15-Feb-2019 14:25:56 line_ncc_rule_test MATLAB version: Test line_ncc_rule. 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 1 2 Sum(|W)|) = 2 Newton-Cotes (Closed) Rule #2 I X(I) W(I) 1 -1 1 2 1 1 Sum(|W)|) = 2 Newton-Cotes (Closed) Rule #3 I X(I) W(I) 1 -1 0.333333 2 0 1.33333 3 1 0.333333 Sum(|W)|) = 2 Newton-Cotes (Closed) Rule #4 I X(I) W(I) 1 -1 0.25 2 -0.333333 0.75 3 0.333333 0.75 4 1 0.25 Sum(|W)|) = 2 Newton-Cotes (Closed) Rule #5 I X(I) W(I) 1 -1 0.155556 2 -0.5 0.711111 3 0 0.266667 4 0.5 0.711111 5 1 0.155556 Sum(|W)|) = 2 Newton-Cotes (Closed) Rule #6 I X(I) W(I) 1 -1 0.131944 2 -0.6 0.520833 3 -0.2 0.347222 4 0.2 0.347222 5 0.6 0.520833 6 1 0.131944 Sum(|W)|) = 2 Newton-Cotes (Closed) Rule #7 I X(I) W(I) 1 -1 0.097619 2 -0.666667 0.514286 3 -0.333333 0.0642857 4 0 0.647619 5 0.333333 0.0642857 6 0.666667 0.514286 7 1 0.097619 Sum(|W)|) = 2 Newton-Cotes (Closed) Rule #8 I X(I) W(I) 1 -1 0.0869213 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 0.0869213 Sum(|W)|) = 2 Newton-Cotes (Closed) Rule #9 I X(I) W(I) 1 -1 0.0697707 2 -0.75 0.415379 3 -0.5 -0.0654674 4 -0.25 0.740459 5 0 -0.320282 6 0.25 0.740459 7 0.5 -0.0654674 8 0.75 0.415379 9 1 0.0697707 Sum(|W)|) = 2.90243 Newton-Cotes (Closed) Rule #10 I X(I) W(I) 1 -1 0.0637723 2 -0.777778 0.351362 3 -0.555556 0.0241071 4 -0.333333 0.431786 5 -0.111111 0.128973 6 0.111111 0.128973 7 0.333333 0.431786 8 0.555556 0.0241071 9 0.777778 0.351362 10 1 0.0637723 Sum(|W)|) = 2 Newton-Cotes (Closed) Rule #11 I X(I) W(I) 1 -1 0.0536683 2 -0.8 0.355072 3 -0.6 -0.162087 4 -0.4 0.909893 5 -0.2 -0.87031 6 0 1.42753 7 0.2 -0.87031 8 0.4 0.909893 9 0.6 -0.162087 10 0.8 0.355072 11 1 0.0536683 Sum(|W)|) = 6.12959 Newton-Cotes (Closed) Rule #12 I X(I) W(I) 1 -1 0.0498665 2 -0.818182 0.309711 3 -0.636364 -0.0743385 4 -0.454545 0.579317 5 -0.272727 -0.220356 6 -0.0909091 0.355801 7 0.0909091 0.355801 8 0.272727 -0.220356 9 0.454545 0.579317 10 0.636364 -0.0743385 11 0.818182 0.309711 12 1 0.0498665 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 2.71828 1 2 1.85914 0.140859 3 1.71886 0.000579323 4 1.71854 0.000258325 5 1.71828 8.59466e-07 6 1.71828 4.84531e-07 7 1.71828 1.05866e-09 8 1.71828 6.47826e-10 9 1.71828 1.50013e-11 10 1.71828 9.1511e-11 11 1.71828 4.75217e-10 12 1.71828 1.41685e-09 13 1.71828 1.31693e-08 14 1.71828 2.22914e-08 15 1.71828 1.33324e-06 16 1.71828 2.23641e-06 17 1.71825 2.68474e-05 18 1.7183 1.40187e-05 19 1.72013 0.00184715 20 1.72148 0.00319798 21 1.76478 0.0464972 22 2.16025 0.441972 line_ncc_rule_test Normal end of execution. 15-Feb-2019 14:25:57