3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Ok my approach is wrong, but im not sure how to fix it. $\endgroup$ – user113353 Dec 5 '13 at 8:14
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.