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Posted

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]

Posted

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? :rolleyes:

Posted

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.

Posted

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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