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geerky42
 3 years ago
\[\Large \int\limits_{1}^{2}(x1)\sqrt{2x}\space dx\]
geerky42
 3 years ago
\[\Large \int\limits_{1}^{2}(x1)\sqrt{2x}\space dx\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0im thinking integration by parts ... have you tried that?

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

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1I don't see how I can use usubstitution for this problem...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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 \[\int\limits_{?}^{?} f(x) *g(x) dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the equation is given as: \[\int\limits_{?}^{?} u*dv = uv \int\limits_{?}^{?}v*du\] where u is f(x) and dv is g(x)

geerky42
 3 years ago
Best ResponseYou've already chosen the best response.1Ok, I'm giving it a try.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no 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 \[u = 2x\] \[du = dx\] \[\rightarrow \int\limits_{?}^{?}(1u) \sqrt{u} du = \int\limits_{?}^{?} \sqrt{u}  u \sqrt{u}\]
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