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This example is confusing me

  1. Oct 27, 2012 #1
    So, um, I am getting confused on integration problems where you have to do something with "a constant factor of n". Like, this example...

    [itex]\int\sqrt{1 + e^{4x^{3}}}e^{4x^{3}}x^{2}dx[/itex]

    Then the example says to match it to the formula [itex]\int u^{n}du[/itex]

    Okay... so it does that, but then... something I don't quite understand happens. It says that "du = [itex]e^{4x^{3}}(12x^{2})[/itex]" WAIT? WHERE DID THE 12 COME FROM? Then it says that "our integrand contains all of du except for the constant factor of 12" Then it does this...

    [itex]\frac{1}{12}\int(1 + e^{4x^{3}})^{1/2} e^{4x^{3}}(12x^{2})dx[/itex]

    Then it integrates like normal... but... WHERE DID THE 12 COME FROM? I don't know why, it just isn't obvious where this "constant factor of 12" came from?
     
  2. jcsd
  3. Oct 27, 2012 #2

    arildno

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    What is the derivative of 4x^3?
     
  4. Oct 27, 2012 #3
    So wait, I have to find the derivative of the exponent then add it to the terms in du? Like, there was an x^2 so I add 12x^2 to that and then take the reciprocal of 12 and move it outside the integral?
     
  5. Oct 27, 2012 #4
    Seriously, somebody, I feel dumb because I am not getting this like am I supposed to find the derivative of u then add/multiply by the other thing?? UHHHHH...
     
  6. Oct 27, 2012 #5

    AlephZero

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    If you compared the integrals, presumably the example then wants to substitute ##u = \sqrt{1 + e^{4x^3}}##

    So what does ##du/dx## equal?
     
  7. Oct 27, 2012 #6

    arildno

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    No.
    What is the derivative of 1+e^(4x^3)??
     
  8. Oct 27, 2012 #7
    Um I think this is the derivative..

    [itex]\frac{du}{dx} = 12x^{2}e^{4x^{3}}[/itex]

    OH OH OH! I'M SUPPOSED TO TAKE THE DERIVATIVE OF U THEN ADD IT TO THE OTHER PART? Right? And when do I have to move a constant factor outside the integral?
     
  9. Oct 27, 2012 #8

    arildno

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    Now, I hope you see that 1=1/12*12.

    Thus, we recognize that the expression in your original integral, "e^(4x^3)x^2dx"=1/12du"
    Agreed?
    Furthermore, the square root thing is now to be written as sqrt(u). Agreed?
     
  10. Oct 27, 2012 #9
    Yes, I know 1/12(12)=1 and sure, sqrt(u). Oh wait.... du means... derivative of u... okay but what happens to the things that aren't part of u?
     
  11. Oct 27, 2012 #10

    arildno

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    So, then you have no further problems? :smile:
     
  12. Oct 27, 2012 #11
    Well actually one last thing, what happens to stuff that isnt a part of u? Like the x^2 outside of sqrt(u)?
     
  13. Oct 27, 2012 #12

    arildno

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    Read your own post 7 again.
     
  14. Oct 27, 2012 #13
    Oh wait I don't add it... I replace it? OR DO I ADD IT? Lol I'm confusing myself now.
     
  15. Oct 27, 2012 #14
    Okay I figured it out and worked a practice problem, the answers are at the back of the book but... I got:

    [itex]-\frac{(cos2x + 1)^{3/2}}{3/2} + C [/itex]

    However, the answer to the practices in the back of the book says it is over 3 not 3/2? What happened here, I'm sure it's algebraic but still...

    Wait... is it because du = -2sin2x? Then the -2 would be in front and the 2's cancel? Is that right?
     
  16. Oct 28, 2012 #15
    How do you expect us to verify your answer without first knowing the initial question?

    If your answer is true though, then the question must be this:

    [tex]\int 2\sin(2x)\sqrt{\cos(2x)+1}\,dx[/tex]

    which is easily integrable (what is the derivative of [itex]\cos(2x)+1[/itex]?)
     
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