Question

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

Answer 1

Couple of ways to go about it, I suppose. Can you write an expression that represents the average value on the interval?

Answer 2

how?

Answer 3

Area = Height * Width or, in this case Area = (Average Height) * Width

Are you seeing how to do it?

Answer 4

so its true?

Answer 5

?? 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?

Answer 6

we don't have real values so i just used a and b

Answer 7

Every continuous function has an antiderivative right?

Answer 8

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.

Answer 9

yes to what?

Answer 10

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?

Answer 11

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?

Answer 12

yea

Answer 13

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?

Answer 14

yea

Answer 15

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?