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anonymous
 5 years ago
I need to evaluate this integral..... x/(2x1)^1/2 with an upper =5 and the lower =1. Can anyone Help?
anonymous
 5 years ago
I need to evaluate this integral..... x/(2x1)^1/2 with an upper =5 and the lower =1. Can anyone Help?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let's rewrite it in a better form first: \[\int\limits_{1}^{5}x/\sqrt{2x1}dx\] where you can take : \[ f' =({2x1} )^{1/2}\] g = x so now you must find g' and f in this case :) you can take the sqrt of (2x1) up and solve using the following rule : \[\int\limits_{}^{}f'gdx = fg  \int\limits_{}^{}fg'dx\] give it a try ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0substitute and solve :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How do you take the antiderivative of (2x1)^1/2???
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