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Evaluate integral by area

  1. Dec 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Evaluate the integral by intrepreting it in terms of areas

    from -3 to 0

    (1+ sqrt(9-x2)dx


    2. Relevant equations

    integral = F(0)-F(-3)

    3. The attempt at a solution

    first find F

    F = x - [(9-x2)3/2]/3

    I solved using integral = F(0)-F(-3)

    and I got the incorrect answer I think it is because I am finding the definite integral and not the area. If this was my problem how would I find the area? Graphing?
     
    Last edited: Dec 5, 2009
  2. jcsd
  3. Dec 5, 2009 #2

    Mark44

    Staff: Mentor

    Yes - graph it. The function defines a fairly simple geometric object.

    BTW, your work finding an antiderivative, besides not being the direction you're supposed to go, is incorrect. If you take the derivative of the function you found, you don't get 1 + sqrt(9 -x^2)
     
  4. Dec 5, 2009 #3
    ok i changed it i think its right now and i will try to graph it
     
  5. Dec 5, 2009 #4

    Mark44

    Staff: Mentor

    If you want to get credit for your work, the first thing you should do is graph your function. This problem has nothing to do with finding antiderivatives.
     
  6. Dec 5, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It would help to think of this as two separate integrals:
    [tex]\int_{-3}^0 1+ \sqrt{9- x^2} dx= \int_{-3}^0 1 dx+ \int_{-3}^0 \sqrt{9- x^2} dx[/tex]
    Graph each, if necessary, to recognise that those are simple figures and use basic area formulas.
     
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