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Definite integral problem

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  1. Mar 24, 2015 #1
    1. The problem statement, all variables and given/known data
    integrate from 1 to 2 x(x^2-3)^(1/2) with respect to x.

    2. Relevant equations


    3. The attempt at a solution
    i attempted using numerical approximations but at x=1, the function is not defined so is there a way to combine improper integrals with this?

    Aceix.
     
  2. jcsd
  3. Mar 24, 2015 #2
    I read your problem as $$\int_{1}^{2} x\sqrt{x^{2} - 3}\,dx$$ & used the substitution ##u = x^{2} - 3## then ##\frac{1}{2}du = x\,dx## so it's no improper integral. I'm not sure why you say the function is not defined at x=1.
     
  4. Mar 24, 2015 #3

    Dick

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    It's not defined because at x=1 because 1-3 is negative and square root of a negative is not defined. You would have to properly define the square root as a complex number to be able to integrate.
     
  5. Mar 24, 2015 #4
    So how do I define the square root as a complex number?
     
  6. Mar 24, 2015 #5
    argh sorry about that but I did notice that when I tried ##\int^{2}_{1} \frac{x\,dx}{\sqrt{x^2 - 3}}## just in case I misread the original post
     
  7. Mar 24, 2015 #6

    Dick

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    If ##x^2-3## is negative then the complex square roots are either ##i \sqrt{3-x^2}## or ##-i \sqrt{3-x^2}##. You have to pick which one you want. This is called 'choosing a branch'. Why are you doing this problem?
     
  8. Mar 24, 2015 #7
    Saw it in a book(preparing for an exam).
     
  9. Mar 24, 2015 #8

    Dick

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    The problem has two possible answers since you need to make a branch choice. If you aren't really doing complex numbers, then possibly i) they just expect you to say it's not defined or ii) it's a typo.
     
  10. Mar 24, 2015 #9
    thanks!
     
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