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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)
In a parallelogram ABCD AB=20cm and AD=12cm.The bisector of angle A meets DC at E and BC produced at F.Find the length of CF.
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To answer each question, use the function t(r) = d , where t is the time in hours, d is the distance in miles, and r is the rate in miles per hour. a. Sydney drives 10 mi at a c
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project on shares and dividends
Let u = sin(x). Then du = cos(x) dx. So you can now antidifferentiate e^u du. This is e^u + C = e^sin(x) + C. Then substitute your range 0 to pi. e^sin (pi)-e^sin(0) =0-0 =0
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