Question

What is the average value of the function `h(r) = 3 / (1+r)^2` on the interval `[1,6]`?

Answer 1

It is `3/14`

Explanation:

The average valus is given by

#h_(avg)=1/(b-a)*int_a^b h(r)dr=> h_(avg)=1/(6-1)int_1^6 3/(1+r)^2dr=> h_(avg)=1/5*15/14=3/14#