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Analysis proof 2

Note: If you have already answered this exact question please do not answer it again. I would like an answer from a different T.A. Thanks

Say abs = absolute value.
Suppose that the function f:[a,b]->R is Lipschitz;
that is , there is a number c such that:

abs(f(u) - f(v)) <= (c)abs(u-v)

for all u and v in [a,b]. Let P be a partition of [a,b] and R(f,P) be a Riemann sum based on P. Prove that

abs((R(f,P)) - (the integral from a to b of f)) <= ||P||(b-a)