Timeline for Why the integral is zero
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2015 at 14:38 | comment | added | Arsenal | @Chu I can't say much more, maybe it helps if you think how large the area of a half cycle is with increasing frequency. It will get smaller and smaller. (Maybe this helps a bit) | |
| Dec 3, 2015 at 12:38 | comment | added | Chu | @Arsenal, I'm not sure. sin(wT) could occur any part of the sinusoid so the accumulated area could as much as +/- a half cycle. Without a 1/T coefficient, there's no averaging | |
| Dec 3, 2015 at 12:19 | comment | added | Arsenal | @Chu yeah that was a bit short-sighted in hindsight. I've updated my answer to make the point clear. It cannot be a long way from zero for higher omegas. | |
| Dec 3, 2015 at 12:18 | history | edited | Arsenal | CC BY-SA 3.0 |
update because of comments
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| Dec 3, 2015 at 11:02 | comment | added | Chu | @Arsenal, but you've assumed a value for T. There is no such specification in the original question - both w and T are free to wander. So the integral could be a long way from zero | |
| Dec 3, 2015 at 10:36 | comment | added | Arsenal | @Chu I'm not saying that it will be 0, it just tends to be very close to 0, so close that for practical purposes it can be neglected (this is a common simplification to make things solvable for humans). FMarazzi has actually given a better analysis of the upper bound of the result. | |
| Dec 3, 2015 at 10:00 | comment | added | Chu | If T is arbitrary, the area under sin(wt) will, generally, be non-zero. There must be another constraint. | |
| Dec 3, 2015 at 9:11 | comment | added | user59419 | Thanks. Your question definitely makes sense and that's exactly my problem because range of T and w is not given and only condition that wT>>1 is mentioned. I was thinking what if T=1000 and w=1 then the integral is not zero. | |
| Dec 3, 2015 at 9:07 | history | answered | Arsenal | CC BY-SA 3.0 |