# Evaluating integrals are common in engineering problems. Sometimes, it is difficult to evaluate complex integrals. The need for Gauss Quadrature Therefore,

## Presentation on theme: "Evaluating integrals are common in engineering problems. Sometimes, it is difficult to evaluate complex integrals. The need for Gauss Quadrature Therefore,"— Presentation transcript:

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.

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.

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

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

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

Table for finding Weighting factors and Function Arguments

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

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