Gavin's Calc II Class Clarification: Sep 13

There are two possibilities for how we choose n, the number of subdivisions to use in a numerical approximation of an integral: either we pick a value based on practical restrictions (e.g., if we're doing it by hand we probably don't want 100s of boxes) or intuition (e.g., we may be willing to say "2000 rectangles darn well ought to be good enough").

The other possibility is that we start off saying "we ought to have an answer accurate to within (say) 0.01 of the actual answer." In this case, we can figure out approximately how large n will have to be just by doing two numerical calculations and knowing how the error for the numerical method changes with increasing n. Let's do an example: suppose we use Simpson's rule, for which error drops like n⁴, with n=10 find the value of the integral is 0.613 and with n=50 that it's 0.638. If the error in the first calculation is E, then the error in the second ought to be E/5⁴, and adding the respective errors to both should give the actual value of the integral:

E + 0.613 = E/625 + 0.638 Solving gives E=0.025, so that the error in the second calculation is approximately 0.00004, and we can be pretty sure that it's within 0.01 of the actual value.