MATH1151 Chapter Notes - Chapter 5: Mean Value Theorem, Intermediate Value Theorem, Riemann Integral
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Xk =1 n (ak ak 1)1 = an a0 = . Here the set is so messy that you just can"t put polygons inside and outside the region which have nearly equal area. Riemann"s de nition fails to determine an area for this region, and we say that this function is not riemann integrable. 1 n3 f k = k 2 n2 n (k 1)2. Xk =1 f k = (k 1)2 n2 n. Note: f has in nitely many discontinuities, but they are more spread out now. Let f : [0, 1] r be de ned by f (x) =(1 if x = 1. 0 otherwise. k for k = 1, 2, . There is only one number a such that 0 a for all > 0. 8 + and hence f is riemann integrable and z 1. Suppose that f , g : [a, b] r are riemann integrable.
