1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrating by parts?

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]\text {Evaluate } \int^m_1 x^{3}ln{x}\,dx[/tex]

    2. Relevant equations



    3. The attempt at a solution

    Integrating by parts, but not sure which term to substitute out...it's not turning out clean...argh I've done every other problem except for this one, can someone just provide the first step? Much appreciated.
     
  2. jcsd
  3. Aug 18, 2008 #2

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    I don't know what you mean by "I'm not sure which term to substitute out?" What is the method for integration by parts: you should have a formula, no? What specifically are you confused with, with respect to this formula?
     
  4. Aug 18, 2008 #3

    Defennder

    User Avatar
    Homework Helper

    You should try substituting for v and du, the terms for which vdu is easy to integrate. It's harder to integrate ln than it is to differentiate it, so that provides an obvious choice.
     
  5. Aug 18, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since you have x3ln x dx, and want u(x) dv, there are really just two choices: either u(x)= x3 and dv= ln(x)dx or u(x)= ln(x) and dv= x3dv. Try both and see which gives you a decent integral!
     
  6. Aug 18, 2008 #5
    Thanks, I integrated it successfully, but since this is a definite integral, at the end I'm not quite sure what to do with the boundaries, namely with the m term. Is it okay to leave the answer as an expression of both m and x?

    It comes out to this:

    [tex]\frac {1}{4} (x^{4} ln{x} - \int^m_1 x^{3} \,dx) [/tex]
     
  7. Aug 18, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi avr10! :smile:
    Evaluate the first part between 1 and m:

    [tex]\left[\frac {1}{4} x^{4} ln{x} \right]^m_1 [/tex] :smile:
     
  8. Aug 18, 2008 #7
    Hm? Why is that?
     
  9. Aug 18, 2008 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Because that's what the little numbers at top and bottom of the integral sign mean!
    [tex]\int_a^b f(x)dx= F(b)- F(a)[/tex]
    where F is an anti-derivative of f. The result of a definite integral is a number, not a function of x.
     
  10. Aug 18, 2008 #9
    Integrating by parts using the standard formula gives me
    [tex]\left[ln(x)*\frac {1}{3} x^{3} -\frac {1}{9}x^{3}\right]^m_1 [/tex]

    Following up by the formula presented by HallsofIvy, I get:

    [tex]ln(m)-\frac{3}{9}*m^{3}-\frac{1}{9}*m^{3}+\frac{1}{9}[/tex]

    Which is pulled together to:

    [tex]ln(m)+\frac {1}{9}-\frac {4}{9}m^3[/tex]

    Voila?

    Edit: HallsofIvy, in this case the answer is not a number, but yes, it is a constant.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integrating by parts?
  1. Integration by parts (Replies: 6)

  2. Integration by parts (Replies: 8)

  3. Integration by Parts (Replies: 8)

Loading...