Question

Integral with U sub!

Answer 1

integral y^2 (1+y)^2 dy

Answer 2

My work:

u = 1+y
du/dy = 1
du=dy?

Answer 3

then integral (1+y)^2 dy

integral u^2 du

Answer 4

(1_y)^3/3 +c

Answer 5

*(1+y)^3

Answer 6

do i keep the y^2?

Answer 7

ohh ok.. but what about u sub?

Answer 8

will u now be y^2+y^4?

Answer 9

well my teacher made this worksheet specifically for u sub :(

Answer 10

but i got 1 for u sub anyways. so i just keep y^2 ?

Answer 11

no

Answer 12

it's y^2 (1+y)^2

Answer 13

the square is outside the parenthesis

Answer 14

I see my mistake. Let me reevaluate the problem and I will come at you with a response.

Answer 15

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.

Answer 16

\[y^2(1+y)^2dy=y^2(y^2+2y+1)dy=(y^4+2y^3+y^2)dy\]

Answer 17

i didn;t know i had to lol oops

Answer 18

is there an indication to do so?

Answer 19

No, you are simplifying your y terms and happen to end up with \[ (y^4+2y^3+y^2)dy \]Integrate each term dy

Answer 20

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

Answer 21

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.

Answer 22

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.

Answer 23

OHH ok! I'll make sure to tell my teacher to clarify this to the class. I did not know that! hahaha

Answer 24

oh but in the case you're presenting it's (1+y^2)
not (1+y)^2

Answer 25

u sub can be used when the exponent is outside right?