I found a problem in a paper which has asked below two questions.


Let $f : [0,1] -> \mathbb{R} $ be the fucntion given by $f(x) = x^2+x $ for all $x\in [0,1]$

1) Use the partition $P=[0,1/4,1/2,3/4,1]$ to find an Upper bound and a lower bound for the area of the region lying between the graph $f(x)$ and the $x-$ axis.

2) Without calculating lower and upper Riemann integrals, can you check the riemann integrability of $f(x)$ on $[0,1]$? Give reasons.


As question has asked I tried to find rimann left and right sums which will give the upper riemann integral and lower one. After the calculation has done I found that Upper rimann sum is $1.09375$ and lower as $0.59375$.

MY Problem

There is only one partition is given. So obviously Upper rimann sum is equals to the Upper riemann integral and lower rimann sum is equals to lower riemann integral. (If I'm wrong correct me). But that says this function is not riemann integrable. Which makes me confused about my result. (I also tried to verify result by calculating sums using online calculator and got the same result)

And also Since this function is monotonically increasing in $[0,1]$ interval, definitely this should be riemann integrable (By Theory)

So please can someone suggest or explain me why is that or what I've done incorrectly? Thank you.


This is your problem:

But that says this function is not riemann integrable.

You correctly found that the upper and lower sums for this one partition are not equal (assuming your arithmetic is right - I didn't check). But when you look at other partitions you will find different bounds. When you consider all the partitions you'll find that the smallest number larger than all the lower sums is the same as the largest number smaller than all the upper sums. That's the definition of Riemann integrability.

  • $\begingroup$ So When the question asked to calculate upper and lower bounds I have to get riemann upper and lower integrals right? Please correct me if I'm wrong. $\endgroup$ – Samitha Nanayakkara Oct 22 '16 at 14:33
  • $\begingroup$ You answered question (1) correctly by finding those two numbers. You answered question (2) correctly by quoting "theory". My answer explains why there is no contradiction. $\endgroup$ – Ethan Bolker Oct 22 '16 at 14:39

The upper Riemann sum in your question is definitely not equal to the upper Riemann integral. Upper Riemann sums are merely upper bounds to the upper Riemann integral, which is in fact defined to be the infimum of all the upper Riemann sums over all possible partitions. Similarly for lower Riemann sums and the lower Riemann integral which by definition is the supremum of all the lower Riemann sums.

You cannot reach any conclusions to this problem by considering only one partition, you will have to more deeply study the upper and lower Riemann sums of more arbitrary partitions (or, perhaps, apply the theory).

  • $\begingroup$ So If I truely want to find the riemann integral then I have to take several some other partitions too. Then what I get is correct right? $\endgroup$ – Samitha Nanayakkara Oct 22 '16 at 14:31
  • $\begingroup$ QUESTION: Do not you think StackExchange (S.E.) removal in the publication of answers is inconvenient and should be returned to the previous mode in which each time that came an answer it was published as "answered by" in a "new" post ?. As things are now, after this recent change, few learn that a question unanswered before has been responded. What, in particular, with the unanswered questions (there are many ones and listed as "unanswered" in S.E.) if someone finds a solution? Who will read it? My respects. (Sorry for the bad English). $\endgroup$ – Piquito Oct 22 '16 at 14:47
  • $\begingroup$ @Piquito: I'm not clear on what you are asking. But from its format your question seems more relevant to meta.math.stackexchange.com than to math.stackexchange.com. $\endgroup$ – Lee Mosher Oct 22 '16 at 15:10
  • $\begingroup$ Thank you very much for "meta.math" link where i will publish the same question (with bad English...). $\endgroup$ – Piquito Oct 22 '16 at 15:15

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