The Average value of f(x)=25−x^2 on the interval [0,1] is 74/3

Question

Find a value c in the interval [0,1] such that f(c) is equal to the average value?

Answer 1

the average value is
(∫[0,1] (25-x^2) dx)/(1-0)

or, since the area of the semicircle is 25/4 π. the average value is, as always,
area / width