Question

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]

Answer 1

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

Answer 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

Answer 3

so | f '' (x) | < 2/1^3

Answer 4

you do need to when you cant find the antiderivative

Answer 5

like integral e^(-x^2) or integral sin (x^2)

Answer 6

That's when you reach for your handy TI-83, since it is an approximation anyway...

Answer 7

when you want the definite integral

Answer 8

suppose you work in industry and they need more precision than a TI 83

Answer 9

you want to bound your error

Answer 10

midpoint formula is not more precise than TI-83

Answer 11

yes it is , if n is large enough

Answer 12

Why not Simpson's rule, then?

Answer 13

i have no clue what the precision of TI 83 is, but i know the error of midpoint approximation

Answer 14

they all work, im just using midpoint at the moment

Answer 15

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

Answer 16

how precise to what decimal

Answer 17

9, I think

Answer 18

see what i mean

Answer 19

you have no idea. thats why we have those exact bound formulas , no guess work here

Answer 20

maybe 10 , maybe 8 , who knows?

Answer 21

Ti 83 is fine for quick work, not for sensitive industrial stuff

Answer 22

Okay, so how do you do this?

Answer 23

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)

Answer 24

ahhh, theres a formula http://cims.nyu.edu/~kiryl/teaching/c2/Calculus_II_NYU_Kiryl_Tsishchanka/Section_6.5--Approximate_Integration/Approximate_Integration.pdf

Answer 25

2/x^-3 on [1,2] = 2
So K<2

Answer 26

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

Answer 27

no, that would make the error more precise than it actually is