Question

What is the net area between `f(x) = x-cosx` and the x-axis over `x in [0, 3pi ]`?

Answer 1

`int_0^3pi (x-cos)dx = (x^2/2 -sinx)|_0^(3pi)`
`(3pi)^2/2`
Note the `sin3pi` evaluated at `0 and 3pi` is zero

Explanation:

graph{x^2/2 - sinx [-9, 9, -15, 15]}
insignificant the `sin` function adds a maximum of 1 and minimum of 0,so for large number the effect of `sinx` is insignificant and this integral approaches `x^2/2`