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the second derive is 2/x^3, which is decreasing on [ 1, 2 ]

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

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?

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