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anonymous
 4 years ago
If f(x) ≤ g(x) on the interval [a, b], then the average value of f is less than or equal to the average value of g on the interval [a, b].
True
or
False
anonymous
 4 years ago
If f(x) ≤ g(x) on the interval [a, b], then the average value of f is less than or equal to the average value of g on the interval [a, b]. True or False

This Question is Closed

tkhunny
 4 years ago
Best ResponseYou've already chosen the best response.0Couple of ways to go about it, I suppose. Can you write an expression that represents the average value on the interval?

tkhunny
 4 years ago
Best ResponseYou've already chosen the best response.0Area = Height * Width or, in this case Area = (Average Height) * Width Are you seeing how to do it?

tkhunny
 4 years ago
Best ResponseYou've already chosen the best response.0?? How did you get to that? We haven't done anything, yet. Generally, we call this guessing, but you may be going on evidence that you are not sharing. How do you find the area under a curve on a Cartesian Coordinate system?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we don't have real values so i just used a and b

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Every continuous function has an antiderivative right?

tkhunny
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. I'm pretty excited to see where you are going with that. It's important to point out that we may not be able to write the antiderivative. We just need to know that it exists.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If F(x) is an antiderivative of f(x) and G(x) = F(x) + 2, then G(x) is an antiderivative of f(x) right? or am i wrong?

tkhunny
 4 years ago
Best ResponseYou've already chosen the best response.0So close. G and F are not necessarily related. Let's back up just a hair. We can calcualte the average value of f(x) on the interval [a,b] using the integral and that rectangle formula. Average Value of f(x) on [a,b] = \(\dfrac{\int\limits_a^b f(x)\;dx}{ba}\). Do you see how htat works?

tkhunny
 4 years ago
Best ResponseYou've already chosen the best response.0And, as you very correctly pointed out, \(\int\limits_a^b f(x)\;dx = F(b)  F(b)\), where \(F(x)\) is an antiderivative of \(f(x)\). Still making sense?

tkhunny
 4 years ago
Best ResponseYou've already chosen the best response.0If f(x) ≤ g(x) on the interval [a, b], then the average value of f is less than or equal to the average value of g on the interval [a, b]. Well, then, all we know is \(f(x) \le g(x)\) and we are trying to prove that \[\dfrac{F(b)  F(a)}{ba} \le \dfrac{G(b)  G(a)}{b  a}\]. Can we do it?
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