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Area under the curve

  1. Dec 5, 2009 #1
    how do you find the integral of the equation sqrt(36-.22x^2) between x=0 and x=9
     
  2. jcsd
  3. Dec 5, 2009 #2

    arildno

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    Well, we have:
    [tex]\int_{0}^{9}\sqrt{36-0.22x^{2}}dx=6\int_{0}^{9}\sqrt{1-(\frac{\sqrt{0.22}x}{6})^{2}}dx[/tex]

    Set:
    [tex]u=\frac{\sqrt{0.22}x}{6}}\to{dx}=\frac{6}{\sqrt{0.22}}[/tex]
    And our integral may be rewritten as:
    [tex]\frac{36}{\sqrt{0.22}}\int_{0}^{\frac{3\sqrt{0.22}}{2}}\sqrt{1-u^{2}}du[/tex]

    Do you have any ideas how to proceed from here?
     
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