x = 1 + 3t^2 y = 4 + 2t^3 0<= t >= 1 k so i got that it is the integral from 0 to 1 of the sqrt((6t)^2+(6t^2)^2) then i mess up somewhere when im solving that
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Originally Posted by amiv4 x = 1 + 3t^2 y = 4 + 2t^3 0<= t >= 1 k so i got that it is the integral from 0 to 1 of the sqrt((6t)^2+(6t^2)^2) then i mess up somewhere when im solving that $\displaystyle L = \int_{0}^{1} 6t \sqrt{1 + t^2} \, dt$. Make the substitution $\displaystyle u = 1 + t^2$.
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