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swin2013Best ResponseYou've already chosen the best response.0
integral y^2 (1+y)^2 dy
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
My work: u = 1+y du/dy = 1 du=dy?
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
then integral (1+y)^2 dy integral u^2 du
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
ohh ok.. but what about u sub?
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
will u now be y^2+y^4?
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
well my teacher made this worksheet specifically for u sub :(
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
but i got 1 for u sub anyways. so i just keep y^2 ?
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
the square is outside the parenthesis
 one year ago

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

zepdrixBest ResponseYou've already chosen the best response.1
Yah 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.
 one year ago

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

swin2013Best ResponseYou've already chosen the best response.0
i didn;t know i had to lol oops
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
is there an indication to do so?
 one year ago

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

swin2013Best ResponseYou've already chosen the best response.0
oh 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
 one year ago

stampBest ResponseYou've already chosen the best response.1
Had 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.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
To 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.
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
OHH ok! I'll make sure to tell my teacher to clarify this to the class. I did not know that! hahaha
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
oh but in the case you're presenting it's (1+y^2) not (1+y)^2
 one year ago

swin2013Best ResponseYou've already chosen the best response.0
u sub can be used when the exponent is outside right?
 one year ago
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