# Basic Integration question

• Dec 11th 2012, 09:22 AM
asilvester635
Basic Integration question
Heres what i got so far....
I got
= u^2(u^3 + 2)^1/2
= (u^5 + 2u^2)^1/2 (DID I DO IT RIGHT?)
• Dec 11th 2012, 09:30 AM
ebaines
Re: Basic Integration question
Quote:

Originally Posted by asilvester635
Heres what i got so far....
I got
= u^2(u^3 + 2)^1/2
= (u^5 + 2u^2)^1/2 (DID I DO IT RIGHT?)

No - not right. when you bring u^2 inside the radical it becomes u^4, so the result is $\displaystyle u^2 \sqrt{u^3+2} = \sqrt{u^7+2u^4}$. However - this is not the way to proceed to solve this integral. Try this instead: substitute $\displaystyle w = u^3+2$ and see what you get.
• Dec 11th 2012, 09:34 AM
asilvester635
Re: Basic Integration question
i got this (u^2)(w^1/2) is that what you meant?
• Dec 11th 2012, 09:58 AM
ebaines
Re: Basic Integration question
If $\displaystyle w = u^3+2$, then $\displaystyle dw = 3u^2 du$, or $\displaystyle \frac 1 3 dw = u^2 du$. So $\displaystyle \int u^2 \sqrt{u^3+2}du = \int \frac 1 3 \sqrt w dw$

Can you take it from here?
• Dec 11th 2012, 10:04 AM
asilvester635
Re: Basic Integration question
sorry im lost... this is not how i learned it....
• Dec 11th 2012, 10:46 AM
skeeter
Re: Basic Integration question
Quote:

Originally Posted by asilvester635
sorry im lost... this is not how i learned it....

it's called integration by substitution ... what method are you speaking about?
• Dec 11th 2012, 11:14 AM
asilvester635
Re: Basic Integration question
oh snappp yessss integration by substitution.... okay lets start again from scratch..... so i pick which one to differentiate which is

w = u^3 + 2
dw = 3u^2 + 0 = 3u^2du
1/3dw = u^2du

= √w (1/3du)
= 1/3(2/3u^3/2 + c)
= 2/9u^3/2 + c
• Dec 11th 2012, 07:32 PM
voyance
Re: Basic Integration question
Well, I'll give it a shot for you:

So, we have integ((u^2)sqrt((u^3) + 2))du)
By substitution, you can make w = u^3 + 2 and if you get the derivative of that, you get dw = 3u^2 du
By substituting what you have back into the integral, you get integ(((u^2)sqrt(w))du)
Now, since you have a u^2 and your dw is 3u^2, you have multiply the inside of the integral by 3 and multiply the outside by 1/3.

(1/3)integ(((3u^2)sqrt(w)du)) -> (1/3)integ(sqrt(w)dw)
From here, you get the anti derivative -> (1/3)(2/3)(w^3/2) -> (2/9)(w^3/2) + c
c is a constant...

In other words, you got it right
• Dec 12th 2012, 05:59 AM
ebaines
Re: Basic Integration question
Quote:

Originally Posted by voyance
From here, you get the anti derivative -> (1/3)(2/3)(w^3/2) -> (2/9)(w^3/2) + c
c is a constant...

In other words, you got it right

Well actually the OP didn't get it right because what he had for an answer was 2/9u^3/2 + c. In any event the next (and last )step is to replace w with u^3 +2, to yield $\displaystyle \frac 2 9 (u^3+2)^{3/2} + C$. It's always a good idea to check the answer by taking the derivative to verify that it yields $\displaystyle u^2(u^3+2)^{1/2}$ as expected.
• Dec 12th 2012, 06:15 AM
asilvester635
Re: Basic Integration question
i knew that i just didn't care to put it was pretty obvious that it was the last step.... but thank you