> For a function $f(x)$, its average value on an interval $[a, b]$ is given as; $$f_{avg} =\frac{1}{b-a}\int_a^b f(x)dx$$
Suppose the given distance function respect to time ($t$) is given by $s(t)$. Then the velocity would be given as $v(t)=s'(t)$
Now if we were to calculate the average velocity, we would compute; $$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt$$
Using the fundamental theorem of calculus, $$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt=\frac{1}{b-a}[s(t)]_a^b$$
which is equal to $$\frac{s(b)-s(a)}{b-a}$$
Implication: Subtract initial position from the final position and divide it by whole time. So method 1 would be correct for calculating the average **velocity**.