Gryno formulė: Skirtumas tarp puslapio versijų

369 pridėti baitai ,  prieš 7 metus
:Pasinaudodami internetiniu integratoriumi, [http://integrals.wolfram.com/index.jsp?expr=x-%28R%5E2+-+x%5E2%29%5E%281%2F2%29&random=false gauname], [http://integrals.wolfram.com/index.jsp?expr=x%2B%28R%5E2+-+x%5E2%29%5E%281%2F2%29&random=false kad]
:<math>\oint_L(x-y)dx+(x+y)dy=\int_0^R (x-\sqrt{R^2-x^2}) dx + \int_0^R (\sqrt{R^2-y^2}+y) dy=</math>
:<math>\left(-\frac{1}{2}x\sqrt{R^2-x^2}+\frac{1}{2} R^2 \arctan\frac{x\sqrt{R^2-x^2}}{x^2-R^2}+\frac{x^2}{2} \right)|_0^R+\frac{1}{2}\left( y(\sqrt{R^2-y^2}+y) +R^2\arctan\frac{y}{R^2-y^2} \right).|_0^R=</math>
:Riba <math>\lim_{xleft(-\tofrac{1}{2} R \sqrt{R^2-R^2}+\frac{1}{2} R^2 \arctan\frac{xR\sqrt{R^2-xR^2}}{xR^2-R^2}=+\lim_frac{x\to R^2}{2} \arctanright)+\frac{x1}{2}\left( R(\sqrt{xR^2-R^2}}=\lim_{z\to+R) 0}+R^2\arctan\frac{R}{zR^2-R^2}= \arctan(\inftyright)=\pi.</math>
:<math>\left(\frac{1}{2} R^2 \arctan(\infty)+\frac{R^2}{2} \right)+\frac{1}{2}\left( R(0+R) +R^2\arctan(\infty) \right)=</math>
:Riba <math>\lim_{x\to R}\arctan\frac{x\sqrt{R^2-x^2}}{x^2-R^2}=\lim_{x\to R}\arctan\frac{x}{\sqrt{x^2-R^2}}=\lim_{z\to 0}\arctan\frac{R}{z}=\arctan(\infty)=\pi.</math> Beje, <math>\arctan(0)=0.</math>
 
 
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