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I posted this question yesterday, but it could have been written better and it was a bit "disorderly". So I try to post it again:

Given

$$f(x) = \begin{cases} 5 & \quad \text{if } x \text{ <3}\\ 7 & \quad \text{if } x \geq3 \end{cases}$$ were the partitioning is $$Pn=[0,3−\frac{1}{n},3+\frac{1}{n},4]$$ where $$n∈N$$ and $$I=[0,4]$$ make the graph to f and calculate $$L(fPn),U(f,Pn)$$Find out if f(x) is integrable or not

WHAT I HAVE DONE:

$$L(f,P_n)=5(3-\frac{1}{n}-0)+5(3+\frac{1}{n}-(3-\frac{1}{n}))+7(4-(3+\frac{1}{n}))=22-\frac{2}{n}$$

$$U(f,P_n)=5(3-\frac{1}{n}-0)+7(3+\frac{1}{n}-(3-\frac{1}{n}))+7(4-(3+\frac{1}{n}))=22+\frac{2}{n}$$

To prove that a function is integrable on [a,b] you use this inequality $$U(f,P_n)-L(f,P_n)<\epsilon$$

So I get that

$$U(f,P_n)-L(f,P_n)=\frac{4}{n}$$ But I don't know how to continue from here. I have only done tasks where I have been given a partition expressed with epsilon beforehand in the text. But here I get an expression with n. And I don't understand how to prove if this inequality holds or not$$\frac{4}{n}<\epsilon$$

Mathomat55
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    by the Archimedean property, for all positive $\epsilon\in\mathbb R$, there exists $n$ such that $\dfrac4n<\epsilon$ – J. W. Tanner Feb 18 '20 at 19:39
  • Hmm, I read the link. But I don't understand so much. I do find it hard to read math, so that's probably why. So you mean by that property that the inequality hold for every $$\epsilon>0$$ ?And therefor the the function is integrable on I? – Mathomat55 Feb 18 '20 at 19:58

1 Answers1

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You should choose $ n$ such that $$\frac 4n<\epsilon$$

or equivalently,

$$n>\frac{4}{\epsilon}$$

so, you can take for example

$$n = \lfloor \frac{4}{\epsilon}\rfloor +1$$

for example, if $ \epsilon =10^{-2}$, then $n = 401$.

$f$ is then integrable at $ [0,4]$ since for any given $\epsilon>0$, there exists a Partition $P_n$ (you gave), with $n = \lfloor \frac{4}{\epsilon}\rfloor +1$, such that

$$U(f,P_n)-L(f,P_n)<\epsilon$$

hamam_Abdallah
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