Question

How do you find the average value of the function for `f(x)=1+x, -1<=x<=1`?

Answer 1

`1`

Explanation:

`"average of f(x) in interval [a,b] is : "`
`(1/(b-a)) int_a^b f(x)`
`"So here we have"`
`a = -1`
`b = 1`
`f(x) = 1+x`
`=> avg = (1/2) int_{-1}^1 (1+x)(dx)`
`= (1/2) [x + x^2/2 + C]_{-1}^1`
`= (1/2) [1 + 1^2/2 + cancel(C) - (-1) - (-1)^2/2 - cancel(C)]`
`= (1/2) [1 + cancel(1/2) + 1 - cancel(1/2)]`
`= (1/2) [1 + 1]`
`= 1`