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The area of a definite integral

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data

    The problem requires the computation of an indefinite integral but says to "interpret as areas" which basically changes the question to the following;

    What is the area under the curve of the function y = root( 6 - 2x2 + 4x) from x = -1 to x = 3

    2. Relevant equations

    Relevant equations would include the area of a circle, A=pi*r2

    3. The attempt at a solution

    To begin with, I am not sure if the function is just a semi circle or also extends beyond just half of a circle. That being said, with the assumption that the function is a semi circle, the diameter is 4 (3 - (-1)), thus the radius is 2, making the total area (1/2)pi*22. However, with a graphing calculator, I can see that the area is 8.89.

    Therefore I am led to believe that the area under the curve is a semicircle + a rectangular area. However, I am unsure as to how I can find the radius of the semicircle and the height of the rectangle.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 28, 2009 #2

    zcd

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    try completing the square
    [tex]6+4x-2x^{2}=2(3+2x-x^{2})=2(4-(x-1)^{2})[/tex]
     
  4. Nov 28, 2009 #3
    Unfortunately, integration is not allowed. The question must be completed by interpreting the area under the function as a sum of areas (a semicircle and perhaps a rectangle) but I have no idea how to, using the given function, find the semi circle's radius to be able to find its area. Unless completing the square puts the equation in a form where the semicircle's radius is more apparent, I'm not sure how it would help.
     
  5. Nov 28, 2009 #4

    zcd

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    But it does; it gives you the location of the center and from there you can figure out the radius.
     
  6. Nov 28, 2009 #5
    :surprised

    Really?

    This is the first time in my life I've seen a semicircle function. With that manipulation, what is the location of the centre?

    Thank you so much, I've been stumped on this one for about an hour.
     
  7. Nov 28, 2009 #6

    zcd

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    [tex]y=\sqrt{2(4-(x-1)^{2})}\implies y^{2}=2(4-(x-1)^{2})[/tex] with some minor caveats in the range. [tex]y^{2}+2(x-1)^{2}=8[/tex] is an equation of?
     
  8. Nov 28, 2009 #7
    It seems like the equation of a circle, the only equation of a circle I've ever seen is y^2 + x^2 = 9 which has a radius of 3 and center of (0, 0). But I'm still not quite sure how to obtain the location of the center of the circle from the newly manipulated one you gave me. It seems as the y coordinate of the center would still be 0, but I don't know about the x coordinate of the origin.

    After some googling, I think that the y coordinate is 0 and the x coordinate is 1, but i'm not sure how the factor of 2 affects it.
     
  9. Nov 28, 2009 #8
    Ah, apparently it's an equation of a semi-ellipse. Any idea as to how I can find its area? Either if a formula for it exists or if I must split it up into common geometric shapes and take their sum?
     
    Last edited: Nov 28, 2009
  10. Nov 28, 2009 #9

    zcd

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    [tex]A=ab\pi[/tex] where a and b are the lengths of semi axes
     
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