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Integrating the square root of a linear function.

  1. Jul 4, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int\sqrt{2*x-1}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data
    Obviously this can be fixed with the antiderivative of a linear function with proper constants i.e. [tex]\int A\times f(bx + c) = \frac{A}{b} F(bx + c)[/tex], however my instructor provided me with a "clue" which I somehow cant seem to work out.

    He claims that [tex]\int\sqrt{2*x-1}[/tex] is equivilant to [tex]\int u^2[/tex] using "the right" substitution. My idea usually is:
    1) [tex]u = \sqrt{2*x - 1}[/tex]
    2) [tex]x = u^2 [/tex]
    3) [tex]dx = 2u du[/tex]

    Which would lead to [tex]\int u dx = \int u 2u du = \int 2u^2 du[/tex].
    Obviously this is NOT identical to what he suggests, so can anyone point me in the right direction? I'm quite keen to know how he see such a problem.
     
  2. jcsd
  3. Jul 4, 2009 #2
    If u = sqrt(2x-1)
    then how x = u^2?
     
  4. Jul 4, 2009 #3
    I don't think he quite knew what he was talking about; there are some problems with that which I wouldn't even try to work with or fix.
    What part of the integrand can u be that will easily let you integrate with respect to u?
     
  5. Jul 5, 2009 #4

    HallsofIvy

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    The teacher not being here to defend himself, I might suggest that cptrsn did not quite remember what the teacher had suggested. Any time you have a linear term, ax+ b, an obvious substitution is just u= ax+ b so that du= adx.
     
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