1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Definite Integration using U-Substitution

  1. Apr 3, 2010 #1
    Hi,
    I'm I on the right track here?

    1. The problem statement, all variables and given/known data
    Evaluate the definite integral, if it exists


    2. Relevant equations
    [tex]\int^{2}_{0}\stackrel{dx/}{(2x-3)^2}[/tex]


    3. The attempt at a solution
    Let u = 2x-3

    du = 2dx

    1/2du = dx

    [tex]1/2\int^{2}_{0}\stackrel{du/}{u^2}[/tex]

    going back to our original u substitution: u = 2x - 3

    = 2(0) - 3 = -3
    = 2(2) - 3 = 1

    [tex]1/2\int^{-3}_{1}\stackrel{du/}{u^2}[/tex]

    [tex]1/2\int^{-3}_{1}\stackrel{du/}{(u^3)/3}[/tex]

    (1/6) x du/(u)^3

    [(1/6) x du/(1)^3] - [(1/6) x du/(-3)^3]

    [(1/6) x du] - [(1/6) x du/(-27)]
     
  2. jcsd
  3. Apr 3, 2010 #2

    Mark44

    Staff: Mentor

    No, if I understand what you're trying to integrate. That stackrel command is not helpful. frac{}{} is what you should use for rational expressions.
    Is this the integral?
    [tex]\int_0^{2}\frac{dx}{(2x-3)^2}[/tex]

    It's a major problem that the integrand is undefined at x = 1.5.
     
  4. Apr 3, 2010 #3
    Yes this is the correct problem I have to work out.
    Ok, correct me if I am wrong but once there is a possibility of getting an undefined answer (because of the possibility of dividing by 0) then I can easily say that the difinite integral does not exist because it can't be defined if x = 1.5?

    So therefore, if x is not 1.5, then the definite integral does exist. So how do I proceed with this question then? Where did I go wrong with my attempted answer?
     
    Last edited: Apr 3, 2010
  5. Apr 3, 2010 #4

    Mark44

    Staff: Mentor

    Just say that this definite integral doesn't exist because the integrand is undefined at x = 1.5.
     
  6. Apr 3, 2010 #5
    Ok, anytime I have a quotient like that, I'll check for this, thanks.
     
  7. Apr 3, 2010 #6
    @Mark44,

    On second thought, I was looking at the following question in the book:

    1. The problem statement, all variables and given/known data
    Evaluate [tex]\int\stackrel{2}{1}\frac{dx}{(3-5x)^{2}}[/tex]

    and they gave the answer as [tex]\frac{1}{14}[/tex]

    My question is, isn't this the same situation where some number would make this also undefined? Or is the question asking something different?
     
  8. Apr 3, 2010 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    How is 1/(3-5x)^2 in any way undefined on the interval [1,2]? Just integrate it.
     
    Last edited: Apr 3, 2010
  9. Apr 4, 2010 #8
    Ok, I think I finally get it now, thanks
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook