Question

How do you find the average value of the function for `f(x)=4-x^2, -2<=x<=2`?

Answer 1

The average value of the function `f(x)` on `a<=x<=b` is:

`A=1/(b-a)int_a^bf(x)dx`

So here the average value is:

`A=1/(2-(-2))int_(-2)^2(4-x^2)dx`

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

`A=1/4(4(2)-1/3(2)^3)-1/4(4(-2)-1/3(-2)^3)`

`A=1/4(8-8/3)-1/4(-8+8/3)`

`A=(2-2/3)+(2-2/3)`

`A=4-4/3`

`A=8/3`