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Dieter · 06-06-2018, 11:40 AM

First of all: I have absolute no experience with RPL, so I cannot comment on your code. But maybe this helps get you on the right track.

(06-05-2018 07:51 PM)Vtile Wrote: I have tested this against wolfram alpha and so far it does give approx. right values, but only in 1 decimal accuracy. Another option is that Wolfram Alpha does calculate incorrectly as it gives "Standard computational time exceeded..." message.

I just entered "arc length of 2x*sin(10*x) x=1..2" and got a result with the integral of sqrt(1 + 4 (10 x cos(10 x) + sin(10 x))^2) dx and a numeric value of 19,628792649... The "Standard computation time exceeded..." message appears below the results, but this does not mean that these are incorrect.

(06-05-2018 07:51 PM)Vtile Wrote: I also did misuse the Parisses XCAS and copypasted the derivative of the f(x)=2x*Sin(10*x) from WA to XCAS and used integral there from 1 to 2 and the values between XCAS and WA do agree, while 50g with above program do not after the first digit. Any ideas?

What does the 50g return as the derivative?
What exactly do you get as a result for the integral?
What do you get if you manually calculate the integral of sqrt(1 + 4 (10 x cos(10 x) + sin(10 x))^2) ?

For the record: my 35s returns 19,64 ±0,20 in FIX 2 mode and 19,642 ±0,020 in FIX 3. Trying to get more accurate results with FIX 4 or more causes a very long calculation which I finally interrupted. Here the display setting controls the accuracy, but I don't know how this is handled on the 50g.

So the integral seems to be a bit tricky. ;-) I tried it on the WP34s. Here FIX 2 yielded 19,62876... with an error estimate of ~0,0002 while FIX 4 returned 19,6287948... with an error estimate of ~0,00003. Finally, FIX 6 returned 19,628792649 which agrees with the WA value above. So the 34s performed better here. The last values took a few seconds on the emulator, so a hardware 34s would have required quite some time to come up with these results – and it is much faster than the 35s.

(06-05-2018 07:51 PM)Vtile Wrote: EDIT1: Above program also generates at times IERR variable "Integral ERROR" maybe, unfortunately at this time of night I can not recall.

IERR is the estimated error of the result of a numeric integration, actually more like an upper bound of this error. So this tells you how accurate the 50g thinks the returned integral is. What value do you get here?

Dieter