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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)
Logarithmic Differentiation : There is one final topic to discuss in this section. Taking derivatives of some complicated functions can be simplified by using logarithms. It i
Proof of: ∫ f(x) + g(x) dx = ∫ f(x) dx + ∫g(x) dx It is also a very easy proof. Assume that F(x) is an anti-derivative of f(x) and that G(x) is an anti-derivative of
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show that one of the straight lines given by ax2+2hxy+by2=o bisect an angle between the co ordinate axes, if (a+b)2=4h2
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