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the second derive is 2/x^3, which is decreasing on [ 1, 2 ]
Man, instead of counting rectangles, ask your teacher why you should count rectangles when you won't need to again after a week when you are introduced to integrals
so | f '' (x) | < 2/1^3
you do need to when you cant find the antiderivative
like integral e^(-x^2) or integral sin (x^2)
That's when you reach for your handy TI-83, since it is an approximation anyway...
when you want the definite integral
suppose you work in industry and they need more precision than a TI 83
you want to bound your error
midpoint formula is not more precise than TI-83
yes it is , if n is large enough
Why not Simpson's rule, then?
i have no clue what the precision of TI 83 is, but i know the error of midpoint approximation
they all work, im just using midpoint at the moment
you asked why not just use TI 83, because TI 83 cannot be made as precise as you want. it has a limit, which i dont even know what it is , what decimal
how precise to what decimal
9, I think
see what i mean
you have no idea. thats why we have those exact bound formulas , no guess work here
maybe 10 , maybe 8 , who knows?
Ti 83 is fine for quick work, not for sensitive industrial stuff
Okay, so how do you do this?
i dont know, in math we like to talk about making the error as small as we please, thats important because you want or expect the error to go to zero . if the error does not go to zero then there is no definitive answer (divergent)
ahhh, theres a formula http://cims.nyu.edu/~kiryl/teaching/c2/Calculus_II_NYU_Kiryl_Tsishchanka/Section_6.5--Approximate_Integration/Approximate_Integration.pdf
2/x^-3 on [1,2] = 2 So K<2
and K (b-a)^3 / (24n^2) = 2(1)/(24n^2) = 10^-3 --> 10^3/12 = n^2 n = sqrt(250/3) = 9.128... so 10
no, that would make the error more precise than it actually is