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Math Help - Basic Integration question

  1. #1
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    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?)
    Attached Thumbnails Attached Thumbnails Basic Integration question-screen-shot-2012-12-11-8.19.09-am.png  
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Basic Integration question

    Quote Originally Posted by asilvester635 View Post
    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 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  w = u^3+2 and see what you get.
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    Re: Basic Integration question

    i got this (u^2)(w^1/2) is that what you meant?
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    MHF Contributor ebaines's Avatar
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    Re: Basic Integration question

    If w = u^3+2, then dw = 3u^2 du, or \frac 1 3 dw = u^2 du. So  \int u^2 \sqrt{u^3+2}du = \int \frac 1 3 \sqrt w dw

    Can you take it from here?
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    Re: Basic Integration question

    sorry im lost... this is not how i learned it....
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    Re: Basic Integration question

    Quote Originally Posted by asilvester635 View Post
    sorry im lost... this is not how i learned it....
    it's called integration by substitution ... what method are you speaking about?
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    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
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  8. #8
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    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
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  9. #9
    MHF Contributor ebaines's Avatar
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    Re: Basic Integration question

    Quote Originally Posted by voyance View Post
    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  \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 u^2(u^3+2)^{1/2} as expected.
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  10. #10
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    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
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