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Evaluating integrals are common in engineering problems. Sometimes, it is difficult to evaluate complex integrals. The need for Gauss Quadrature Therefore, techniques are devised in order to simplify these.

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From Trapezoidal rule, Trapezoidal rule evaluates an integral by approximating the function of the curve to a straight line. And then finding the area under the straight line.

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Larger error Reduced error (Using Trapezoidal rule) (Taking an improved estimate)

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Method of Undetermined Coefficients Constant function Linear function Quadratic function Cubic function

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For x d =-1, For x d =1, Then, Therefore, When differentiated, Then, the above two equations can be substituted in to the equation to be integrated. Assuming that a new variable x d is related to the original x in a linear fashion

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Table for finding Weighting factors and Function Arguments

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Example 1: Use three-point formula to evaluate it. Solution: Given: a=-1 and b=1; f(x) = x+3x 2 +x 5 From table for three points, C 0 =0.5555556 C 1 =0.8888889 C 2 =0.5555556 X 0 =-0.774596669 X 1 =0 X 2 =0.774596669 I=C 0 f(X 0 ) +C 1 f(X 1 ) +C 2 f(X 2 ) f (X 0 )=.7465 f(X 1 )=0 f(X 2 )=2.8537762 I=[0.5555556*.7465+.8888889*0+0.5555556*2.8537762]= 2.001

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Example 2: Given: a=-1 and b=1; f(x) = x+3x 2 +x 5 From table for three points, C 0 =0.5555556 C 1 =0.8888889 C 2 =0.5555556 X 0 =-0.774596669 X 1 =0 X 2 =0.774596669 I=C 0 f(X 0 ) +C 1 f(X 1 ) +C 2 f(X 2 ) X 0 `= ((b-a)/2+ X 0 (b+a)/2 ) =4.6762 X 1 `= ((b-a)/2+ X 1 (b+a)/2) =7 X 2 `= ((b-a)/2+ X 2 (b+a)/2)=9.3238 f (X 0 `)=2306.24373 f (X 1 `)=16961 f (X 2 `)=70733.71 I=3[0.5555556*2306.24373+.8888889*16961+0.5555556*70733.71] I= 166,962.5999

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Error Analysis for Gauss Quadrature

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Lecture 16 - Approximation Methods CVEN 302 July 15, 2002.

Lecture 16 - Approximation Methods CVEN 302 July 15, 2002.

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