If anyone could explain how the following is done, it would be greatly appreciated!

Use Simpson's rule with delta x = .1 to obtain an approximation for the integral of cos(x^2) dx from .6 to 1.

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- Apr 25th 2007, 03:07 PMfaure72Simpson's Rule (Calculus)
If anyone could explain how the following is done, it would be greatly appreciated!

Use Simpson's rule with delta x = .1 to obtain an approximation for the integral of cos(x^2) dx from .6 to 1. - Apr 26th 2007, 02:00 PMCaptainBlack
Simpsons rule tells us that the integral from a to b of f(x) may be

approximated as follows.

Divide the interval [a,b] into n equal parts (n even) with end points:

a, a+h, a+2h, ... a+nh

where h=(b-a)/n. Then:

integral_{x=a to b} f(x) dx ~= [h/3] (f(a) + 4f(a+h) + 2f(a+2h) + 4f(a+3h) +

.................. ... 4f(a+(n-1)h) + f(a+nh))

In the case here we have a=0.6, b=1, h=0.1, so n=4

integral_{x=a to b} cos(x^2) dx = [0.1/3] [cos(0.6^2) + 4cos(0.7^2) + 2cos(0.8^2) + 4cos(0.9^2) + cos(1^2)]

............... = [0.033333..][0.936+4(0.882)+2(0.802)+4(0.689)+0.540]

...............~= 0.312

RonL