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Joe Horn · (This post was last modified: 03-08-2018 08:54 PM by Joe Horn.)

(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.

[Image: int10.png]