28 March 2023 7:53:05.163 AM LINE_NCC_RULE_TEST(): FORTRAN90 version: Test LINE_NCC_RULE(). TEST01 LINE_NCC_RULE computes the Newton-Cotes (closed) rule using N equally spaced points for an interval [A,B]. Newton-Cotes (Closed) Rule # 1 I X(I) W(I) 1 0.00000 2.00000 Sum(|W)|) = 2.00000 Newton-Cotes (Closed) Rule # 2 I X(I) W(I) 1 -1.00000 1.00000 2 1.00000 1.00000 Sum(|W)|) = 2.00000 Newton-Cotes (Closed) Rule # 3 I X(I) W(I) 1 -1.00000 0.333333 2 0.00000 1.33333 3 1.00000 0.333333 Sum(|W)|) = 2.00000 Newton-Cotes (Closed) Rule # 4 I X(I) W(I) 1 -1.00000 0.250000 2 -0.333333 0.750000 3 0.333333 0.750000 4 1.00000 0.250000 Sum(|W)|) = 2.00000 Newton-Cotes (Closed) Rule # 5 I X(I) W(I) 1 -1.00000 0.155556 2 -0.500000 0.711111 3 0.00000 0.266667 4 0.500000 0.711111 5 1.00000 0.155556 Sum(|W)|) = 2.00000 Newton-Cotes (Closed) Rule # 6 I X(I) W(I) 1 -1.00000 0.131944 2 -0.600000 0.520833 3 -0.200000 0.347222 4 0.200000 0.347222 5 0.600000 0.520833 6 1.00000 0.131944 Sum(|W)|) = 2.00000 Newton-Cotes (Closed) Rule # 7 I X(I) W(I) 1 -1.00000 0.976190E-01 2 -0.666667 0.514286 3 -0.333333 0.642857E-01 4 0.00000 0.647619 5 0.333333 0.642857E-01 6 0.666667 0.514286 7 1.00000 0.976190E-01 Sum(|W)|) = 2.00000 Newton-Cotes (Closed) Rule # 8 I X(I) W(I) 1 -1.00000 0.869213E-01 2 -0.714286 0.414005 3 -0.428571 0.153125 4 -0.142857 0.345949 5 0.142857 0.345949 6 0.428571 0.153125 7 0.714286 0.414005 8 1.00000 0.869213E-01 Sum(|W)|) = 2.00000 Newton-Cotes (Closed) Rule # 9 I X(I) W(I) 1 -1.00000 0.697707E-01 2 -0.750000 0.415379 3 -0.500000 -0.654674E-01 4 -0.250000 0.740459 5 0.00000 -0.320282 6 0.250000 0.740459 7 0.500000 -0.654674E-01 8 0.750000 0.415379 9 1.00000 0.697707E-01 Sum(|W)|) = 2.90243 Newton-Cotes (Closed) Rule #10 I X(I) W(I) 1 -1.00000 0.637723E-01 2 -0.777778 0.351362 3 -0.555556 0.241071E-01 4 -0.333333 0.431786 5 -0.111111 0.128973 6 0.111111 0.128973 7 0.333333 0.431786 8 0.555556 0.241071E-01 9 0.777778 0.351362 10 1.00000 0.637723E-01 Sum(|W)|) = 2.00000 Newton-Cotes (Closed) Rule #11 I X(I) W(I) 1 -1.00000 0.536683E-01 2 -0.800000 0.355072 3 -0.600000 -0.162087 4 -0.400000 0.909893 5 -0.200000 -0.870310 6 0.00000 1.42753 7 0.200000 -0.870310 8 0.400000 0.909893 9 0.600000 -0.162087 10 0.800000 0.355072 11 1.00000 0.536683E-01 Sum(|W)|) = 6.12959 Newton-Cotes (Closed) Rule #12 I X(I) W(I) 1 -1.00000 0.498665E-01 2 -0.818182 0.309711 3 -0.636364 -0.743385E-01 4 -0.454545 0.579317 5 -0.272727 -0.220356 6 -0.909091E-01 0.355801 7 0.909091E-01 0.355801 8 0.272727 -0.220356 9 0.454545 0.579317 10 0.636364 -0.743385E-01 11 0.818182 0.309711 12 1.00000 0.498665E-01 Sum(|W)|) = 3.17878 TEST02 Use a sequence of NCC rules to compute an estimate Q of the integral: I = integral ( 0 <= x <= 1 ) exp(x) dx. The exact value is: I = 1.71828 N Q |Q-I| 1 1.64872 0.695606E-01 2 1.85914 0.140859E+00 3 1.71886 0.579323E-03 4 1.71854 0.258325E-03 5 1.71828 0.859466E-06 6 1.71828 0.484531E-06 7 1.71828 0.105856E-08 8 1.71828 0.650492E-09 9 1.71828 0.150013E-10 10 1.71828 0.559703E-10 11 1.71828 0.244095E-09 12 1.71828 0.464967E-08 13 1.71828 0.113316E-07 14 1.71828 0.137356E-06 15 1.71828 0.853077E-06 16 1.71828 0.223641E-05 17 1.71825 0.268474E-04 18 1.71836 0.802021E-04 19 1.71663 0.165085E-02 20 1.70869 0.959075E-02 21 1.91569 0.197412E+00 22 3.09691 0.137863E+01 LINE_NCC_RULE_TEST Normal end of execution. 28 March 2023 7:53:05.164 AM