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Integral by interpreting it in terms of area

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data

    ∫(a= -3 , b= 0) (1 + √9 - x^2) dx

    2. Relevant equations
    ∫(a,b) f(x) dx = lim as n → [itex]\infty[/itex] [itex]\sum[/itex] f(xi) delta x


    3. The attempt at a solution
    I tried plugging in my a and b value into the function just as I would with any other function to find the area and i get a number but the answer is a ∏ so I am not sure with the pi comes from
     
  2. jcsd
  3. Feb 13, 2014 #2
    What did you plug your numbers into? That's not an elementary integral. I think the intention was for you to think about what the graph of 1 + sqrt(9 - x^2) looks like, and then use geometry to calculate it.
     
  4. Feb 13, 2014 #3
    I've been doing some research and i figure that 9-x^2 is 1/4 of a circle
     
  5. Feb 13, 2014 #4

    haruspex

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    There is certainly a quarter circle involved, but the region is a bit more than that.
     
  6. Feb 13, 2014 #5
    how do i know its a quarter circle though?
     
  7. Feb 13, 2014 #6
    A couple different ways. One is to rearrange the equation a bit.

    Replace f(x) with y, and ignore the + 1 in front, and you have this:
    $$y = \sqrt{9 - x^{2}}$$
    With a little rearranging (square both sides, move x^{2} over), you get this:
    $$x^{2} + y^{2} = 9$$
    You can recognize that as the equation of a circle, with radius 3. So if plotted at the origin, it would stretch from -3 to 3 on both axes. Since the limits on integration are from x = -3 to x = 0, that's half of the circle. Then, since you take the positive square root, it's the half of the circle above the x-axis: thus, a quarter circle in the second quadrant. Look at the addition of 1 to the front of the original equation (f(x)), and you get that same quarter circle moved up one.
     
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