Suppose that g: [2,8] --> R is continuous in [2,8] and g(x) greater than or equal to 0 for all x in [2,8].
Prove that if the integral from 2 to 8 of g dx =0, then g(x) =0 for all x in [2,8]
This is a special case of a more general theorem.
Theorem: Let be a continous function with and then .
Proof: Assume by contradiction that there is a point so that . The reason why we can ignore the endpoints is because if it was the case that on and then that would make non-continous at either or . Now if then there is a so that on we have i.e. . Since is continous on this closed subinterval it means it attains a mimimum . Now define the function in the following way: if then otherwise . Note is integrable. It follows that .
But would would mean a contradiction.