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Math Help - definite integral of square root of 1-x^2 +2 using geometric consideration

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    Question definite integral of square root of 1-x^2 +2 using geometric consideration

    Question states, using geometric considerations, what does \int\limits_0^1 {\sqrt {1 - x^2 } }  + 2dx equal

    I can see that between zero and one that sqrt (1-x^2) is quadrant of a circle so the area would be 1/4*pi*r^2 but where does the +2 come in, I see it would move the circle up the y axis by 2. But I'm stuck
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    Quote Originally Posted by Craka View Post
    Question states, using geometric considerations, what does \int\limits_0^1 {\sqrt {1 - x^2 } }  + 2dx equal

    I can see that between zero and one that sqrt (1-x^2) is quadrant of a circle so the area would be 1/4*pi*r^2 but where does the +2 come in, I see it would move the circle up the y axis by 2. But I'm stuck
    If y= \sqrt{1- x^2}+ 2 then y- 2= \sqrt{1- x^2} so (y-2)^2= 1- x^2 and, finally, x^2+ (y-2)^2= 1. that is a circle with center at (0, 2) and radius 1. Since the square root is positive, you have the upper half of that circle and since x is from 0 to 1, rather than -1 to 1, you have only the upper right quadrant of that circle. The integral is, of course, \pi/4, the area of 1/4 of a circle of radius 1.
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  3. #3
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    wow some things look so simple when laid out. Thanks
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