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January 26, 2015

January 26, 2015

Posted by **anonymous** on Wednesday, January 30, 2008 at 9:58pm.

- CALC -
**drwls**, Wednesday, January 30, 2008 at 11:17pmThat is a slowly varying function over that interval. It increases from 1.3438 at x=3 to 1.6151 at x = 5. An approximate value for the integral is the mean value times 2, or 2.96.

I suggest you review and apply Simpson's Rule for numerical integration and use it with ten intervals (h = 0.2). There is a description of the method and a formula for its accuracy at

http://mathworld.wolfram.com/SimpsonsRule.html

If you get about the same value with four intervals as you get with ten, you can be quite sure the error will be comparable to the difference betrween the two calculations.

Using Simpson's Rule with four intervals, I get 2.9378 for the integral.

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