proof of if fx gt gx for a lt x lt b then aintb


Proof of: if f(x) > g(x) for a < x < b then ab  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,

ab f(x) - g(x) dx > 0

We know as well from Property 4,

ab f(x) - g(x) dx = ab f(x) dx - ab g(x) dx

Therefore, we then get,

ab f(x) dx - ab g(x) dx > 0

ab f(x) dx > ab g(x) dx

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

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

ab m dx < ab f(x) dx ≤ ab M dx

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

m(b -a) < ab f(x) dx ≤ M (b -a)

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Mathematics: proof of if fx gt gx for a lt x lt b then aintb
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