Show that f is integrable on [0,2] and calculate the integral.

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SUMMARY

The function f defined on the interval [0,2] is integrable despite its discontinuity at x=1, where f(1) = 0 and f(x) = 1 for all other x. The supremum and infimum of the lower and upper Riemann sums converge to 1 for all intervals excluding x=1, confirming integrability. The integral of f over [0,2] is calculated to be 2, as the contribution from the point of discontinuity does not affect the overall area under the curve.

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Homework Statement



Let f:[0,2] →ℝ be defined by f(x):= 1 if x ≠ 1 and f(1) :=0. Show that f is integrable on [0,2] and calculate its integral.

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The Attempt at a Solution


i am thinking that the sup{L(p,f)} and inf{U(p,f)} is 1 at every where but where x=1. And I would assume that it is 0 where x=1. So Do I have to Rieman Summations [0,1) and (1,2]. I am confused as to how to approach this when it is not continuous.
 
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Well, just one Riemann sum won't do it- to show a function is integrable, you have to show that any sequence of Riemann sums, with the maximum length of an interval going to 0, converges to the same thing.

If "P" is any partition of [0, 2] we can always make a "finer" partition, P', by adding 1 as a partition point so that L(P', f) is the sum of "x< 1" and "x> 1".
 

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