## Ldaniel Group Title 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 one year ago one year ago

1. tkhunny Group Title

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

2. Ldaniel Group Title

how?

3. tkhunny Group Title

Area = Height * Width or, in this case Area = (Average Height) * Width Are you seeing how to do it?

4. Ldaniel Group Title

so its true?

5. tkhunny Group Title

?? 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?

6. Ldaniel Group Title

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

7. Ldaniel Group Title

Every continuous function has an antiderivative right?

8. tkhunny Group Title

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.

9. Ldaniel Group Title

yes to what?

10. Ldaniel Group Title

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?

11. tkhunny Group Title

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?

12. Ldaniel Group Title

yea

13. tkhunny Group Title

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?

14. Ldaniel Group Title

yea

15. tkhunny Group Title

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?