Question

What is the average value of the function `f(x)=(x-1)^2` on the interval from x=1 to x=5?

Answer 1

The average value is `16/3`

Explanation:

The average value of a function `f` on an interval `[a,b]` is

`1/(b-a) int_a^b f(x) dx`

So the value we seek is

`1/(5-1) int_1^5 (x-1)^2 dx = 1/4[(x-1)^3/3]_1^5`

`= 1/12[(4)^3-(0)^3]`

`= 16/3`