Originally Posted by

**Reckoner** Some calculus *might *make it a bit clearer, if you know that the average value of a function $\displaystyle f$ on an interval $\displaystyle [a,\,b]$ is $\displaystyle \overline f = \frac1{b - a}\int_a^bf(x)\,dx$.

Suppose at a given time $\displaystyle t$ the Sun is a distance $\displaystyle s(t)$ miles from the Earth, and further suppose that the Earth revolves around the Sun exactly once between times $\displaystyle t = a$ and $\displaystyle t = b$.

Then any light ray hitting the Earth at a time $\displaystyle t$ will have traveled for $\displaystyle \frac{s(t)}{186\cdot10^3}$ seconds. As $\displaystyle s$ should be a continuous function of $\displaystyle t,$ we may determine the average time to be

$\displaystyle \overline t = \frac1{b - a}\int_a^b\frac{s(t)}{186\cdot10^3}\,dt$

$\displaystyle =\frac1{\left(186\cdot10^3\right)(b - a)}\int_a^b s(t)\,dt$

$\displaystyle =\frac1{186\cdot10^3}\left(\frac{\int_a^b s(t)\,dt}{b - a}\right)$

$\displaystyle =\frac1{186\cdot10^3}\cdot\overline s$

$\displaystyle =\frac{\overline s}{186\cdot10^3}$

$\displaystyle =\frac{93\cdot10^6}{186\cdot10^3} = \frac{10^3}2 = 500$

Carrying the units through, the answer is in seconds, and this agrees with angel.white's solution.