Intermediate value theorem on Mean Value Theorem for Integrals

[flyingpig]

Homework Statement

Prove the Mean Value Theorem for Integrals

Proof

Let f(x) be defined on [a,b]

Let M be the max of f(x) and m be the min of f(x)

Then

$$m \leq f(x) \leq M$$

$$\int_{a}^{b}m \;dx\leq \int_{a}^{b} f(x)\;dx \leq \int_{a}^{b} M\;dx$$

$$m(b-a) \;dx\leq \int_{a}^{b} f(x)\;dx \leq M(b-a)$$

$$m \leq\frac{\int_{a}^{b} f(x)\;dx}{b-a} \leq M$$

Then by the Intermediate Value Theorem there exists a $$c\in (a,b)$$ such that

$$f(c) = \frac{\int_{a}^{b} f(x)\;dx}{b-a}$$

$$f(c)(b-a) = \int_{a}^{b} f(x)\;dx$$

Question

How do we know that f(c) is the AVERAGE value function? It is just some value between the max and min, it doesn't have to be the AVERAGE right?

 

[Char. Limit]

But isn't the point c defined such that f(c) IS the average value?

 

[Sethric]

In this case you have defined it as the average value, but it doesn't have to be. I feel like there needs to be an "f is continuous" statement somewhere, since we are using the IVT. Anyway, the IVT guarantees that a function will take every value between its max and min. Since the average happens to be between the max and min, the function has to take that value somewhere as well.

 

[chiro]

If you look at it geometrically, your LHS is just a box with length (b-a) and height f(c). Visually the idea of f(c) being an "average value" makes more sense (and is more strongly reinforced) than just the algebraic expression.

 

[flyingpig]

But isn't the point c defined such that f(c) IS the average value?

I thought that's only if we assumed it to be

In this case you have defined it as the average value, but it doesn't have to be. I feel like there needs to be an "f is continuous" statement somewhere, since we are using the IVT. Anyway, the IVT guarantees that a function will take every value between its max and min. Since the average happens to be between the max and min, the function has to take that value somewhere as well.

If f is integrations, then it must be differentiable and hence continuous right?

 

[Sethric]

If f is integrations, then it must be differentiable and hence continuous right?

No. The Riemann Integral only requires that the set of points of discontinuity has measure zero. A step function would work. And would mess up your IVT.

 

[flyingpig]

What's a step function? If the function is discontinuous, the reinman sum would simply stop "summing"

 

[Sethric]

What's a step function?

http://en.wikipedia.org/wiki/Step_function

If the function is discontinuous, the reinman sum would simply stop "summing"

What would make it do that? If you think of the sum as a bunch of rectangles, then encountering a step discontinuity will just make the rectangles jump in size.

 

[flyingpig]

Then getting back to the topic here...

I'll just add the word "continuous"?

 

[Sethric]

Hah, yeah, I was just curious why it was missing from the original assumptions of the proof, since it was needed. Sorry for drawing that out more than necessary.

 

[flyingpig]

I copied off from our notes from our professor

 

[flyingpig]

Also, I am guessing the MVT also only works for continuous functions?

 

[Sethric]

Actually the MVT requires it to be differentiable on the open interval as well.

 

[flyingpig]

And hence continuous...

 

[Sethric]

Correct.