Answer
$\dfrac{125-13^{3/2}}{48}$
Work Step by Step
Here, we have
$ ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
and $ds= t^2 \sqrt {9+16t^2} dt$
This implies that
$L=\int_{1/2}^{1} ( t^2 \sqrt {9+16t^2} ) dt$
Plug $9+16t^2 =p \implies t dt =\dfrac{dp}{32}$
or, $ \int_{13}^{25} \sqrt p (\dfrac{dp}{32})= \dfrac{1}{32} \int_{13}^{40} p^{1/2}$
Thus, $\dfrac{1}{48} [p^{3/2}]_{13}^{25}=\dfrac{125-13^{3/2}}{48}$
