Analysis 2- Riemann integrable functions

perlawin
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1. If abs(f) is Riemann integrable on [a,b], then f is Riemann integrable on [a,b]. True or false (show work)



2. A function f is Riem Int iff f is bounded on [a,b], and for every epsilon>0 there is a partition P of [a,b] s.t. U(f,P)-L(f,P)<epsilon



3. I believe that this is true. So, what I want to do is show that f is bounded don [a,b], and I also want to show the second part of the definition. To show it was bounded, I used the fact that abs(f) was bounded and eventually got sup(abs(f))<=sup(f) and inf(f)>=-sup(abs(f)), which *I think* proved that f is bounded.

My question is how do I show that it fits the second part of the definition?
 
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You can't show the second part. Because it isn't true. Try to find a counterexample.
 
I'm not sure what the question is and what's your answer. However the statement "if abs(f) is Riemann integrable on [a,b] then f is Riemann integrable on [a,b] isn't true."
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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