The product rule says that d(uv)/dx= u dv/dx+ v du/dx
Convert that into "differential form" (if you really have trouble understanding "dv and dx and all these d stuff standfor in integrals" you might want to go back and review the connection between derivatives and differentials): d(uv)= u dv+ v du. Integrating both sides of that gives "integration by parts": uv= int(u dv)+ int(v du) so
int(u dv)= uv- int(vdu).
Typically, an integration by parts problem involves a product of functions to be integrated. You need to select one of them (that you can integrate easily) to be "dv" (with the "dx" from the integral included- integrating dv give you v) and the other (which hopefully you can differentiate) to be "u"- differentiating u gives you du. Notice that integration by parts does not immediately give you the integral- what happens is that it gives you a new integral to do: int(v du). If you can do that, then you can finish the problem so: you need to choose "dv" that you can integrate, "u" that you can differentiate, and, hopefully, so that you can integrate "vdu".
Remember, most elementary functions cannot be integrated in terms of elementary functions!
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