anonymous
  • anonymous
how large should n be so that the midpoint rule approximation of definite integral 1 / x on [1,2] is accurate to within 10^-3 . hint | Em| < = K (b-a)^3 / (24n^2) where |f '' (x) | < = k on [ a, b]
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
the second derive is 2/x^3, which is decreasing on [ 1, 2 ]
anonymous
  • anonymous
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
anonymous
  • anonymous
so | f '' (x) | < 2/1^3

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anonymous
  • anonymous
you do need to when you cant find the antiderivative
anonymous
  • anonymous
like integral e^(-x^2) or integral sin (x^2)
anonymous
  • anonymous
That's when you reach for your handy TI-83, since it is an approximation anyway...
anonymous
  • anonymous
when you want the definite integral
anonymous
  • anonymous
suppose you work in industry and they need more precision than a TI 83
anonymous
  • anonymous
you want to bound your error
anonymous
  • anonymous
midpoint formula is not more precise than TI-83
anonymous
  • anonymous
yes it is , if n is large enough
anonymous
  • anonymous
Why not Simpson's rule, then?
anonymous
  • anonymous
i have no clue what the precision of TI 83 is, but i know the error of midpoint approximation
anonymous
  • anonymous
they all work, im just using midpoint at the moment
anonymous
  • anonymous
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
anonymous
  • anonymous
how precise to what decimal
anonymous
  • anonymous
9, I think
anonymous
  • anonymous
see what i mean
anonymous
  • anonymous
you have no idea. thats why we have those exact bound formulas , no guess work here
anonymous
  • anonymous
maybe 10 , maybe 8 , who knows?
anonymous
  • anonymous
Ti 83 is fine for quick work, not for sensitive industrial stuff
anonymous
  • anonymous
Okay, so how do you do this?
anonymous
  • anonymous
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)
anonymous
  • anonymous
ahhh, theres a formula http://cims.nyu.edu/~kiryl/teaching/c2/Calculus_II_NYU_Kiryl_Tsishchanka/Section_6.5--Approximate_Integration/Approximate_Integration.pdf
anonymous
  • anonymous
2/x^-3 on [1,2] = 2 So K<2
anonymous
  • anonymous
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
anonymous
  • anonymous
no, that would make the error more precise than it actually is

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