(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.