<p>Chapter 07.05 Gauss Quadrature Rule of Integration</p><p>After reading this chapter, you should be able to:</p><p>1. derive the Gauss quadrature method for integration and be able to use it to solve problems, and 2. use Gauss quadrature method to solve examples of approximate integrals.</p><p>What is integration? Integration is the process of measuring the area under a function plotted on a graph. Why would we want to integrate a function? Among the most common examples are finding the velocity of a body from an acceleration function, and displacement of a body from a velocity function. Throughout many engineering fields, there are (what sometimes seems like) countless applications for integral calculus. You can read about some of these applications in Chapters 07.00A-07.00G. Sometimes, the evaluation of expressions involving these integrals can become daunting, if not indeterminate. For this reason, a wide variety of numerical methods has been developed to simplify the integral. Here, we will discuss the Gauss quadrature rule of approximating integrals of the form b I   f xdx a where f (x) is called the integrand, a  lower limit of integration b  upper limit of integration</p><p>07.05.1 07.05.2 Chapter 07.05</p><p>Gauss Quadrature Rule Background: To derive the trapezoidal rule from the method of undetermined coefficients, we approximated b f (x)dx  c f (a)  c f (b)  1 2 (1) a Let the right hand side be exact for integrals of a straight line, that is, for an integrated form of b a  a x dx   0 1  a So b b  x 2  a  a x dx  a x  a   0 1   0 1  a  2  a  b 2  a 2     a0 b  a a1   (2)  2  But from Equation (1), we want b a  a x dx  c f (a)  c f (b)   0 1  1 2 (3) a to give the same result as Equation (2) for f (x)  a0  a1 x . b a  a x dx  c a  a a  c a  a b   0 1  1  0 1  2  0 1  a</p><p> a0 c1  c2  a1 c1a  c2b (4) Hence from Equations (2) and (4),  b 2  a 2    a0 b  a a1   a0 c1  c2  a1 c1a  c2b  2 </p><p>Since a0 and a1 are arbitrary constants for a general straight line</p><p> c1  c2  b  a (5a) b 2  a 2 c a  c b  (5b) 1 2 2 Multiplying Equation (5a) by a and subtracting from Equation (5b) gives b  a c  (6a) 2 2</p><p>Substituting the above found value of c2 in Equation (5a) gives b  a c1  2 (6b) Therefore Gauss Quadrature Rule 07.05.3</p><p> b f (x)dx  c f (a)  c f (b)  1 2 a b  a b  a  f (a)  f (b) (7) 2 2</p><p>Derivation of two-point Gauss quadrature rule Method 1: The two-point Gauss quadrature rule is an extension of the trapezoidal rule approximation where the arguments of the function are not predetermined as a and b , but as unknowns x1 and x2 . So in the two-point Gauss quadrature rule, the integral is approximated as b I   f (x)dx a</p><p> c1 f (x1 )  c2 f (x2 )</p><p>There are four unknowns x1 , x2 , c1 and c2 . These are found by assuming that the formula gives exact results for integrating a general third order polynomial, 2 3 f (x)  a0  a1 x  a2 x  a3 x . Hence b b f (x)dx  a  a x  a x 2  a x 3 dx    0 1 2 3  a a b  x 2 x3 x 4   a0 x  a1  a2  a3  2 3 4   a  b 2  a 2   b3  a 3   b 4  a 4         a0 b  a a1    a2    a3   (8)  2   3   4  The formula would then give b f (x)dx  c f (x )  c f (x )   1 1 2 2 a 2 3 2 3 c1 a0  a1x1  a2 x1  a3 x1  c2 a0  a1x2  a2 x2  a3 x2  (9) Equating Equations (8) and (9) gives  b2  a 2   b3  a 3   b4  a 4  a0 b  a  a1   a2    a3    2   3   4  2 3 2 3  c1 a0  a1x1  a2 x1  a3 x1  c2 a0  a1x2  a2 x2  a3 x2   a c  c  a c x  c x  a c x 2  c x 2  a c x 3  c x 3 0  1 2  1  1 1 2 2  2  1 1 2 2  3  1 1 2 2  (10)</p><p>Since in Equation (10), the constants a0 , a1 , a2 , and a3 are arbitrary, the coefficients of a0 , a1 , a2 , and a3 are equal. This gives us four equations as follows.</p><p> b  a  c1  c2 07.05.4 Chapter 07.05</p><p> b 2  a 2  c x  c x 2 1 1 2 2 b3  a 3  c x 2  c x 2 3 1 1 2 2 b 4  a 4  c x 3  c x 3 (11) 4 1 1 2 2</p><p>Without proof (see Example 1 for proof of a related problem), we can find that the above four simultaneous nonlinear equations have only one acceptable solution b  a c  1 2 b  a c  2 2  b  a  1  b  a x1       2  3  2  b  a  1  b  a x2      (12)  2  3  2</p><p>Hence b f (x)dx  c f x  c f x  1  1  2  2  a b  a  b  a  1  b  a  b  a  b  a  1  b  a   f       f      (13) 2  2  3  2  2  2  3  2 </p><p>Method 2: We can derive the same formula by assuming that the expression gives exact values for the b b b b individual integrals of 1dx,  xdx,  x 2dx, and  x 3dx . The reason the formula can also be a a a a derived using this method is that the linear combination of the above integrands is a general 2 3 third order polynomial given by f (x)  a0  a1 x  a2 x  a3 x . These will give four equations as follows b 1dx  b  a  c  c  1 2 a b b2  a 2 xdx   c x  c x  1 1 2 2 a 2 b b3  a 3 x 2dx   c x 2  c x 2  1 1 2 2 a 3 Gauss Quadrature Rule 07.05.5</p><p> b b4  a 4 x3dx   c x 3  c x 3  1 1 2 2 (14) a 4 These four simultaneous nonlinear equations can be solved to give a single acceptable solution b  a c  1 2 b  a c  2 2  b  a  1  b  a x1       2  3  2  b  a  1  b  a x2      (15)  2  3  2</p><p>Hence b b  a  b  a  1  b  a  b  a  b  a  1  b  a       f (x)dx  f       f      (16) a 2  2  3  2  2  2  3  2  Since two points are chosen, it is called the two-point Gauss quadrature rule. Higher point versions can also be developed.</p><p>Higher point Gauss quadrature formulas For example b f (x)dx  c f (x )  c f (x )  c f (x )  1 1 2 2 3 3 (17) a is called the three-point Gauss quadrature rule. The coefficients c1 , c2 and c3 , and the function arguments x1 , x2 and x3 are calculated by assuming the formula gives exact expressions for integrating a fifth order polynomial b a  a x  a x 2  a x3  a x 4  a x 5 dx   0 1 2 3 4 5  . a General n -point rules would approximate the integral b f (x)dx  c f (x )  c f (x )  ......  c f (x )  1 1 2 2 n n (18) a</p><p>Arguments and weighing factors for n-point Gauss quadrature rules In handbooks (see Table 1), coefficients and arguments given for n -point Gauss quadrature rule are given for integrals of the form 1 n g(x)dx  c g(x )   i i (19) 1 i1</p><p>Table 1 Weighting factors c and function arguments x used in Gauss quadrature formulas 07.05.6 Chapter 07.05</p><p>Weighting Function Points Factors Arguments</p><p> c1  1.000000000 x1  0.577350269</p><p> c2  1.000000000 x2  0.577350269</p><p>2 c1  0.555555556 x1  0.774596669</p><p> c2  0.888888889 x2  0.000000000</p><p> c3  0.555555556 x3  0.774596669 3</p><p> c1  0.347854845 x1  0.861136312</p><p> c2  0.652145155 x2  0.339981044</p><p> c3  0.652145155 x3  0.339981044 4 c4  0.347854845 x4  0.861136312</p><p> c1  0.236926885 x1  0.906179846</p><p> c2  0.478628670 x2  0.538469310</p><p> c3  0.568888889 x3  0.000000000 5 c4  0.478628670 x4  0.538469310</p><p> c5  0.236926885 x5  0.906179846</p><p> c1  0.171324492 x1  0.932469514</p><p> c2  0.360761573 x2  0.661209386</p><p> c3  0.467913935 x3  0.238619186 6 c4  0.467913935 x4  0.238619186</p><p> c5  0.360761573 x5  0.661209386</p><p> c6  0.171324492 x6  0.932469514</p><p>1 b So if the table is given for  g(x)dx integrals, how does one solve  f (x)dx ? 1 a The answer lies in that any integral with limits of a, b can be converted into an integral with limits 1, 1. Let x  mt  c (20) If x  a, then t  1 Gauss Quadrature Rule 07.05.7</p><p>If x  b, then t  1 such that a  m(1)  c b  m(1)  c (21) Solving the two Equations (21) simultaneously gives b  a m  2 b  a c  (22) 2 Hence b  a b  a x  t  2 2 b  a dx  dt 2 Substituting our values of x and dx into the integral gives us b 1  b  a b  a  b  a  f (x)dx   f  x   dx (23) a 1  2 2  2</p><p>Example 1</p><p>1 For an integral  f (x)dx, show that the two-point Gauss quadrature rule approximates to 1 1 f (x)dx  c f (x )  c f (x )  1 1 2 2 1 where</p><p> c1  1</p><p> c2  1 1 x1   3 1 x2  3 Solution Assuming the formula 1 f (x)dx  c f x  c f x  1  1  2  2  (E1.1) 1 1 1 1 1 gives exact values for integrals 1dx,  xdx,  x 2 dx, and  x3dx . Then 1 1 1 1 1 1dx  2  c  c  1 2 (E1.2) 1 07.05.8 Chapter 07.05</p><p>1 xdx  0  c x  c x  1 1 2 2 (E1.3) 1 1 2 x 2 dx   c x 2  c x 2  1 1 2 2 (E1.4) 1 3 1 x 3dx  0  c x 3  c x 3  1 1 2 2 (E1.5) 1 2 Multiplying Equation (E1.3) by x1 and subtracting from Equation (E1.5) gives 2 2 c2 x2 x1  x2  0 (E1.6) The solution to the above equation is</p><p> c2  0, or/and</p><p> x2  0, or/and</p><p> x1  x2 , or/and</p><p> x1  x2 .</p><p>I. c2  0 is not acceptable as Equations (E1.2-E1.5) reduce to c1  2, c1 x1  0, 2 c x 2  , and c x3  0 . But since c  2 , then x  0 from c x  0 , but x  0 1 1 3 1 1 1 1 1 1 1 2 conflicts with c x 2  . 1 1 3</p><p>II. x2  0 is not acceptable as Equations (E1.2-E1.5) reduce to c1  c2  2 , c1 x1  0, 2 c x 2  , and c x3  0 . Since c x  0 , then c or x has to be zero but this violates 1 1 3 1 1 1 1 1 1 2 c x 2   0 . 1 1 3</p><p>III. x1  x2 is not acceptable as Equations (E1.2-E1.5) reduce to c1  c2  2 , 2 c x  c x  0, c x 2  c x 2  , and c x 3  c x 3  0 . If x  0 , then c x  c x  0 1 1 2 1 1 1 2 1 3 1 1 2 1 1 1 1 2 1</p><p> gives c1  c2  0 and that violates c1  c2  2 . If x1  0 , then that violates 2 c x 2  c x 2   0 . 1 1 2 1 3</p><p>That leaves the solution of x1  x2 as the only possible acceptable solution and in fact, it does not have violations (see it for yourself)</p><p> x1  x2 (E1.7) Substituting (E1.7) in Equation (E1.3) gives</p><p> c1  c2 (E1.8) From Equations (E1.2) and (E1.8),</p><p> c1  c2  1 (E1.9) Equations (E1.4) and (E1.9) gives 2 x 2  x 2  (E1.10) 1 2 3 Gauss Quadrature Rule 07.05.9</p><p>Since Equation (E1.7) requires that the two results be of opposite sign, we get 1 x1   3 1 x2  3 Hence 1 f (x)dx  c f (x )  c f (x )  1 1 2 2 (E1.11) 1  1   1   f    f    3   3 </p><p>Example 2</p><p> b For an integral  f (x)dx, derive the one-point Gauss quadrature rule. a Solution The one-point Gauss quadrature rule is b f (x)dx  c f x  1  1  (E2.1) a 1 1 Assuming the formula gives exact values for integrals 1dx, and  xdx 1 1 b 1dx  b  a  c  1 a b b 2  a 2 xdx   c x  1 1 (E2.2) a 2</p><p>Since c1  b  a, the other equation becomes b 2  a 2 (b  a)x  1 2 b  a x  (E2.3) 1 2 Therefore, one-point Gauss quadrature rule can be expressed as b  b  a   f (x)dx  (b  a) f   (E2.4) a  2 </p><p>Example 3 What would be the formula for b f (x)dx  c f (a)  c f (b)  1 2 a 07.05.10 Chapter 07.05</p><p> b a x  b x 2 dx, if you want the above formula to give you exact values of   0 0  that is, a linear a combination of x and x 2 . Solution If the formula is exact for a linear combination of x and x 2 , then b b 2  a 2 xdx   c a  c b  2 1 2 a b b3  a 3 x 2dx   c a 2  c b 2  1 2 (E3.1) a 3 Solving the two Equations (E3.1) simultaneously gives b 2  a 2     a b  c1  2  2 2      3 3  a b  c 2  b  a   3  1  ab  b 2  2a 2 c   1 6 a 1 a 2  ab  2b 2 c   (E3.2) 2 6 b So b 1  ab  b 2  2a 2 1 a 2  ab  2b 2  f (x)dx   f (a)  f (b) (E3.3) a 6 a 6 b Let us see if the formula works. 5 Evaluate  2x 2  3xdx using Equation(E3.3) 2 5 2x 2  3x dx  c f (a)  c f (b)    1 2 2 1  (2)(5)  52  2(2)2 1 22  2(5)  2(5) 2   2(2)2  3(2) [2(5) 2  3(5)] 6 2 6 5  46.5 5 The exact value of  2x 2  3xdx is given by 2 5 3 2 5 2 2x 3x   2x  3xdx     2  3 2  2  46.5 Any surprises? 5 Now evaluate  3dx using Equation (E3.3) 2 Gauss Quadrature Rule 07.05.11</p><p>5 3dx  c f (a)  c f (b)  1 2 2 1  2(5)  52  2(2) 2 1 22  2(5)  2(5) 2   (3)  (3) 6 2 6 5  10.35 5 The exact value of  3dx is given by 2 5 3dx 5   3x2 2  9 Because the formula will only give exact values for linear combinations of x and x 2 , it does 5 not work exactly even for a simple integral of  3dx . 2</p><p>Do you see now why we choose a0  a1x as the integrand for which the formula b f (x)dx  c f (a)  c f (b)  1 2 a gives us exact values?</p><p>Example 4 Use two-point Gauss quadrature rule to approximate the distance covered by a rocket from t  8 to t  30 as given by 30  140000   x   2000ln   9.8t dt 8  140000  2100t   Also, find the absolute relative true error. Solution First, change the limits of integration from 8, 30 to 1, 1 using Equation(23) gives 30 30  8 1  30  8 30  8   f (t)dt   f  x  dx 8 2 1  2 2  1  11 f 11x 19dx 1 Next, get weighting factors and function argument values from Table 1 for the two point rule,</p><p> c1  1.000000000 .</p><p> x1  0.577350269</p><p> c2  1.000000000</p><p> x2  0.577350269 Now we can use the Gauss quadrature formula 07.05.12 Chapter 07.05</p><p>1 11 f 11x 19 dx  11 c f 11x 19  c f 11x 19     1  1  2  2  1  11 f 11(0.5773503) 19 f 11(0.5773503) 19  11 f (12.64915)  f (25.35085)  11(296.8317)  (708.4811)  11058.44 m since  140000  f (12.64915)  2000ln   9.8(12.64915) 140000  2100(12.64915)  296.8317  140000  f (25.35085)  2000ln   9.8(25.35085) 140000  2100(25.35085)  708.4811</p><p>The absolute relative true error, t , is (True value = 11061.34 m) 11061.34 11058.44   100 t 11061.34  0.0262%</p><p>Example 5 Use three-point Gauss quadrature rule to approximate the distance covered by a rocket from t  8 to t  30 as given by 30  140000   x   2000ln   9.8t dt 8  140000  2100t   Also, find the absolute relative true error. Solution First, change the limits of integration from 8, 30 to 1, 1 using Equation (23) gives 30 30  8 1  30  8 30  8   f (t)dt   f  x  dx 8 2 1  2 2  1  11 f 11x 19dx 1 The weighting factors and function argument values are</p><p> c1  0.555555556</p><p> x1  0.774596669</p><p> c2  0.888888889</p><p> x2  0.000000000</p><p> c3  0.555555556</p><p> x3  0.774596669 Gauss Quadrature Rule 07.05.13 and the formula is 1 11 f 11x 19 dx  11 c f 11x 19  c f 11x 19  c f 11x 19     1  1  2  2  3  3  1 0.5555556 f 11(.7745967) 19 0.8888889 f 11(0.0000000) 19  11   0.5555556 f 11(0.7745967) 19   110.55556 f (10.47944)  0.88889 f (19.00000)  0.55556 f (27.52056)  110.55556 239.3327  0.88889 484.7455  0.55556 795.1069  11061.31 m since  140000  f (10.47944)  2000ln   9.8(10.47944) 140000  2100(10.47944)  239.3327  140000  f (19.00000)  2000ln   9.8(19.00000) 140000  2100(19.00000)  484.7455  140000  f (27.52056)  2000ln   9.8(27.52056) 140000  2100(27.52056)  795.1069</p><p>The absolute relative true error, t , is (True value = 11061.34 m) 11061.34 11061.31   100 t 11061.34  0.0003%</p><p>INTEGRATION Topic Gauss quadrature rule Summary These are textbook notes of Gauss quadrature rule Major General Engineering Authors Autar Kaw, Michael Keteltas Date 2018 年 4 月 5 日 Web Site http://numericalmethods.eng.usf.edu</p>