1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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).

  4. Mar 21, 2006 #3


    User Avatar
    Science Advisor

    If u= 16- x2, then
    is NOT
    [tex]\sqrt{16- u^2}[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook