Suppose we have: f(x)= 1 if 0[itex]\leq[/itex] x [itex]\leq[/itex] 1 AND 2 if 1[itex]\leq[/itex] x [itex]\leq[/itex] 2(adsbygoogle = window.adsbygoogle || []).push({});

Using the definition, show that f is Riemann integrable on [0, 2] and find its value?

I have a general idea of how to complete this question using partitions and the L(f,P) U(f,P) definition, but am not quite receiving the answer I would like...

My workings, Fix [itex]\epsilon[/itex][itex]\succ[/itex] 0 with partitions Pe = {0, 1-([itex]\epsilon[/itex] / 2), 1 + ([itex]\epsilon[/itex] / 2), 2}. Then, if U(f,P)=2, we can define: L(f,P)=1(1-[itex]\epsilon[/itex]/2)+1(1-[itex]\epsilon[/itex]/2) = 2-[itex]\epsilon[/itex] Then computing, U(f,P)-L(f,P)= 2-(2-[itex]\epsilon[/itex]) =ε[itex]\prec[/itex]ε

This doesn't seem right to me, and I am not quite sure what else to do? Any suggestions?

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# Real Analysis Riemann Integration

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