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Integerable Functions

  1. Sep 24, 2008 #1
    I need some help with this question please. Thanks. ::biggrin:

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


    [​IMG]


    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?

     
  2. jcsd
  3. Sep 24, 2008 #2

    gabbagabbahey

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

    edit- on the interval [0,1]
     
    Last edited: Sep 24, 2008
  4. Sep 24, 2008 #3

    Dick

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    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?
     
  5. Sep 24, 2008 #4
    Hi dick!

    Uh, isn't it [tex]\int_{0}^{1}g(x) dx[/tex]?
     
  6. Sep 24, 2008 #5

    Dick

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    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?
     
    Last edited: Sep 24, 2008
  7. Sep 24, 2008 #6

    gabbagabbahey

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    My Idea was to simply show that on the interval [0,1],
    [tex]\frac{dg}{dx}=f(x)[/tex]
    [tex]\Rightarrow \int_a^b f(x) dx= \int_a^b \frac{dg}{dx} dx =g(b)-g(a)[/tex]
    for [tex](a,b) \epsilon [0,1][/tex] showing that the integral clearly exists and hence that f is integrable.
     
  8. Sep 24, 2008 #7

    Dick

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    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.
     
  9. Sep 24, 2008 #8

    gabbagabbahey

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    Good point, my mistake.
     
  10. Sep 25, 2008 #9
    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?

    [tex]\sum_{k=0}^{1}k[/tex] [tex]\int_{0}^{1}f(x)dx = F(1)-F(0)[/tex] ?

    Roam
     
  11. Sep 25, 2008 #10

    Dick

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    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.
     
  12. Sep 25, 2008 #11
    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?
     
  13. Sep 25, 2008 #12

    HallsofIvy

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    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.
     
  14. Sep 25, 2008 #13

    Dick

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    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??
     
    Last edited: Sep 25, 2008
  15. Sep 25, 2008 #14
    F is called an integral of f on interval I if F'(x) = f(x) for all x[tex]\in[/tex]I
     
  16. Sep 25, 2008 #15

    Dick

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    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.
     
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