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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 (b-a)^3 / (24n^2) where |f '' (x) | < = k on [ a, b]

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  1. anonymous
    • 5 years ago
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    the second derive is 2/x^3, which is decreasing on [ 1, 2 ]

  2. anonymous
    • 5 years ago
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    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

  3. anonymous
    • 5 years ago
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    so | f '' (x) | < 2/1^3

  4. anonymous
    • 5 years ago
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    you do need to when you cant find the antiderivative

  5. anonymous
    • 5 years ago
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    like integral e^(-x^2) or integral sin (x^2)

  6. anonymous
    • 5 years ago
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    That's when you reach for your handy TI-83, since it is an approximation anyway...

  7. anonymous
    • 5 years ago
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    when you want the definite integral

  8. anonymous
    • 5 years ago
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    suppose you work in industry and they need more precision than a TI 83

  9. anonymous
    • 5 years ago
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    you want to bound your error

  10. anonymous
    • 5 years ago
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    midpoint formula is not more precise than TI-83

  11. anonymous
    • 5 years ago
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    yes it is , if n is large enough

  12. anonymous
    • 5 years ago
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    Why not Simpson's rule, then?

  13. anonymous
    • 5 years ago
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    i have no clue what the precision of TI 83 is, but i know the error of midpoint approximation

  14. anonymous
    • 5 years ago
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    they all work, im just using midpoint at the moment

  15. anonymous
    • 5 years ago
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    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

  16. anonymous
    • 5 years ago
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    how precise to what decimal

  17. anonymous
    • 5 years ago
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    9, I think

  18. anonymous
    • 5 years ago
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    see what i mean

  19. anonymous
    • 5 years ago
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    you have no idea. thats why we have those exact bound formulas , no guess work here

  20. anonymous
    • 5 years ago
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    maybe 10 , maybe 8 , who knows?

  21. anonymous
    • 5 years ago
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    Ti 83 is fine for quick work, not for sensitive industrial stuff

  22. anonymous
    • 5 years ago
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    Okay, so how do you do this?

  23. anonymous
    • 5 years ago
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    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)

  24. anonymous
    • 5 years ago
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    2/x^-3 on [1,2] = 2 So K<2

  25. anonymous
    • 5 years ago
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    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

  26. anonymous
    • 5 years ago
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    no, that would make the error more precise than it actually is

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