stuart clark Posted May 22, 2011 Posted May 22, 2011 Consider the function defined by the equation [math]\mathbf{ y^2-2y.e^{sin^{-1}x}+x^2-1+[x]+e^{2sin^{-1}x} = 0}[/math], Where [math] \mathbf{[x] = }[/math] Greatest Integer function. The find area of the curve bounded by the curve and the equation [math]\mathbf{x = -1}[/math]
baxtrom Posted May 24, 2011 Posted May 24, 2011 If real domain is considered then I guess [math]x \in [-1, 1][/math]? I would try numerically. Is [math]\int[/math] close to 3?
Shadow Posted May 24, 2011 Posted May 24, 2011 I got 4.14159 using the following Mathematica code: Abs[NIntegrate[E^ArcSin[x] - Sqrt[1 - x^2 - Floor[x]], {x, -1, 0}]] + NIntegrate[E^ArcSin[x] + Sqrt[1 - x^2 - Floor[x]], {x, -1, 1}] - NIntegrate[E^ArcSin[x] - Sqrt[1 - x^2 - Floor[x]], {x, 0, 1}] Although I've no idea if it's correct.
baxtrom Posted May 24, 2011 Posted May 24, 2011 I got 4.14159 using the following Mathematica code: Abs[NIntegrate[E^ArcSin[x] - Sqrt[1 - x^2 - Floor[x]], {x, -1, 0}]] + NIntegrate[E^ArcSin[x] + Sqrt[1 - x^2 - Floor[x]], {x, -1, 1}] - NIntegrate[E^ArcSin[x] - Sqrt[1 - x^2 - Floor[x]], {x, 0, 1}] Although I've no idea if it's correct. Looks suspiciously close to [math]\pi + 1[/math]..
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