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

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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".
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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