Quiz 2 ANDREI - Quiz 2 Problem 1 Composite Simpson's Rule...

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Quiz 2 Luarsab Imnaishvili Problem 1 Composite Simpson's Rule f(x) = x^3 ln(x) a = 13.5 b = 14.5 Accuracy 0.000010 h < 0.212132 n > 4.714045 h = 0.125000 n = 8.000000 counter node value weight term 0 13.500000 6403.592635 1 6403.592635 1 13.625000 6606.434485 4 26425.737940 2 13.750000 6813.677099 2 13627.354199 3 13.875000 7025.372624 4 28101.490497 4 14.000000 7241.573312 2 14483.146625 5 14.125000 7462.331522 4 29849.326088 6 14.250000 7687.699716 2 15375.399432 7 14.375000 7917.730461 4 31670.921844 8 14.500000 8152.476426 1 8152.476426 Approximation: 7253.726904 Actual Value: 7253.73 Error: 0.00000058
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Problem 2 Simpson Multiple Integral n = 3 m = 3 f(x) = x/y b = 14.5 d(x) = 16.5x h=(b-a)/2n a = 13.5 c(x) = 15.5x 0.166667 grid x0 x1 x2 x3 x4 x5 x 13.500000 13.666667 13.833333 14.000000 14.166667 14.333333 c(x) 209.250000 211.833333 214.416667 217.000000 219.583333 222.166667 d(x) 222.750000 225.500000 228.250000 231.000000 233.750000 236.500000 k(x) = (d-c)/2m 2.250000 2.277778 2.305556 2.333333 2.361111 2.388889 y0 209.250000 211.833333 214.416667
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This note was uploaded on 12/14/2009 for the course BUSINESS bus 203c taught by Professor Milankas. during the Spring '09 term at American University in Bulgaria.

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Quiz 2 ANDREI - Quiz 2 Problem 1 Composite Simpson's Rule...

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