Let $f(x) = 1$ for $x > 0$ and $f(0)=0$.

How do you set up the upper and lower Riemann sums on the domain $[0,1]$?

If you divide it into $n$ partitions with each having a length of $1\over n$, then is the set up the sum of $n* \frac1n$ as $n$ goes from $1$ to infinity?

Are the upper and lower sums supposed to have the same set up?

  • $\begingroup$ Ok my approach is wrong, but im not sure how to fix it. $\endgroup$ – user113353 Dec 5 '13 at 8:14

when we divide it into n partition each of width $\frac{1}{n}$ and taking maximun value in each partition we will $$I=\lim_{n\to \infty}{\sum_{i=1}^{i=n} \frac{1}{n}}=\lim_{n\to \infty}1=1$$Also taking min value in each partion we will get $$I=\lim_{n\to \infty}{\frac{1}{n}*0+\sum_{i=2}^{i=n} \frac{1}{n}}=\lim_{n\to \infty}\frac{n-1}{n}=1$$Hence upper limit and lower limit are equal.Hence Reiman $sum=1$$$$$So basically for function with discontinuty split the sum same as splitting integrals


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