[PDF] [PDF] Solutions to HW due on Mar 11 - Math 432 - Real Analysis II

For L(f), since there are infinitely many irrationals in every subinterval [tk−1,tk] and thus M(f, [tk−1,tk]) = 0 Thus, for any partition, L(f,P) = 0 So, L(f) = 0 Since U(f) = L(f), then f is not integrable



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f(xn) − f(yn) = cos(2nπ) − cos(2nπ + π/2) = 1 ∀n ∈ IN Hence g is not uniformly continuous on (0,1) On the other hand, the function u given by u(x) 



[PDF] Solutions to HW due on Mar 11 - Math 432 - Real Analysis II

For L(f), since there are infinitely many irrationals in every subinterval [tk−1,tk] and thus M(f, [tk−1,tk]) = 0 Thus, for any partition, L(f,P) = 0 So, L(f) = 0 Since U(f) = L(f), then f is not integrable



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g = 1 Thus, sup f > inf g even though f



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10 mai 2017 · Figure 1: A piecewise linear, continuous function together with an antiderivative we can find with- out using the Fundamental Theorem of Calculus 





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1 f and f2 are integrable when f is integrable Lemma 1 1 Let f : [a, b] → R be a bounded function and let P = {x0,x1, ,xn} be a partition of [a, b] Then for each i 



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Prove that f is not Riemann integrable Solution: f is integrable on [1, 3] if and only if it is integrable on [1, 2] and also 



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If a bounded function f is such that f f(x) dx + ( f(x) dx, then f is not Riemann integrable on [a,b] Examples: 1 A constant function is Riemann integrable on [a, b]

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