Gryno formulė: Skirtumas tarp puslapio versijų

:<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], [http://integrals.wolfram.com/index.jsp?expr=%28x%2Bx%5E2%29*+%281%2B4x%5E2%29%5E%281%2F2%29&random=false kad]:

:<math>=\left(\frac{1}{96}\sqrt{4\cdot 5^2+1} (24\cdot 5^3+32\cdot 5^2+3\cdot 5+8)-\frac{1}{64}\text{arcsinh}(2\cdot 5)\right)-\left(\frac{1}{96}\sqrt{4\cdot 0^2+1} (24\cdot 0^3+32\cdot 0^2+3\cdot 0+8)-\frac{1}{64}\text{arcsinh}(2\cdot 0)\right)=</math>

:<math>=\left(\frac{1}{96}\sqrt{101} (24\cdot 125+32\cdot 25+15+8)-\frac{1}{64}\text{arcsinh}(10)\right)-\left(\frac{1}{96}\sqrt{1} \cdot 8-\frac{1}{64}\text{arcsinh}(0)\right)=</math>