Fazemos a substituição u=2+3t^{2}, de onde du=6t\,dt. Quando t=0, u=2; quando t=1, u=5.
Assim,
L =\sqrt{6}\int_{u=2}^{5}\sqrt{u}\,\frac{du}{6} =\frac{\sqrt{6}}{6}\int_{2}^{5}u^{\tfrac12}du =\frac{\sqrt{6}}{6}\,\biggl[\frac{u^{\tfrac32}}{\tfrac32}\biggr]_{2}^{5} =\frac{\sqrt{6}}{9}\bigl(5^{\tfrac32}-2^{\tfrac32}\bigr).