1v_{avg}=\frac{\int_{0}^{T}{v(t)dt}}{T}
then there exist a t in [0, T]
such that
v(t)=\frac{\int_{0}^{T}{v(t)dt}}{T} =v(avg)
its is true for any continous and integrable function
it a fundamental theorem
for more information
http://en.wikipedia.org/wiki/Mean_value_theorem#Proof_of_the_first_mean_value_theorem_for_integration
http://www.sosmath.com/calculus/integ/integ04/integ04.html