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Math Help - Integrability of a function

  1. #1
    No one in Particular VonNemo19's Avatar
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    Integrability of a function

    Prove that if f(x) is continuous on a closed interval [a,b] then it is integrable there.

    I need some help with this one guys. Thanx.

    Is Anyone Up to the challenge?
    Last edited by VonNemo19; April 25th 2009 at 10:51 AM. Reason: refresh
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  2. #2
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    Quote Originally Posted by VonNemo19 View Post
    Prove that if f(x) is continuous on a closed interval [a,b] then it is integrable there.

    I need some help with this one guys. Thanx.

    Here is the basic Idea

    Let P be any partion of [a,b]

    U(P,f)-L(P,f)=\sum_{i=0}^{N}[f(x_M)-f(x_m)][x_{i+1}-x_{i}]

    Where x_M is the max on each sub interval

    x_m is the min on each sub interval

    They exsist by the extreem value theorem.

    Then on each subinterval by the MVT there exists a c \in (x_{i},x_{i+1}) such that

    f(x_m)(x_{i+1}-x_i)=F(c_i) \mbox{ and } f(x_M)(x_{i+1}-x_i)=F(c_{i+1}

    so we get

    U(P,f)-L(P,f)=\sum_{i=0}^{N}[f(x_M)-f(x_m)][x_{i+1}-x_{i}]=\sum_{i=0}^{N}F(c_{i+1})-F(c_i)
     =F(c_{N+1})-F(C_0)=F(b)-F(a)

    This is just a sketch you will need to justify and fill in the details

    TES
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  3. #3
    No one in Particular VonNemo19's Avatar
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    HUH?

    How did you derive this from the MVT?:

    Then on each subinterval by the MVT there exists a such that

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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    Prove that if f(x) is continuous on a closed interval [a,b] then it is integrable there.

    I need some help with this one guys. Thanx.

    Is Anyone Up to the challenge?
    I will leave the above proof here but it is NOT what you asked for. I should have read the question more carefully. That is my mistake.

    To show that this is integrable we need to show that there exists a partition of [a,b] such that U(f,p)-L(f,p)< \epsilon

    let \epsilon > 0 be given

    Since f is continous of a closed bounded (i.e compact ) interval f is uniformly continous.

    Now by the uniform cont. of f there exists a \delta > 0 such that for |x-y|< \delta \implies |f(x)-f(y)| < \frac{\epsilon}{b-a}

    Choose a partition P of [a,b] such that the mesh (each subdivsion) is less that \delta P=\{ a=x_0,x_1,...x_{N+1}=b \}

    Now by the extreem value theeorem and the continuity of f in each subinterval [x_i,x_{i+1}] there exists points

    x_m,x_M such that f(x_m) is a min and f(x_M) is a max on each sub interval

    Then for this partition note that since the mesh is less that delta f(x_M)-f(x_m) < \frac{\epsilon}{b-a}

    U(f,p) -L(f,p)=\sum_{i=0}^{N}(f(x_M)-f(x_m))(x_{i+1}-x_1)< \frac{\epsilon}{b-a}\sum_{i=0}^{N}(x_{i+1}-x_i)=
    \frac{\epsilon}{b-a}(b-a)=\epsilon
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  5. #5
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  6. #6
    No one in Particular VonNemo19's Avatar
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    Now for an Example

    Now, can I apply this (by epsilon-delta proof) to the function, let's say... 2x, and show that this function is integrable?

    I'm kind of at a loss, because in most cases of the epsilon-delta proofs of ordinary limits of functions ( as in taking the derivative for instance), the limit L is either known, or can be found through algebraic methods. But, in the case of integration, The value of the integral must remain a mystery until the function has, in fact, been integrated.

    Now, if I choose an epsilon, how can I show that the Riemann Sum is approaching some value A without having previous knowlege of what that value may be?

    Why does this stuff make my head spin?
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