Question

What is the average value of a function `y=x^2-2x+4` on the interval `[0,8]`?

Answer 1

`= 52/3`

Explanation:

By definition

`y_{ave} = (int_a^b \ y \ dx)/(b - a)`

so here

`y_{ave} = (int_0^8 \ x^2 - 2x + 4 \ dx)/(8-0)`

`= 1/8 [ (x^3)/3 - x^2 + 4x]_0^8`

`= 52/3`