$x=3t^2+2$ and $y=2t^3-1$ on $[1,3]$
The formula for parametric arc length is $\int\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$
$x'(t) = 6t,y'(t)=6t^2$
Under the radical, I get $36t^2+36t^4$, which will factor into $36t^2(t^2+1)$, and square rooting that.. I get $$6\int_1^3{t\sqrt{t^2+1}}\;dt$$ Obviously this is a simple u-substitution integration, for which I get $2{u}^{3/2}$ and integrate from 1 to 3, changing the limits to 2 and 10 because it is a function of $u$... which gives me $$2({10}^{3/2}-{2}^{3/2})$$ Is this correct?