Answer
$L$ = $2(2\sqrt {2}-1)$
Work Step by Step
$x$ = $1+3t^{2}$
$y$ = $4+2t^{3}$
$\frac{dx}{dt}$ = $6t$
$\frac{dy}{dt}$ = $6t^2$ so
$\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2$ = $36t^{2}+36t^{4}$
Thus
$L$ = $\int_0^1{\sqrt {36t^{2}+36t^{4}}}dt$
$L$ = $\int_0^1{6t\sqrt {1+t^{2}}}dt$
Use substitution:
$[u = 1+t^{2}], du = 2tdt$
$L$ = $6\int_1^2{\sqrt u}{\frac{1}{2}du}$
$L$ = $3\left[\frac{2}{3}u^{\frac{3}{2}}\right]_1^2$
$L$ = $2(2\sqrt {2}-1)$
