Question

How do you find the average value of `x^3` as x varies between -1 and 2?

Answer 1

`= 5/4`

Explanation:

Average value of `f(x)` over interval `[a,b]` is `bar (f) (x) = (int_a^b f(x) dx)/(b-a)`

Here that means that:

`bar (f) (x) = (int_(-1)^2 x^3 dx)/(2 - (-1))`

`= ([ x^4/4]_(-1)^2)/3 = 5/4`