Proving the Integrability of a Given Function f on Interval [0,1]

[roam]

I need some help with this question please. Thanks. :

a) Let f: [0,1] $$\rightarrow R$$ be given by


http://img148.imageshack.us/img148/5435/10187431tc4.gif


Argue that f is integerable.




The attempt at a solution

I don't get it. A function is integerable if its integral exists. How can we integrate this? There is no actual function given here...

So how are we supposed to argue that it is integerable?

 

[gabbagabbahey]

What is the derivative of the following function?
$$g(x)= \left\{ \begin{array}{rl} 2x & x<1 \\ 3x & x=1 \end{array}$$

edit- on the interval [0,1]

 

[Dick]

What is the derivative of the following function?
$$g(x)= \left\{ \begin{array}{rl} 2x & x<1 \\ 3x & x=1 \end{array}$$

edit- on the interval [0,1]

I don't think this problem is about antiderivatives. It's about the definition of the integral in terms of Riemann sums. roam, it is a function. It just doesn't happen to be continuous. That doesn't mean it's not integrable. What's the area under the curve?

 

[roam]

Hi dick!

What's the area under the curve?

Uh, isn't it $$\int_{0}^{1}g(x) dx$$?

 

[Dick]

Hi! Sure it is, but what's the value of that? Think about Riemann upper sums and lower sums. You have to deal with the definition of the integral. If f(x)=2 the area is 2. Does moving only the single point at x=1 to f(1)=3 change that?

 

[gabbagabbahey]

I don't think this problem is about antiderivatives. It's about the definition of the integral in terms of Riemann sums. roam, it is a function. It just doesn't happen to be continuous. That doesn't mean it's not integrable. What's the area under the curve?

My Idea was to simply show that on the interval [0,1],
$$\frac{dg}{dx}=f(x)$$
$$\Rightarrow \int_a^b f(x) dx= \int_a^b \frac{dg}{dx} dx =g(b)-g(a)$$
for $$(a,b) \epsilon [0,1]$$ showing that the integral clearly exists and hence that f is integrable.

 

[Dick]

But g(x) isn't differentiable at x=1, it's not even continuous. It doesn't make sense to say dg/dx=f at x=1. There are more basic definitions of an integral existing than that. Besides g(1)-g(0)=3. That's just plain silly. The area under the curve f(x) between 0 and 1 is 2.

 

[gabbagabbahey]

But g(x) isn't differentiable at x=1, it's not even continuous. It doesn't make sense to say dg/dx=f at x=1. There are more basic definitions of an integral existing than that. Besides g(1)-g(0)=3. That's just plain silly. The area under the curve f(x) between 0 and 1 is 2.

Good point, my mistake.

 

[roam]

Yes, functions that are not continuous can be integrable.
But Dick I'm not quite sure, we haven't studied some of that yet.

Btw, which definitions do I need to use for proving that f is integrable?

$$\sum_{k=0}^{1}k$$ $$\int_{0}^{1}f(x)dx = F(1)-F(0)$$ ?

Roam

 

[Dick]

You haven't studied Riemann sums yet? Qualitatively, the idea is just that over ALMOST ALL of the interval the function is just f(x)=2. The bit where it jumps up to 3 only makes an 'infinitely small' (whatever that means) contribution to the area. If you don't have a rigorous definition of an integral (not antiderivatives - that's a theorem, not a definition), then you'll have to wave your hands and say stuff like that.

 

[roam]

This function has two values i.e.,
f(x) = 2 for all values of x below 1, so this value is integrable

The second value is
f(x) = 3 for x equal to 1, as there is only one number and not a range therefore this can not be integrated and it will have only one value.
thus the original function is integrable.

Does that make sense?

 

[HallsofIvy]

I would recommend partitioning [0, 1] into n intervals with the last interval starting at 1- 1/n, ending of course at x= 1 so it has base length 1-(1- 1/n)= 1/n. You can take the maximum height of that interval to be 3 (the value of f at the right end point) so the maximum area is 3(1/n) which goes to 0 as n goes to infinity. The minimum height would be 2, the function value at the right end point (or anywhere else in the interval except x= 1) so the minimum area is 3(1/n) which also goes to 0 as n goes to infinity. Any other value for the height must lie between 2 and 3 and the area also goes to 0. That is sufficient to show that "f(x)= 3 for x= 1" contributes nothing to the integral. Use whatever method you like to show that f(x)= 2, for x between 0 and 2, is integrable.

 

[Dick]

This function has two values i.e.,
f(x) = 2 for all values of x below 1, so this value is integrable

The second value is
f(x) = 3 for x equal to 1, as there is only one number and not a range therefore this can not be integrated and it will have only one value.
thus the original function is integrable.

Does that make sense?

Sort of. You can integrate f(x)=3 from x=1 to x=1. The result is zero. What I really wanted was a DEFINITION of what an integral is. Nothing in the book or notes??

 

[roam]

What I really wanted was a DEFINITION of what an integral is. Nothing in the book or notes??

F is called an integral of f on interval I if F'(x) = f(x) for all x$$\in$$I

 

[Dick]

F is called an integral of f on interval I if F'(x) = f(x) for all x$$\in$$I

Then that leaves you in a pretty tough spot. The integrals F(x) of f(x) on the interval [0,1) (not including 1) are 2x+C for any constant. If you define F(1)=2+C then F'(1)=2 not 3. If you define it to be anything else then F doesn't have a derivative since it's not even continuous. Frustrating, huh? I think you are just stuck with arguing the area under the curve should be 2.