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
what was a?
missing info
which rule do you like best?
that's all the info
simpsons is a quadratic fit trap is an average rectangle fit midpoint is a rectangle fit
midpoint rule i guess..whichever one is easier
midpoint is simplest, is this spose to be limits, or are you spose to do an approximation?
what choice di you use of a
did*
to approximate, we would want to define a subinterval length
it says experiment with the number of subintervals until you obtain successive approximations that differ by less than 10^(-3)
a is a simpson rule choice
ah, then we will want to do this a few times :)
since the interval is of length 1, then any division of it into subintervals is /n for n subintervals, agreed?
*is 1/n for n ......
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
hmmm, thats the trap on second thought
|dw:1411584938262:dw|
yes agreed
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
once we know the areas, we add them up
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
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
if you have excel, you can create the values quite easily
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