Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Average Value

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data
    http://img37.imageshack.us/img37/1237/63391287.jpg [Broken]


    2. Relevant equations
    [tex]$\displaystyle \Large \int udv$[/tex] = uv - [tex]$\displaystyle \Large \int vdu$[/tex]


    3. The attempt at a solution
    can i take this 3 out of the integral as well and make it 3/8 *[tex]$\displaystyle \Large \int _0^8 sqrt(64 + 24x^2)dx$[/tex]?

    then i do this:
    let u = sqrt(64 + 24x2)
    du = [tex]$\displaystyle \Large \int _0^8 sqrt(64 + 24x^2)dx$[/tex]
    let v = x
    dv = dx

    is this right so far?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 9, 2010 #2

    Mark44

    Staff: Mentor

    Yes, you can always pull a constant out of the integrand.
    No, not right. You went backwards by integrating instead of differentiating. If u = sqrt(64 + 24x2), then du = (1/2)(64 + 24x2)-(1/2)(48x). This can be cleaned up, but I left it as I did so you can see the chain rule at work.
     
    Last edited by a moderator: May 4, 2017
  4. Feb 9, 2010 #3
    okay so i have made some changes. i think it would be easier if i do this:
    [tex] $ 3/8 \displaystyle \Large \int _0^8 x*sqrt(64 + 24x^2)dx$ [/tex]
    f=x
    g'=sqrt(64 + 24x^2)
    f'=1
    g=[tex] $ \displaystyle \Large \int sqrt(64 + 24x^2)dx$ [/tex]

    to find g,
    let u =sqrt(64 + 24x2)
    [tex] $ \displaystyle \Large \int u dx$ [/tex]
    du = 1/2 (64 + 24x2)-1/2 * 48x dx

    now im lost..
    i know that i have to use the forumla
    [tex] $ \displaystyle \Large \int f(x)g'(x) dx = f(x)g(x) - \int g(x)f'(x) dx$ [/tex]

    but how do i get g? its too complicated..
     
  5. Feb 9, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You don't need any integration by parts. Substitute u=64+24x^2. du=48*x*dx. So replace x*dx in your original integrand with du/48 and the sqrt with sqrt(u). Now there's no more x's. It's just a u-substitution.
     
  6. Feb 9, 2010 #5

    Mark44

    Staff: Mentor

    You want to evaluate this integral, right?
    [tex]3/8 \int_0^8 x \sqrt{64 + 24x^2}dx[/tex]

    This is actually a pretty easy one. Use an ordinary substitution, with u = 64 + 24x2. So du = ?
     
  7. Feb 9, 2010 #6
    oh wow... this is pissing me off how easy that was and i did it the long and hard way. i guess the chapter just wants us to do it the hard way >:(

    yeah i got the answer is 333.3333

    thanks a lot for all the help! :)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook