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    Calculus : Partitions and Riemann Integral

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    Let f be the function:
    1/4 x=o,
    x 0<x<1
    3/4 x=1

    Using standard partition Pn (0,1) where n greater or equal to 4

    L(f, Pn) = 2n(squared) -3n+4 all divided by 4n(squared)

    U(f,Pn) = 2n(squared) +3n+4 all divided by 4n(squared)

    and deduce that f is intergrable on (0,1) and evaluate
    (intergral sign with 1 at top and 0 at bottom) of f

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    Solution Preview

    dL/dn =0=> n = 8/3 and d^2L/dn^2 = +ve
    => L has minima at n =8/3 < 4 => increasing fn for n>= 4
    suprema(L) = lim(n->infinity) L = lim(n->inf) {2/4 -3/4n +4/4n^2}
    => ...

    Solution Summary

    A function is proven to be integrable over a given range.