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swin2013
 one year ago
Best ResponseYou've already chosen the best response.0integral y^2 (1+y)^2 dy

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0My work: u = 1+y du/dy = 1 du=dy?

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0then integral (1+y)^2 dy integral u^2 du

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0ohh ok.. but what about u sub?

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0will u now be y^2+y^4?

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0well my teacher made this worksheet specifically for u sub :(

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0but i got 1 for u sub anyways. so i just keep y^2 ?

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0the square is outside the parenthesis

stamp
 one year ago
Best ResponseYou've already chosen the best response.1I see my mistake. Let me reevaluate the problem and I will come at you with a response.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Yah you wouldn't use a `U sub` for this one I'm afraid swin. You would just expand out the brackets and integrate each term individually.

stamp
 one year ago
Best ResponseYou've already chosen the best response.1\[y^2(1+y)^2dy=y^2(y^2+2y+1)dy=(y^4+2y^3+y^2)dy\]

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0i didn;t know i had to lol oops

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0is there an indication to do so?

stamp
 one year ago
Best ResponseYou've already chosen the best response.1No, you are simplifying your y terms and happen to end up with \[ (y^4+2y^3+y^2)dy \]Integrate each term dy

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0oh ok. that makes sense. usually i don't have to expand a problem or distribute so i didn't know if using u sub for 1+y is correct

stamp
 one year ago
Best ResponseYou've already chosen the best response.1Had you done some usub, you would had \[(y^2u)du\] and would not really have gotten anywhere as far as the integral goes. Be sure to have + C to your integral term since the integral is undefined.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1To apply a `u sub`, you want to look for a suitable `u` and `u'` somewhere in your problem. \(2y(1+y^2)\) See how if \(u=1+y^2\) then \(u'=2y\) ? In the problem you were given, letting your inner function be `u` did NOT result in the outer function being u'. So that is usually your indicator.

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0OHH ok! I'll make sure to tell my teacher to clarify this to the class. I did not know that! hahaha

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0oh but in the case you're presenting it's (1+y^2) not (1+y)^2

swin2013
 one year ago
Best ResponseYou've already chosen the best response.0u sub can be used when the exponent is outside right?
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