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geerky42
Group Title
\[\Large \int\limits_{1}^{2}(x1)\sqrt{2x}\space dx\]
 one year ago
 one year ago
geerky42 Group Title
\[\Large \int\limits_{1}^{2}(x1)\sqrt{2x}\space dx\]
 one year ago
 one year ago

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dumbcow Group TitleBest ResponseYou've already chosen the best response.1
im thinking integration by parts ... have you tried that?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
wELL, i'M NEW TO INTEGRAL, i KNOW SOME TECHNIQUES BUT NOT COMPETELY CONFIDENT WITH IT. i'M CURRENTLY WORK WITH usUBSTITUTION. Sorry about caps.
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
I don't see how I can use usubstitution for this problem...
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.1
oh ok ... yeah substitution may work but i don't see it right away integration by parts is a technique when you have a product of 2 distinct functions \[\int\limits_{?}^{?} f(x) *g(x) dx\]
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.1
the equation is given as: \[\int\limits_{?}^{?} u*dv = uv \int\limits_{?}^{?}v*du\] where u is f(x) and dv is g(x)
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Ok, I'm giving it a try.
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
I don't think it will work. While I'm attempting to solve it, it's getting uglier.
 one year ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.1
no it will work but yes it can get uglier in the process ... anyway it takes some getting used to i found how usubstitution will work \[u = 2x\] \[du = dx\] \[\rightarrow \int\limits_{?}^{?}(1u) \sqrt{u} du = \int\limits_{?}^{?} \sqrt{u}  u \sqrt{u}\]
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.1
Hmm clever! Thanks.
 one year ago
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