Timeline for Prove $\int_{a}^{b}{f(x)dx}>0$.
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 27, 2015 at 20:20 | comment | added | Aaron Maroja | @Timbuc No problem, note taken. Even though I'm pretty sure the word hint gives you that. | |
| Mar 27, 2015 at 20:19 | comment | added | Timbuc | @AaronMaroja Fair request from you, Aaron. But maybe next time you will also tell that to whomever asks you something. Had you written that you're waiting some time to pass for the OP to think on this or that and I would have waited. Thank you. | |
| Mar 27, 2015 at 20:16 | comment | added | Aaron Maroja | @Timbuc I just don't agree. Using the hypothesis is essential on that part. You wouldn't have a $\delta$ to work with in the first place, and for some bad sketching of yours you just couldn't see that and were asking something completely out of question. All I'm asking is: be patient next time and don't spoil all the fun. | |
| Mar 27, 2015 at 20:10 | comment | added | Timbuc | @AaronMaroja Oh, don't worry about Meitar: she seems to be out of the site for quite some time now. And I don't think your hint would have been hurt if you had answered my question before | |
| Mar 27, 2015 at 20:08 | comment | added | Aaron Maroja | @Timbuc You know, it's really important for the student to think over a problem for quite some time, that's why I just gave a hint, and that's precisely the reason I didn't address to your question. It makes you mature faster, and you totally took this learning time from Meitar. Anyways, it's cool. | |
| Mar 27, 2015 at 19:56 | comment | added | Timbuc | @AaronMaroja So sorry for your points, but I almost never downvote (not even if the answer has some big mistake) , and neither did this time. I think downloading is mostly for losers, in particular if they do it without explaining. And thanks for the invitation to analysis 101, though it took you quite some time to address my question. | |
| Mar 27, 2015 at 19:35 | comment | added | Aaron Maroja | @MeitarAbarbanel It's the lower sum. | |
| Mar 27, 2015 at 19:17 | comment | added | Aaron Maroja | @Timbuc Okay, let's get back to real analysis 101. If we take $\epsilon = \frac{f(x_0)}{2} > 0$ then there surely exists $\delta > 0$ (due to the continuity of $f$ at $x_0$) such that $$x \in (x_0 -\delta, x_0 + \delta) \implies f(x) > f(x_0) -\frac{f(x_0)}{2} = \frac{f(x_0)}{2}$$ Are you pleased now? Can I have my points back? | |
| Mar 27, 2015 at 19:05 | comment | added | Timbuc | @AaronMaroja You still continue to assume $\;m=f(x_0)\;$ is some kind of minimal point of $\;f\;$ somewhere. Do you care to explain this? | |
| Mar 27, 2015 at 19:01 | history | edited | Aaron Maroja | CC BY-SA 3.0 | deleted 2 characters in body |
| Mar 27, 2015 at 18:57 | comment | added | Aaron Maroja | Make sure you have tried something first, before looking the answer. | |
| Mar 27, 2015 at 18:56 | history | edited | Aaron Maroja | CC BY-SA 3.0 | added 463 characters in body |
| Mar 27, 2015 at 18:55 | comment | added | Timbuc | @AaronMaroja I just can't understand that. What if $\;m:=f(x_0)\;$ is the maximum of $\;f\;$ on $\;[a,b]\;$ ? Or even better: what if $\;f(x_0)\;$ is not a local extreme point of $\;f\;$ at all?! | |
| Mar 27, 2015 at 18:50 | comment | added | Meitar Abarbanel | I don't really understand what $L(f:P). | |
| Mar 27, 2015 at 18:40 | comment | added | Timbuc | You seem to be assuming $\;x_0\;$ is a minimal point of $\;\frac{f(x)}2\;$ . Why? | |
| Mar 27, 2015 at 18:34 | history | answered | Aaron Maroja | CC BY-SA 3.0 |