Answer
$\displaystyle \int g(t)dt$ represents the volume of rocket fuel burned during $t$ seconds of flight.
Work Step by Step
The unit of $t$ is "seconds of flight".
If the unit of $g(t)$ is "volume of rocket fuel burned per second", then $g(t)$ is the derivative of some $G(t)$, whose units are "volume of rocket fuel burned", such that $G'(t)=g(t)$ (g is the instantaneous rate of change of $G$).
This means that $G$ is an antiderivative of $g$, that is,
$G(x)$ is one of the solutions of $\displaystyle \int g(t)dt$
Thus,
$\displaystyle \int g(t)dt$ represents the volume of rocket fuel burned during $t$ seconds of flight.
