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410 eilutė:
:''Sprendimas''.
:<math>y'=(x^2)'=2x;</math>
:<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 [Wolframo internetiniu integratoriumi http://integrals.wolfram.com/index.jsp?expr=x*%281%2Bx%29*+sqrt%280.25%2Bx%5E2%29&random=false] [gauname http://integrals.wolfram.com/index.jsp?expr=x*%281%2Bx%29*+sqrt%281%2F4%2Bx%5E2%29&random=false]:
 
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