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I am having trouble with this problem. Can anyone help me? I think I have to show that any sequence of Riemann sums, with the maximum length of an interval going to 0, converges to the same thing.

Let f:[0,2] -> be defined by f(x) =: 1 if x does not equal 1 and f(1):=0. Show that f is integrable on [0.2] and calculate its integral.


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Given a maximum mesh size $\epsilon$, can you calculate the maximum upper sum? – Antonio Vargas Nov 14 '12 at 1:20

To show Riemann integrability, it suffices for any $\epsilon > 0$ to find a partition of $[0,2]$ over which the upper and lower sums differ by at most $\epsilon$.

In this case, $[0, 1 - \frac{\epsilon}{2}], [1 - \frac{\epsilon}{2}, 1 + \frac{\epsilon}{2}], [1 + \frac{\epsilon}{2}, 2]$ will work, since the lower sum is zero and the upper $\epsilon$.

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do you know how to actually calculate the integral? – Jackson Hart Nov 14 '12 at 3:48
yeah it's 0, since the lower sums were all 0, and the upper sums got arbitrarily small! (maybe draw a picture if this intuitive yet...basically the function is flat at 0 except for one blip with negligible volume...the fiddling with $\epsilon/2$ is exactly capturing the intuition of negligible volume) – uncookedfalcon Nov 16 '12 at 0:40

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