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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.

Physical Mathematics
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BesMath
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1 Answers1

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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.

Physical Mathematics
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  • but here the function is continuous. the page you refered me to is discussing of points of discontinuoity. – BesMath Apr 22 '20 at 13:26
  • 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. – Physical Mathematics Apr 22 '20 at 13:47
  • 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. – BesMath Apr 22 '20 at 13:51
  • See here. – Physical Mathematics Apr 22 '20 at 13:54
  • 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 – BesMath Apr 22 '20 at 14:01
  • Integrals are unchanged if you change the function by a point. Just extend the domain to be $[-1,1]$. – Physical Mathematics Apr 22 '20 at 14:05