Question

The average value of the function y = x² – 1 on [0, 12] is

Answer 1

47.

Explanation:

1) the rule:

`f_(avr)=(1)/(b-a) \int\limits^b_a {f(x)} \, dx ,`

where a;b are 0 and 12, f(x)=x²-1.

2) according to the rule above:

`f_(avr)=(1)/(12-0) \int\limits^(12)_0 {(x^2-1)} \, dx=(1)/(12)((x^3)/(3)-x)|^(12)_0=48-1=47.`