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Integral Question

  1. May 30, 2005 #1
    I seem to be having trouble on this one integral.Any help would be much appreciated.

    [tex] \int 3x \sqrt{5-2x} dx [/tex]

    I'm assuming the substitution rule applies to this so I have u=5-2x.
    Then du=-2dx

    And now I'm stuck. How can I get 3xdx to equal -2dx? Please and thanks.
     
  2. jcsd
  3. May 30, 2005 #2

    shmoe

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    Your substitution is good, try writing x in terms of u to deal with the "x" in 3xdx
     
  4. May 30, 2005 #3

    dextercioby

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    Part integration,maybe...?

    Daniel.
     
  5. May 30, 2005 #4
    U-substitution won't work in the normal way of " u =, du = " ... Like Daniel said, you'll need to do integration by parts. Are you familiar with this method?

    [tex]\int udv = uv - \int vdu[/tex]

    Jameson
     
  6. May 30, 2005 #5
    yep, do this.

    it'll be MUCH easier than integration by parts, which you may not even know yet.
     
  7. May 30, 2005 #6

    shmoe

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    Carrying out his substitution works just fine. By parts also work of course, but it's not necessary.
     
  8. May 30, 2005 #7
    Fine, we'll do things your way.

    [tex]\int 3x\sqrt{5-2x}dx[/tex]

    [tex]u = 5 - 2x[/tex]

    [tex]du = -2dx[/tex]

    [tex]-\frac{du}{2} = dx[/tex]

    That takes care of the second term, now for the first.

    [tex]u = 5 - 2x[/tex]

    [tex]2x = 5 - u[/tex]

    [tex]x = \frac{5-u}{2}[/tex]

    So now we have...

    [tex]\int 3x\sqrt{5-2x}dx = -\frac{3}{4} \int (5-u) \sqrt{u} du[/tex]

    I don't call that easier than integration by parts...
     
    Last edited: May 30, 2005
  9. May 30, 2005 #8

    shmoe

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    The relative difficulty is a subjective thing. I object to telling the OP that he'd need integration by parts when his substitution idea is good.

    By the way, you lost a pair of brackets in your last line and also a minus sign went missing when you solved for x.
     
  10. May 30, 2005 #9
    Ah, I fixed them both. Now it's correct :)
     
  11. May 30, 2005 #10

    shmoe

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    Brackets around 5-u in the integral.
     
  12. May 30, 2005 #11
    Yeah, fixed 'em. Thanks.

    But my point is, even after all of that subsitution, the integral still can't be evaluated in one step. More substitution is needed.
     
  13. May 30, 2005 #12

    shmoe

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    No more substitution is needed to integrate that. Distribute over those brackets and go to town.
     
  14. May 30, 2005 #13
    Ah, crap. Good point. Thought I had you. Alas, substitution will work. Can't win all of them.

    Jameson
     
  15. May 30, 2005 #14

    dextercioby

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    I didn't claim it was necessary for this example,just gave a suggestion,an advice.He could take or not. :wink:

    Daniel.
     
  16. May 30, 2005 #15

    shmoe

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    Of course! The necessary part wasn't aimed at you. Both methods are worth a stab, and proving they give the same answer is a nice bit of algebra practice (they look a bit different).
     
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