Answer
$\dfrac{40^{3/2}-13^{3/2}}{27}$
Work Step by Step
Here, we have
$ ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
and $ds= t \sqrt {4+9t^2} dt$
This implies that
$L=\int_{1}^{2} (t \sqrt {4+9t^2}) dt$
Plug $4+9t^2 =p \implies t dt =\dfrac{dp}{18}$
or, $ \int_{13}^{40} \sqrt p (\dfrac{dp}{18})= \dfrac{1}{18} \int_{13}^{40} p^{1/2}$
Thus, $\dfrac{1}{27} [p^{3/2}]_{13}^{40}=\dfrac{40^{3/2}-13^{3/2}}{27}$
