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Okay, so I have a hard time understanding this math problem:

Given

$$[f(x)] = \begin{cases} 5 & \quad \text{if } x \text{ <3}\\ 7 & \quad \text{if } x \geq3 \end{cases} $$ were the partitioning is $$P_n=[0,3-\frac{1}{n},3+\frac{1}{n},4]$$ where $$n\in N$$ and $$I=[0,4]$$ make the graph to f and calculate $$L(f P_n) , U(f,P_n)$$ But I don't understand how the formula for $$P_n$$ works. Could someone give me an example of how to use it?

EDIT: How do I find out if f(x) is integrable or not? Is this right:

Since epsilon must be bigger than $$0$$ (or I assume so), and $$U(f,P_n)-L(f,P_n)=-\frac{4}{n}<\epsilon$$ the function must be integrable on [0,4]

Mathomat55
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    Find the $\sup$ and $\inf$ of $f$ on each subinterval of the partition, for example, on the first interval $[1,3-{ 1\over n}]$ we have the $\sup$ and $\inf$ equal to $5$. – copper.hat Feb 17 '20 at 20:39
  • I have only started with the Riemann integral. My problem is understand how the formula works. I guess n is the number of intervalls I want? So if I have $$P_6$$. What will this formula then tell us to do? – Mathomat55 Feb 17 '20 at 20:44
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    No. each $n$ defines a partition $P_n$ as above. There is a formula for $L(f,P)$ and $U(f,P)$. – copper.hat Feb 17 '20 at 20:50
  • Ohh, so are there only 4 intervals? But the partition of the intervals depend on the n you choose? – Mathomat55 Feb 17 '20 at 21:09
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    That is the case with this example. Usually one needs a sequence of partitions that have more and more points as it refines. In this particular example, because the function is piecewise constant, a partition with a fixed number of points is sufficient. But that is unusual. – copper.hat Feb 17 '20 at 21:25
  • Okay, I see. Thank you so much!!! But I do get confused with the n. I have only worked with tasks where I was given the partition in the text. The partition always included epsilon. So it was easy to calculate afterwards that $$U(f,P_n)-L(f,P_n)<\epsilon$$ But how do I work around the $$n$$? – Mathomat55 Feb 17 '20 at 22:05

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Let $x_0=0,\,x_1=3-\frac{1}{n},\,x_2=3+\frac{1}{n},\,x_3=4$. Then by definition $$L(P_n,f)=\sum_{m=1}^3\inf_{x\in[x_{m-1},x_m]} f(x)(x_m-x_{m-1}),\text{ and } U(P_n,f)=\sum_{m=1}^3\sup_{x\in[x_{m-1},x_m]} f(x)(x_m-x_{m-1})$$. Hence computing the supremum and infimum accordingly, we get $$L(P_n,f)=5\Big(3-\frac{1}{n}\Big)+5\Big(\frac{2}{n}\Big)+7\Big(1-\frac{1}{n}\Big)=22-\frac{2}{n},$$ $$U(P_n,f)=5\Big(3-\frac{1}{n}\Big)+7\Big(\frac{2}{n}\Big)+7\Big(1-\frac{1}{n}\Big)=22+\frac{2}{n}.$$

Surajit
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  • I saw that I wrote down the first digit wrong, So it is supposed to be $$P_n=[0,3-1/n..]$$ But I have done the exact same thing as you with the right numbers. So I got 22 instead of 17 but also have 2/n. But when proving that the function is intgrabel or not, I have to prove by using $$U(f,P_n)-L(f,P_n)<\epsilon$$. But I don't have epsilon. – Mathomat55 Feb 17 '20 at 22:17
  • Since epsilon must be bigger that $$0$$ (or I assume so), and $$U(f,P_n)-L(f,P_n)=-\frac{4}{n}<\epsilon.$$ Is that enough to say that the function is integrable? – Mathomat55 Feb 17 '20 at 22:22
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    Sorry to say, but if you want to prove the integrability, then this is not the correct approach! Please edit the question accordingly. – Surajit Feb 17 '20 at 22:23
  • Off course I will! I got carried away. In the beginning I just wanted to understand the formula. And when I tried to solve the rest of the task to see if it was integrable or not, I forgot to edit my equestion before asking more! – Mathomat55 Feb 17 '20 at 22:28