Evaluate the integral of f(x) below between x=1.5 and x=5.3 using Gauss-Legendre formulas for 2, 3,...

Evaluate the integral of f(x) below between x=1.5 and x=5.3 using Gauss-Legendre formulas for 2, 3, and 4 points.

Compare with analytical integration, calculate the % error of the numerical method.

f(x)=4+8x−21x² +16x³ −5x⁴ +7x⁵

Expert Solution

The values above calculations are obtained using Excel. The intermediate calculations are as follow

t	3.4+1.9t	(3.4+1.9t)^2	(3.4+1.9t)^3	(3.4+1.9t)^4	(3.4+1.9t)^5	f(3.4+1.9t)	Coefficients	f(3.4+1.9t)*coefficients
-0.5773502	2.303035	5.303968461	12.21522299	28.13208143	64.78915748	419.3482	1	419.3482023
0.5773502	4.496965	20.22269763	90.94077113	408.9574994	1839.067717	11899.04	1	11899.03793
-0.7745966	1.928266	3.718211541	7.169702605	13.82509706	26.65847097	173.5427	0.5555555	96.41262508
0	3.4	11.56	39.304	133.6336	454.35424	2929.616	0.8888888	2604.102566
0.7745966	4.871734	23.73378768	115.6246895	563.2926779	2744.211832	17787.58	0.5555555	9881.987229
-0.8611363	1.763841	3.111135179	5.487547879	9.679162103	17.07250325	111.6894	0.3478584	38.85208478
-0.339981	2.754036	7.58471484	20.88857848	57.52789921	158.4339112	1022.368	0.6521451	666.7325522
0.339981	4.045964	16.36982388	66.23171647	267.9711339	1084.201534	7001.864	0.652145	4566.23056
0.8611363	5.036159	25.36289717	127.7315821	643.2765529	3239.642982	21016.49	0.347858	7310.754824

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Colby Messinger answered 1 year ago

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