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If $f(x)= \begin{cases} x &-1<x\le0 \\ x^2 &0<x\le1\\ \end{cases} $ then prove $f$ is integrable on $[-1,1]$

I am studying this topic, and my textbook has no example how to partition a piecewise function. I know the formulas for lower sums and upper sums, but I need some help with partitioning and deciding for component intervals, please.

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You can use this result to easily prove your proposition. It is worth thinking about this construction in general, but also how it would be specifically applied to you problem.

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  • $\begingroup$ but here the function is continuous. the page you refered me to is discussing of points of discontinuoity. $\endgroup$
    – BesMath
    Apr 22 '20 at 13:26
  • $\begingroup$ Huh, I didn't even notice that it was continuous (I assumed the different pieces introduced a discontinuity). You should try to prove that every continuous function is integrable. $\endgroup$ Apr 22 '20 at 13:47
  • $\begingroup$ Well that is what I had in mind. even choosing a function g on a broader interval to show f is differentable on all [-1,1]. but I believe I am supposed to use upper and lower sums. so there is no way to partition such interval? I ve been googling this for hours and I can not see any similar problem to see how this proof might look like. $\endgroup$
    – BesMath
    Apr 22 '20 at 13:51
  • $\begingroup$ See here. $\endgroup$ Apr 22 '20 at 13:54
  • $\begingroup$ the behaviour of function here is know on (-1,1], the question is asking wether it is integrable on [-1,1]. am I supposed to consider this at all? thanks $\endgroup$
    – BesMath
    Apr 22 '20 at 14:01

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