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I'm using the latest version : 2018.02.12 1.4.1.13441

1. Put the calculator in degree mode
2. In numeric mode evaluate an integral from 0 to 10 of sin(X^2)dX
3. The calculator reboot nearly instantaneous
   

It also crashes with the virtual calculator on Windows.

Hi!

Same for me... my calculator reboot.

Marcel

Funny enough that you tried to integrate a trigonometric function in degrees.
Arno

Hi,
Here, the angular mode is not the problem..
The prime don't have to reboot on this simple calculation.
Marcel

Mine crashes also. Interestingly, the same integral dosen't crash my 49G+ in degree
mode, but it does give an incorrect answer - 4.66682...
But the fact remains that the Prime should not crash simply because the degree
mode is used instead of radian mode. Definitely a bug.

John Colvin. My HP 50g got the right answer of 4.66829167156. I believe the 49G+ should get the right answer as well. Did you accidentally put in the lower and upper bound in the wrong order?

(03-08-2018 04:48 AM)Carsen Wrote: [ -> ]My HP 50g got the right answer of 4.66829167156.

Your answer is what the 50g gets in FIX 4 mode, leaving a pretty big value stored in IERR. STD mode takes a few seconds longer but returns 4.66829104623 with a much smaller IERR.

This bug is already fixed in source code. Until it is available in a new firmware, you can run int(sin(x^2),x,0,10.0)

(03-08-2018 05:04 AM)Joe Horn Wrote: [ -> ]

(03-08-2018 04:48 AM)Carsen Wrote: [ -> ]My HP 50g got the right answer of 4.66829167156.

Your answer is what the 50g gets in FIX 4 mode, leaving a pretty big value stored in IERR. STD mode takes a few seconds longer but returns 4.66829104623 with a much smaller IERR.

Huh. That's neat. I did not know about Integration Error (IERR) variable. I also didn't know (or forgot) that the number format changes the precision of the answer. Like the 15C. Learn something new everyday. Thanks Joe Horn.

I tried in on my SM42. It went into an indefinite loop. Even with accuracy of 0.1

(03-08-2018 02:41 PM)DA74254 Wrote: [ -> ]I tried in on my SM42. It went into an indefinite loop. Even with accuracy of 0.1

"SM42" or "DM42"?

Anyway, it is always better to experience a machine reset than an infinite loop, so in this regard the HP Prime wins hands down

(03-08-2018 03:04 PM)jebem Wrote: [ -> ]

(03-08-2018 02:41 PM)DA74254 Wrote: [ -> ]I tried in on my SM42. It went into an indefinite loop. Even with accuracy of 0.1

"SM42" or "DM42"?

Anyway, it is always better to experience a machine reset than an infinite loop, so in this regard the HP Prime wins hands down

SM DM42
Anyway, I was a bit quick as I set up sin (x^3) which went on and on. With the correct integration it spent abt 4 sec. to get 0.5836... in RAD and almost instantly 4.6682... in DEG mode. (And 4.3825... in GRAD mode)

(03-08-2018 04:48 AM)Carsen Wrote: [ -> ]John Colvin. My HP 50g got the right answer of 4.66829167156. I believe the 49G+ should get the right answer as well. Did you accidentally put in the lower and upper bound in the wrong order?

Am I missing something here? How is 4.6668.... the correct answer? If I convert
10 deg. to pi/18 red. in the upper boundary, I get a result of 0.001772.... on my
50G as well. A graph of sin(x^2) clearly indicates that in this interval, the area
under the curve is quite small.

(03-08-2018 08:25 PM)John Colvin Wrote: [ -> ]

(03-08-2018 04:48 AM)Carsen Wrote: [ -> ]John Colvin. My HP 50g got the right answer of 4.66829167156. I believe the 49G+ should get the right answer as well. Did you accidentally put in the lower and upper bound in the wrong order?

Am I missing something here? How is 4.6668.... the correct answer? If I convert
10 deg. to pi/18 red. in the upper boundary, I get a result of 0.001772.... on my
50G as well. A graph of sin(x^2) clearly indicates that in this interval, the area
under the curve is quite small.

Yes, 10_deg = pi/18_rad, but sin((10_deg)^2) is not the same as sin((pi/18_rad)^2). Plot the sin(x^2) from 0_deg to 10_deg and you'll see it. The integral from 9 to 10 alone is almost 1.

(03-08-2018 08:47 PM)Joe Horn Wrote: [ -> ]

(03-08-2018 08:25 PM)John Colvin Wrote: [ -> ]Am I missing something here? How is 4.6668.... the correct answer? If I convert
10 deg. to pi/18 red. in the upper boundary, I get a result of 0.001772.... on my
50G as well. A graph of sin(x^2) clearly indicates that in this interval, the area
under the curve is quite small.

Yes, 10_deg = pi/18_rad, but sin((10_deg)^2) is not the same as sin((pi/18_rad)^2). Plot the sin(x^2) from 0_deg to 10_deg and you'll see it. The integral from 9 to 10 alone is almost 1.

That''s what I missed, Joe. Thanks.