anonymous
  • anonymous
Integral with U sub!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
integral y^2 (1+y)^2 dy
anonymous
  • anonymous
My work: u = 1+y du/dy = 1 du=dy?
anonymous
  • anonymous
then integral (1+y)^2 dy integral u^2 du

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.