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Intergration by substitution

  1. Feb 24, 2006 #1

    Hootenanny

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    The question is to find the following intergal:
    [tex]\int x\cdot u^{\frac{1}{2} [/tex] where [itex]u = 2x -1[/itex].
    [tex]= \int x\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du [/tex]
    [tex] u = 2x -1 \Rightarrow x = \frac{u+1}{2} [/tex]
    [tex]= \int \frac{1}{2}(u+1)\cdot u^{\frac{1}{2}} \;\; \frac{1}{2} du \;\; = \int \frac{1}{4}\left( u^{\frac{3}{2}} + u^{\frac{1}{2}}\right) \;\; du [/tex]
    [tex] = \frac{2}{20} u^{\frac{5}{2}} + \frac{2}{12} u^{\frac{3}{2}} = \frac{1}{10} u^{\frac{5}{2}} + \frac{1}{6} u^{\frac{3}{2}} [/tex]

    But this isnt going to give me the correct answer. Can anybody see where I've gone wrong? Thank's
     
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  3. Feb 24, 2006 #2

    HallsofIvy

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    If u= 2x- 1 then 2x= u+1 so x= (1/2)(u+1). Substitute THAT for x and the integral becomes
    [tex]\frac{1}{2}\int(u+1)u^{\frac{1}{2}}du= \frac{1}{2}\int (u^{\frac{3}{2}}+ u^{\frac{1}{2}})du[/tex]
    That should be easy to integrate.
     
  4. Feb 24, 2006 #3

    Hootenanny

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    [tex]\frac{1}{2}\int (u^{\frac{3}{2}}+ u^{\frac{1}{2}})du = \frac{1}{2} \left[ \frac{2}{5} u ^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}} \right] [/tex]

    Is that correct?
     
  5. Feb 24, 2006 #4

    Hootenanny

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    Or would i have to make a further substitution for [itex](u^{\frac{3}{2}}+ u^{\frac{1}{2}}) [/itex] ?
     
  6. Feb 24, 2006 #5

    TD

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    That's correct, but don't forget the constant of integration :smile:
     
  7. Feb 24, 2006 #6

    Hootenanny

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    Thank's TD +c
     
  8. Feb 24, 2006 #7

    TD

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    You're welcome.

    To get back to your last question, there is no need for an extra substitution since the integral is lineair, which means that:

    [tex]\int {\alpha f\left( x \right) + \beta g\left( x \right)dx} = \alpha \int {f\left( x \right)dx} + \beta \int {g\left( x \right)dx}[/tex]

    So in your case, that gives:

    [tex]\int {u^{\frac{3}{2}} + u^{\frac{1}{2}} du} = \int {u^{\frac{3}{2}} du} + \int {u^{\frac{1}{2}} du} [/tex]

    Then you can apply the standard exponent-rule, as you did.
     
  9. Feb 24, 2006 #8

    0rthodontist

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    What? He had it completely right the first time. 1/4 is the correct thing to multiply by because dx = 1/2 du. I'm assuming there was a dx in the original integral because otherwise he wouldn't have included the 1/2 du in his second step.
     
    Last edited: Feb 24, 2006
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