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Infinite definite integral - lrdheat - 11-08-2014 11:40 PM
On the WP 34S, if consecutive screen updates increase by what appears to be a constant increment, can I take this to mean that the integral is infinite? If the screen updates are by smaller and smaller increments, does this mean that the integral coverges? Or are there exceptions to both cases? What prompts this is int(x/(1-sqrt(x)) from 0 to 1. This should be infinite. My WP 34S (at least my WP 34S app on my I-pad) stops at 37.8151, was increasing with increments of ~2.77. I'm amazed at how close to 1 one can get, and still have a fairly small area, far from infinity....from 0 to 1-1EE12 produces something 50 or 60ish! RE: Infinite definite integral - lrdheat - 11-08-2014 11:51 PM
The 52 ish answer for that interval from my Prime... RE: Infinite definite integral - Paul Dale - 11-09-2014 12:03 AM
(11-08-2014 11:40 PM)lrdheat Wrote: On the WP 34S, if consecutive screen updates increase by what appears to be a constant increment, can I take this to mean that the integral is infinite? If the screen updates are by smaller and smaller increments, does this mean that the integral coverges? Or are there exceptions to both cases? No, no and yes - Pauli RE: Infinite definite integral - Thomas Klemm - 11-09-2014 05:22 AM
(11-08-2014 11:40 PM)lrdheat Wrote: What prompts this is int(x/(1-sqrt(x))) from 0 to 1. \(\int\frac{x}{1-\sqrt{x}}dx=-\frac{1}{3}\sqrt{x}(2x+3\sqrt{x}+6)-2\log(1-\sqrt{x})+C\) Quote:I'm amazed at how close to 1 one can get, and still have a fairly small area, far from infinity.... That's due to the last term: \(-2\log(1-\sqrt{x})\) For \(x=1-\epsilon\) we get: \(-2\log(1-\sqrt{1-\epsilon})\approx-2\log(1-(1-\frac{1}{2}\epsilon))=-2\log(\frac{1}{2}\epsilon)\) This grows slowly as \(\epsilon \to 0\). Quote:from 0 to 1-1EE12 produces something 50 or 60ish! With \(\epsilon=10^{-12}\) we get approximately: \(-\frac{11}{3}+2(\log(2)+12\log(10))\approx52.982\) Kind regards Thomas |