Answer
$\dfrac{15}{32}[e^{16}-e^{64}]$
Work Step by Step
Here, we have
$ ds=\sqrt {(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$
and $ds= 5 dt$
This implies that
$L=\int_{-1}^{2} (-3t) e^{16t^2}( 5 dt)$
Plug $16t^2 =p \implies t dt =\dfrac{dp}{32}$
or, $(-15) \int_{16}^{64} e^{p}(\dfrac{dp}{32})= \dfrac{-1 5}{32} \int_{16}^{64} e^{p}$
Thus, $\dfrac{-1 5}{32} [e^u]_{16}^{64}=\dfrac{15}{32}[e^{16}-e^{64}]$
