Gryno formulė: Skirtumas tarp puslapio versijų

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:<math>m=\int_0^5 \gamma \sqrt{1+[y']^2} dx=\int_0^5 (x+y)\sqrt{1+[y']^2} dx=\int_0^5 (x+x^2) \sqrt{1+4x^2} dx=2\int_0^5 x(1+x) \sqrt{\frac{1}{4}+x^2} dx;</math>
:Toliau pasinaudodami [http://integrals.wolfram.com/index.jsp?expr=x*%281%2Bx%29*+sqrt%280.25%2Bx%5E2%29&random=false Wolframo internetiniu integratoriumi] [http://integrals.wolfram.com/index.jsp?expr=x*%281%2Bx%29*+sqrt%281%2F4%2Bx%5E2%29&random=false gauname], [kad http://integrals.wolfram.com/index.jsp?expr=%28x%2Bx%5E2%29*+%281%2B4x%5E2%29%5E%281%2F2%29&random=false]:
:<math>m=2\int_0^5 (x+x^2) \sqrt{1+4x^2} dx=\left(\frac{1}{38496}\left(2\sqrt{4x^2+1} (24x^3+32x^2+3x+8)-\frac{1}{64}\text{arcsinh}(2x)\right)|_0^5=</math>
 
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