Real Analysis Riemann Integration

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SUMMARY

The function f(x) defined as f(x) = 1 for 0 ≤ x ≤ 1 and f(x) = 2 for 1 ≤ x ≤ 2 is Riemann integrable on the interval [0, 2]. By applying the definition of Riemann integrability using partitions, it is established that the upper sum U(f,P) is 2 and the lower sum L(f,P) approaches 2 - ε as ε approaches 0. The integral can be computed using the additivity property, yielding the result ∫02 f(x) dx = ∫01 f(x) dx + ∫12 f(x) dx, effectively resolving the issue at x = 1.

PREREQUISITES
  • Understanding of Riemann integration concepts
  • Familiarity with partitions and upper/lower sums
  • Knowledge of the additivity property of integrals
  • Basic calculus and limit concepts
NEXT STEPS
  • Study the properties of Riemann integrable functions
  • Learn about the construction and significance of partitions in integration
  • Explore the concept of upper and lower sums in detail
  • Investigate the implications of discontinuities in Riemann integration
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Students of real analysis, mathematics educators, and anyone seeking to deepen their understanding of Riemann integration and its applications in calculus.

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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
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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You could simply use additivity: ##\int_0^2=\int_0^1+\int_1^2## and the problem at ##x=1## will be solved.
 

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