Proof of: if f(x) > g(x) for a < x < b then _a∫^b  f(x) dx > g(x).

Because we get f(x) ≥ g(x) then we knows that f(x) - g(x) ≥ 0 on a ≤ x ≤ b and therefore by Property 8 proved as above we know that,

_a∫^b f(x) - g(x) dx > 0

We know as well from Property 4,

_a∫^b f(x) - g(x) dx = _a∫^b f(x) dx - _a∫^b g(x) dx

Therefore, we then get,

_a∫^b f(x) dx - _a∫^b g(x) dx > 0

_a∫^b f(x) dx > _a∫^b g(x) dx

Proof of: If m ≤ f(x) ≤ M for a ≤ x ≤ b then m (b - a)≤ _a∫^b f(x) dx ≤ M (b - a).

 Provide m ≤ f(x) ≤ M we can utilize Property 9 on each inequality to write,

_a∫^b m dx < _a∫^b f(x) dx ≤ _a∫^b M dx

So by Property 7 on the left and right integral to find,

m(b -a) < _a∫^b f(x) dx ≤ M (b -a)