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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
     
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