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anonymous
 5 years ago
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 (ba)^3 / (24n^2) where f '' (x)  < = k on [ a, b]
anonymous
 5 years ago
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 (ba)^3 / (24n^2) where f '' (x)  < = k on [ a, b]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the second derive is 2/x^3, which is decreasing on [ 1, 2 ]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Man, 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
 5 years ago
Best ResponseYou've already chosen the best response.0so  f '' (x)  < 2/1^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you do need to when you cant find the antiderivative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like integral e^(x^2) or integral sin (x^2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's when you reach for your handy TI83, since it is an approximation anyway...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when you want the definite integral

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0suppose you work in industry and they need more precision than a TI 83

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you want to bound your error

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0midpoint formula is not more precise than TI83

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes it is , if n is large enough

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why not Simpson's rule, then?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have no clue what the precision of TI 83 is, but i know the error of midpoint approximation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they all work, im just using midpoint at the moment

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0how precise to what decimal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have no idea. thats why we have those exact bound formulas , no guess work here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0maybe 10 , maybe 8 , who knows?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ti 83 is fine for quick work, not for sensitive industrial stuff

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, so how do you do this?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh, theres a formula http://cims.nyu.edu/~kiryl/teaching/c2/Calculus_II_NYU_Kiryl_Tsishchanka/Section_6.5Approximate_Integration/Approximate_Integration.pdf

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02/x^3 on [1,2] = 2 So K<2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and K (ba)^3 / (24n^2) = 2(1)/(24n^2) = 10^3 > 10^3/12 = n^2 n = sqrt(250/3) = 9.128... so 10

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, that would make the error more precise than it actually is
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