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Math Help - Prove that a function is Riemann integrable?

  1. #1
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    Prove that a function is Riemann integrable?

    Consider the function f(x)=-x+2, 0<=x<=1 on the interval [0,1]

    1)Let Pn be the partition {0,1/n,2/n,...,1}.Calculate L(Pn) (the lower sum) and U(Pn) (the upper sum).
    I know the formula for L(P) and U(P) but I don't really understand it and I'm having trouble applying it. I know it's the sum of m=(i-1)/n and M=(i/n) but I don't know how to adapt this to the formula.Can anyone help please?

    2)Hence prove that f is Riemann Integrable on [0,1] and evaluate the integral?
    Once I have U(P) and L(P) I find the limit as n tends to infinity and if the limits are equal the function is integrable?and the value of the integral is -1/2x^2+2x+c.

    Can someone please correct me where I'm wrong and help me with the rest.

    Thanks in advance.
    Last edited by chocaholic; August 28th 2010 at 07:37 AM. Reason: spelling
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  2. #2
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    U(P) = \sum\limits_{k = 1}^n {f\left( {\frac{k}<br />
{n}} \right)\Delta _x }  = \sum\limits_{k = 1}^n {\left( {\frac{k}<br />
{n} + 2} \right)\frac{1}<br />
{n}}  = \frac{1}<br />
{n}\left( {\frac{{n\left( {n + 1} \right)}}<br />
{2n} + 2n} \right) = \frac{{5n + 2}}<br />
{{2n}}
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  3. #3
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    I understand why it would be like that for a positive function but for a negative straight line wouldn't that be the formula for L(P) and wouldn't it be -(k/n)+2? and the because -x+2 is a straight line wouldn't U(P) be the sum from 0 to (n-1)/n and how would you then derive a formula for the sum where the terms are negative?
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  4. #4
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    Yes, you are correct. I simply did not seen that - sign.
    That is a good reason to learn to post your questions in LaTex.
    As you have written it, it is very hard to read.
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  5. #5
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    I apologise for the misunderstanding but thank you so much for clarifying this for me
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