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Mathematics 15 Online
OpenStudy (anonymous):

use your choice of a) simpsons rule b) the trapezoidal rule or c) the midpoint rule to approximate pi using the equation pi= integral (4/(1+x^2)) from 0 to 1

OpenStudy (anonymous):

what was a?

OpenStudy (anonymous):

missing info

OpenStudy (amistre64):

which rule do you like best?

OpenStudy (anonymous):

that's all the info

OpenStudy (amistre64):

simpsons is a quadratic fit trap is an average rectangle fit midpoint is a rectangle fit

OpenStudy (anonymous):

midpoint rule i guess..whichever one is easier

OpenStudy (amistre64):

midpoint is simplest, is this spose to be limits, or are you spose to do an approximation?

OpenStudy (anonymous):

what choice di you use of a

OpenStudy (anonymous):

did*

OpenStudy (amistre64):

to approximate, we would want to define a subinterval length

OpenStudy (anonymous):

it says experiment with the number of subintervals until you obtain successive approximations that differ by less than 10^(-3)

OpenStudy (amistre64):

a is a simpson rule choice

OpenStudy (amistre64):

ah, then we will want to do this a few times :)

OpenStudy (amistre64):

since the interval is of length 1, then any division of it into subintervals is /n for n subintervals, agreed?

OpenStudy (amistre64):

*is 1/n for n ......

OpenStudy (amistre64):

so the sum of height times width defines the area of a rectangle the width is 1/n, the height is [i(1/n) + (i+1)(1/n)]/2 for some leftside i if that makes any sense

OpenStudy (amistre64):

hmmm, thats the trap on second thought

OpenStudy (amistre64):

|dw:1411584938262:dw|

OpenStudy (anonymous):

yes agreed

OpenStudy (amistre64):

so i + half the subinterval is the midpoint between i and i+1 so i+1/(2n) spose our subinterval is a length of 1/10 the midpoint to use for x in our function for the intercal from 7/10 to 8/10 is 7/10 + 1/20

OpenStudy (amistre64):

once we know the areas, we add them up

OpenStudy (amistre64):

f(x) = 4/(1+x^2) lets try a subinterval of length 1/10 f(0/10+1/20) * 1/10 +f(1/10+1/20) * 1/10 +f(2/10+1/20) * 1/10 +f(3/10+1/20) * 1/10 ... +f(9/10+1/20) * 1/10 ------------------- approx area of integration

OpenStudy (amistre64):

i spose i/10 + (i+1)/10 -------------- 2 is the midpoint/average between 2 x values with in a length of 1/10 2i + 1 ------ 20

OpenStudy (amistre64):

if you have excel, you can create the values quite easily

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