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AmTran_Bus
 one year ago
Is the LIATE rule for integration by parts correct?
AmTran_Bus
 one year ago
Is the LIATE rule for integration by parts correct?

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.2Yes, that is the order

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.2of choosing a function to differentiate

AmTran_Bus
 one year ago
Best ResponseYou've already chosen the best response.0Does it always go? I'm a little hard to believe it always works.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.2Well, it is a suggestion for user that don't know intuitively  to tell them which function should they diffeentiate. It is not a "rule" literally, as the "chain rule"  not to that extent a "rule".

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.2just an advice...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.2I mean if you want: \({\displaystyle \int} wx~dx=(\int w)\cdot(x){\displaystyle \int} (\int w)\cdot(x)'dx\) \({\displaystyle \int} wx~dx=( wx)\cdot(x){\displaystyle \int} (wx)\cdot(1)dx~+C\) by parts  comes from product rule, thus should have +C (because it comes from INTEGRATING both sides of the product of the deriavtive) \({\displaystyle \int} wx~dx=( wx)\cdot(x){\displaystyle \int} wx~dx~+C\) \(2{\displaystyle \int} wx~dx=( wx)\cdot(x)~+C\) \(\displaystyle \int wx~dx=\frac{w}{2}x^2~+\frac{C}{2}\) \(\displaystyle \int wx~dx=\frac{w}{2}x^2~+C\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.2i did by parts on a wx (with respect to x)
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