Question

How do you find the average value of `f(x)=-x^3+7x^2-11x+3` as x varies between `[1,5]`?

Answer 1

`-2`

Explanation:

Average

`=(int f(x) dx)/(5-1)`, between x = 1 and x = 5

`(1/4)[-x^4/4+7x^3/3-11x^2/3+3x],` between the limits

`=(1/4)[-(5^4-1)+7(5^3-1)-11(5^2-1)+3(5-1)]`

`=(1/4)[-624+868-264+12]`

`=(1/4)(-8)`

`=-2`