Math Challenges

Solution:
\( \displaylines{\mathbf{r}^{\prime}\left(t\right)=\lt2t,3t^2\gt\\ L=\int_0^4\sqrt{\left(2t\right)^2+\left(3t^2\right)^2}dt=\int_0^4\sqrt{4t^2+9t^4}dt\\ =\int_0^4\sqrt{4+9t^2}tdt=\frac{1}{18}\int_4^{148}u^{\frac12}du\approx66.3888} \)