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Well defined integrals

  1. Mar 20, 2006 #1
    Here iv to show that integrals are well defined and find their values:

    [tex] a) I = \left\int_{0}{4} \frac{x - 1}{\sqrt{16 - x^2}}dx [/tex]

    this one i can show its well defined ok (the domain will be positive as x will be between 0 and 4) but can't find the value. So far my work goes:

    [tex] I = \int_{0}^{4} ( \frac{x}{\sqrt{16 - x^2}} - \frac{1}{\sqrt{16 - x^2}} [/tex]

    I know the first part will be arcsin x/4 but i can't do the first part. I began doing substitution but wasn't sure:

    [tex] u = 16 - x^2 \left \frac{du}{dx} = -2x \left So \left xdx = \frac{-du}{2} [/tex] so the intergral is
    [tex] \frac{-1}{2} \int_{u = 16}^{u = 0} \frac{du}{\sqrt{16 - u^2}} [/tex]
    I dont feel this is right. especially the limits i got, theyr impossible right? To get them i said:
    [tex]u = 16 - x^2 \left So \left u = 16 - (0)^2 = 16 \left and \ u = 16 - 4^2 = 0 [/tex]

    b) [tex]\int_{0}^{1} (lnx + 2)[/tex]
    This one i can do but cant say why its well defined. Is it because the value of lnx will be 0 but the addition of 2 will keep keep it non-zero?
     
  2. jcsd
  3. Mar 20, 2006 #2

    Try this substitution: [tex] x = 4sin\theta[/tex] for both integrals of part (a).

    -LD
     
  4. Mar 21, 2006 #3

    HallsofIvy

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    If u= 16- x2, then
    [tex]\sqrt{16-x^2}[/tex]
    is NOT
    [tex]\sqrt{16- u^2}[/tex]
    !!!
     
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