**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)