Here's the question you clicked on:
swin2013
Integral with U sub!
integral y^2 (1+y)^2 dy
My work: u = 1+y du/dy = 1 du=dy?
then integral (1+y)^2 dy integral u^2 du
ohh ok.. but what about u sub?
will u now be y^2+y^4?
well my teacher made this worksheet specifically for u sub :(
but i got 1 for u sub anyways. so i just keep y^2 ?
the square is outside the parenthesis
I see my mistake. Let me reevaluate the problem and I will come at you with a response.
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.
\[y^2(1+y)^2dy=y^2(y^2+2y+1)dy=(y^4+2y^3+y^2)dy\]
i didn;t know i had to lol oops
is there an indication to do so?
No, you are simplifying your y terms and happen to end up with \[ (y^4+2y^3+y^2)dy \]Integrate each term dy
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
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.
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.
OHH ok! I'll make sure to tell my teacher to clarify this to the class. I did not know that! hahaha
oh but in the case you're presenting it's (1+y^2) not (1+y)^2
u sub can be used when the exponent is outside right?