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Math Help - MVT and Reimann Integral

  1. #1
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    Exclamation MVT and Reimann Integral

    Can someone please help me prove this? I've tried it to work it out in different ways and I'm beat. Will appreciate your help.

    The problem reads:
    Let f be a continuous function on [a,b]. The length l of the curve y=f(x) on the interval [a,b] is defined by

    l=sup_{\substack{P}}[\lambda_P(f)]
    (don't know how to the _P under the sup)

    where

    \lambda_P(f)=\sum_{k=0}^N\sqrt{(x_k-x_{k-1})^2+f(x_k)-f(x_{k-1}))^2}

    and the supremum is taken over all partition points D=\{a=x_0,x_1,...,x_N=b\} of [a,b].

    Assume that f has continuous derivative on [a,b]. Take any partition D=\{a=x_0, x_1,...,x_n=b\} of [a,b] and derive

    \lambda_P(f) = \sum_{k=0}^N \sqrt{1+\left(\frac{f(x_k)-f(x_k-1)}{x_k-x_{k-1}}\right)^2} (x_k-x_{k-1}).

    Next, use the Mean Value Theorem, and then use the definition of the Reimann integral to establish that

    l=\int_{a}^b\sqrt{1+[f'(x)]^2}dx.

    Thank you so much for your help.
    Last edited by yvonnehr; December 16th 2009 at 07:10 PM. Reason: typo and missing relevant information
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  2. #2
    MHF Contributor matheagle's Avatar
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    This is just arc length.
    I used to prove it in calc 2 all the time.
    You're missing a ( in the first series.

    Use the MVT in each subinterval.
    There is a c_k in each ( x_{k-1},  x_k)

    such that f'(c_k)=(f(x_k)-f( x_{k-1}))/(x_k-x_{k-1})

    Now substitute this in the series and let the mesh (largest subinterval) go to zero.
    (Not n to infinity, all subintervals have to become fine.)
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  3. #3
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    Quote Originally Posted by yvonnehr View Post
    Can someone please help me prove this? I've tried it to work it out in different ways and I'm beat. Will appreciate your help.

    The problem reads:
    Let f be a continuous function on [a,b]. The length l of the curve y=f(x) on the interval [a,b] is defined by

    l=sup_{\substack{P}}[\lambda_P(f)]
    (don't know how to the _P under the sup)

    where

    \lambda_P(f)=\sum_{k=0}^N\sqrt{(x_k-x_{k-1})^2+(f(x_k)-f(x_{k-1}))^2}

    and the supremum is taken over all partition points D=\{a=x_0,x_1,...,x_N=b\} of [a,b].

    Assume that f has continuous derivative on [a,b]. Take any partition D=\{a=x_0, x_1,...,x_n=b\} of [a,b] and derive

    \lambda_P(f) = \sum_{k=0}^N \sqrt{1+\left(\frac{f(x_k)-f(x_k-1)}{x_k-x_{k-1}}\right)^2} (x_k-x_{k-1}).
    Well, this step, at least, is easy- just factor (x_k- x_{k-1})^2 out of the square root.

    Next, use the Mean Value Theorem, and then use the definition of the Reimann integral to establish that

    l=\int_{a}^b\sqrt{1+[f'(x)]^2}dx.

    Thank you so much for your help.
    The mean value theorem, applied to the interval x_{k-1} to x_k says the there exist some x* in that interval such that \frac{f(x_k)- f(x_{k-1})}{x_k- x_{k-1}}= f'(x*). Replace each such fraction in the sum by that derivative. Then recognise it as a Riemann sum that, in the limit, gives the corresponding integral.
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