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Intergration by substitution

  1. Jun 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Using the substitution u² = 2x - 1, or otherwise, find the exact value of
    [tex]\int^{5}_{1} \frac{3x}{\sqrt{2x-1}}dx[/tex]

    3. The attempt at a solution

    Right lets rearrange u in terms of x (i think that's how you say it):

    [tex] x = \frac{u^{2} - 1}{2}[/tex]

    And now get an expression for dx

    [tex]u = (2x-1)^{\frac{1}{2}}[/tex]

    use the chain rule on it to give

    [tex]\frac{dx}{du} (2x-1)^{-\frac{1}{2}}[/tex]

    = [tex]dx = \frac{du}{(2x-1)^{-\frac{1}{2}}}[/tex]

    Right now substitute that in

    [tex]\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}[/tex]

    now according to the mark scheme

    [tex]\int^{5}_{1} \frac{3\frac{u^{2} - 1}{2}}{u}\frac{du}{(2x-1)^{-\frac{1}{2}}}[/tex]

    can be simplified

    [tex]\int^{5}_{1} \frac{3u^{2} - 3}{2u}\frac{du}{(2x-1)^{-\frac{1}{2}}}[/tex]

    shouldn't the bit here:

    [tex]\frac{3u^{2} - 3}{2u}[/tex]


    [tex]\frac{3u^{2} - 3}{6u}[/tex]
    Last edited: Jun 9, 2008
  2. jcsd
  3. Jun 9, 2008 #2
    Should be...

    [tex] x = \frac{u^{2} + 1}{2}[/tex]

    Shouldn't it?

    I managed to get the integral with 2u on the bottom and looking the same, but let me know how you're making out.
    Last edited: Jun 9, 2008
  4. Jun 9, 2008 #3
    woops my bad i did 3 x EVERYTHING On the fraction (including the denominator). didn;t really have my brain in gear


    cheers tho :)
  5. Jun 9, 2008 #4
    Haha no worries, least you found out where you went wrong
  6. Jun 9, 2008 #5


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    It's much simpler not to rearrange u in terms of x (or "solve for x"). Instead, from u2= 2x- 1, 2udu= 2dx so udu= dx.

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