anonymous
  • anonymous
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
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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tkhunny
  • tkhunny
Couple of ways to go about it, I suppose. Can you write an expression that represents the average value on the interval?
anonymous
  • anonymous
how?
tkhunny
  • tkhunny
Area = Height * Width or, in this case Area = (Average Height) * Width Are you seeing how to do it?

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anonymous
  • anonymous
so its true?
tkhunny
  • tkhunny
?? 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
  • anonymous
we don't have real values so i just used a and b
anonymous
  • anonymous
Every continuous function has an antiderivative right?
tkhunny
  • tkhunny
Yes. 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
  • anonymous
yes to what?
anonymous
  • anonymous
If 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
  • tkhunny
So 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] = \(\dfrac{\int\limits_a^b f(x)\;dx}{b-a}\). Do you see how htat works?
anonymous
  • anonymous
yea
tkhunny
  • tkhunny
And, 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?
anonymous
  • anonymous
yea
tkhunny
  • tkhunny
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]. Well, then, all we know is \(f(x) \le g(x)\) and we are trying to prove that \[\dfrac{F(b) - F(a)}{b-a} \le \dfrac{G(b) - G(a)}{b - a}\]. Can we do it?

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