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Problem with Integration

  1. Mar 2, 2005 #1
    I'm trying to integrate this awful expression (by dx):

    {[1-(x-3)^2]/9 - 4[(1-(x-3)^2)/9)^1/2] + 4

    Help! How can I integrate this? It's so complicated. For the first part, would it be [x- (x-3)^3]/27x???? For the last one I know it is 4x, so that's not a problem, but its mainly the second term with the square root that's the worst. Thank you!
  2. jcsd
  3. Mar 2, 2005 #2
    By the way, if you get a nonreal answer for this like I did on the calculator, this problem is part of my bigger problem of finding the volume of the ellipse:
    y = ((1-(x-3)^2)/9)-2, which is supposed to have its center at 3, cross the x-axis at 0 and 6, and have a maximum y value at 2, rotated 360 degrees about the x- axis. Maybe I got up to that point wrong, but what I did was find the volume of a section in the figure, which I said was V= pi(y^2)dx and in order to get y^2, I squared the equation for the ellipse that I gave at the top. That's where my above question comes in. Thanks.
  4. Mar 2, 2005 #3
    Do a "u" substitution to see if that helps you look at it better.

    [tex] u = x-3[/tex]

    [tex]\int\frac{1-u^2}{9}du-\int4\sqrt{\frac{1-u^2}{9}}du + \int4dx [/tex]

    EDIT: Also, remember that [tex] \sqrt{\frac{a}{n}} = \frac{\sqrt{a}}{\sqrt{n}}[/tex]
    Last edited: Mar 2, 2005
  5. Mar 2, 2005 #4
    ok, so I ended up figuring out the first and last terms without substitution, but I'm using it for the middle one. I sad let u= x-3 and then du/dx= 1 so du=dx. But in order to do that, I think I still need to have the derivitive of 1-u^2 in order to apply the rule, which would be 2u, but how can that be possible?
  6. Mar 2, 2005 #5
    For the first term, think of the trig inverse functions.
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