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

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

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

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

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

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

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

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

stamp
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0i didn;t know i had to lol oops

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

stamp
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0u sub can be used when the exponent is outside right?
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