Question

How do you find the average value of `f(x)=-x^5+4x^3-5x-3` as x varies between `[-2,0]`?

Answer 1

The answer is `=-2/3`

Explanation:

The average value is `1/(b-a)int_a^bf(x)dx`

We need `intx^ndx=x^(n+1)/(n+1)+C (n!=-1)`

Therefore,

The average value is `1/(0--2)int_-2^0(-x^5+4x^3-5x-3)dx`

`=1/2[-x^6/6+4x^4/4-5x^2/2-3x]_-2^0`

`=1/2(0-(-2^6/6+16-10+6))`

`=1/2(32/3-12)`

`=-4/6=-2/3`