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Chapter 07.08 Simpson 3/8 Rule for Integration After reading this chapter, you should be able to 1. derive the formula for Simpson’s 3/8 rule of integration, 2. use Simpson’s 3/8 rule it to solve integrals, 3. develop the formula for multiple-segment Simpson’s 3/8 rule of integration, 4. use multiple-segment Simpson’s 3/8 rule of integration to solve integrals, 5. compare true error formulas for multiple-segment Simpson’s 1/3 rule and multiplesegment Simpson’s 3/8 rule, and 6. use a combination of Simpson’s 1/3 rule and Simpson’s 3/8 rule to approximate integrals. Introduction The main objective of this chapter is to develop appropriate formulas for approximating the integral of the form b I f ( x)dx (1) a Most (if not all) of the developed formulas for integration are based on a simple concept of approximating a given function f (x) by a simpler function (usually a polynomial function) f i (x) , where i represents the order of the polynomial function. In Chapter 07.03, Simpsons 1/3 rule for integration was derived by approximating the integrand f (x) with a 2nd order (quadratic) polynomial function. f 2 ( x) f 2 ( x) a0 a1 x a 2 x 2 (2) 07.08.1 07.08.2 Chapter 07.08 ~ Figure 1 f ( x) Cubic function. In a similar fashion, Simpson 3/8 rule for integration can be derived by approximating the given function f (x) with the 3rd order (cubic) polynomial f 3 ( x ) f 3 ( x ) a0 a1 x a 2 x 2 a3 x 3 a0 a 1 2 3 1, x, x , x a 2 a 3 (3) which can also be symbolically represented in Figure 1. Method 1 The unknown coefficients a0 , a1 , a2 and a3 in Equation (3) can be obtained by substituting 4 known coordinate data points {x0 , f x0 },{x1 , f x1 },{x2 , f x2 } and {x3 , f x3 } into Equation (3) as follows. f ( x0 ) a0 a1 x0 a 2 x02 a3 x02 f ( x1 ) a0 a1 x1 a 2 x12 a3 x12 (4) f ( x 2 ) a0 a1 x 2 a 2 x 22 a3 x 22 f ( x3 ) a0 a1 x3 a 2 x32 a3 x32 Equation (4) can be expressed in matrix notation as Simpson 3/8 Rule for Integration 07.08.3 1 x0 x02 x03 a0 f x0 2 3 1 x1 x1 x1 a1 f x1 (5) 1 x 2 x22 x23 a 2 f x 2 2 3 1 x3 x3 x3 a3 f x3 The above Equation (5) can symbolically be represented as A44 a41 f 41 (6) Thus, a1 a 1 (7) a 2 A f a3 a 4 Substituting Equation (7) into Equation (3), one gets 1 (8) f 3 x 1, x, x 2 , x 3 A f As indicated in Figure 1, one has x0 a x1 a h ba a 3 2a b 3 x 2 a 2h (9) 2b 2a a 3 a 2b 3 x3 a 3h 3b 3a a 3 b With the help from MATLAB [Ref. 2], the unknown vector a (shown in Equation 7) can be solved for symbolically. Method 2 Using Lagrange interpolation, the cubic polynomial function f 3 x that passes through 4 data points (see Figure 1) can be explicitly given as 07.08.4 Chapter 07.08 x x1 x x2 x x3 x x0 x x2 x x3 f x0 f x1 x0 x1 x0 x2 x0 x3 x1 x0 x1 x2 x1 x3 x x0 x x1 x x3 x x0 x x1 x x2 f x3 f x3 x2 x0 x2 x1 x2 x3 x3 x0 x3 x1 x3 x2 f 3 x (10) Simpsons 3/8 Rule for Integration Substituting the form of f 3 x from Method (1) or Method (2), b I f x dx a b f 3 x dx a b a f x0 3 f x1 3 f x2 f x3 8 (11) Since ba 3 b a 3h and Equation (11) becomes 3h I f x0 3 f x1 3 f x 2 f x3 8 Note the 3/8 in the formula, and hence the name of method as the Simpson’s 3/8 rule. The true error in Simpson 3/8 rule can be derived as [Ref. 1] (b a) 5 Et f , where a b 6480 Example 1 The vertical distance covered by a rocket from x 8 to x 30 seconds is given by 30 140000 s 2000 ln 9.8 x dx 140000 2100t 8 Use Simpson 3/8 rule to find the approximate value of the integral. h (12) (13) Simpson 3/8 Rule for Integration Solution ba n ba 3 30 8 3 7.3333 3h I f x0 3 f x1 3 f x 2 f x3 8 x0 8 h 140000 f x0 2000 ln 9.8 8 140000 2100 8 177.2667 x x h 0 1 8 7.3333 15.3333 140000 f x1 2000 ln 9.8 15.3333 140000 2100 15.3333 372.4629 x x 2h 0 2 8 2(7.3333) 22.6666 140000 f x 2 2000 ln 9.8 22.6666 140000 2100 22.6666 608.8976 07.08.5 07.08.6 Chapter 07.08 x x 3h 0 3 8 3(7.3333) 30 140000 f x3 2000 ln 9.8 30 140000 2100 30 901.6740 Applying Equation (12), one has 3 I 7.3333 177.2667 3 372.4629 3 608.8976 901.6740 8 11063.3104 The exact answer can be computed as I exact 11061.34 Multiple Segments for Simpson 3/8 Rule Using n = number of equal segments, the width h can be defined as ba h (14) n The number of segments need to be an integer multiple of 3 as a single application of Simpson 3/8 rule requires 3 segments. The integral shown in Equation (1) can be expressed as b I f x dx a b f 3 x dx a x3 xn b x6 f x dx f x dx ........ f x dx 3 x0 a 3 x3 3 (15) xn 3 Using Simpson 3/8 rule (See Equation 12) into Equation (15), one gets 3h f x0 3 f x1 3 f x2 f x3 f x3 3 f x4 3 f x5 f x6 I 8 ..... f xn3 3 f xn2 3 f xn1 f xn n 2 n 1 n 3 3h f x0 3 f xi 3 f xi 2 f xi f xn 8 i 1, 4, 7,.. i 2,5,8,.. i 3, 6, 9,.. Example 2 The vertical distance covered by a rocket from x 8 to x 30 seconds is given by (16) (17) Simpson 3/8 Rule for Integration 07.08.7 140000 s 2000 ln 9.8 x dx 140000 2100t 8 Use Simpson 3/8 multiple segments rule with six segments to estimate the vertical distance. Solution 30 In this example, one has (see Equation 14): 30 8 h 3.6666 6 x0 , f x0 8,177.2667 x1 , f x1 11.6666,270.4104where x1 x0 h 8 3.6666 11.6666 x2 , f x2 15.3333,372.4629 where x2 x0 2h 15.3333 x3 , f x3 19,484.7455 where x3 x0 3h 19 x4 , f x4 22.6666,608.8976 where x4 x0 4h 22.6666 x5 , f x5 26.3333,746.9870 where x5 x0 5h 26.3333 x6 , f x6 30,901.6740 where x6 x0 6h 30 Applying Equation (17), one obtains: n 2 4 n 15 n 33 3 I 3.6666177.2667 3 f xi 3 f xi 2 f xi 901.6740 8 i 1, 4,.. i 2, 5,.. i 3, 6,.. 1.3750177.2667 3270.4104 608.8976 3372.4629 746.9870 2484.7455 901.6740 11,601.4696 Example 3 Compute b 30 I 140,000 2000 ln 140,000 2100x 9.8xdx, a 8 using Simpson 1/3 rule (with n1 4), and Simpson 3/8 rule (with n2 3). Solution The segment width is ba h n ba n1 n2 30 8 4 3 3.1429 07.08.8 Chapter 07.08 x0 a 8 x1 x0 1h 8 3.1429 11.1429 x2 x0 2h 8 23.1429 14.2857 Simpson' s 1/3 rule x3 x0 3h 8 33.1429 17.4286 x4 x0 4h 8 43.1429 20.5714 x5 x0 5h 8 53.1429 23.7143 x6 x0 6h 8 63.1429 26.8571 x7 x0 7 h 8 73.1429 30 140,000 f x0 8 2000 ln 9.8 8 177.2667 140,000 2100 8 Similarly: f x1 11.1429 256.5863 f x2 342.3241 f x3 435.2749 f x4 536.3909 f x5 646.8260 f x6 767.9978 f x7 901.6740 For multiple segments n1 first 4 segments , using Simpson 1/3 rule, one obtains (See Equation 19): n1 1 3 n1 2 2 h I1 f x0 4 f xi 2 f xi f xn1 3 i 1, 3,... i 2 ,... h f x0 4 f x1 f x3 2 f x2 f x4 3 3.1429 177.2667 4256.5863 435.2749 2342.3241 536.3909 3 4364.1197 For multiple segments n2 last 3 segments , using Simpson 3/8 rule, one obtains (See Equation 17): Simpson 3/8 Rule for Integration 07.08.9 n2 2 1 n2 1 2 n2 3 0 3h I 2 f t 0 3 f ti 3 f ti 2 f ti f t n1 8 i 1, 3,... i 2 ,... i 3, 6 ,.. 3h f t 0 3 f t1 3 f t 2 2(no contribution) f (t3 ) 8 3h f x4 3 f x5 3 f x6 f ( x7 ) 8 3 3.1429 536.3909 3646.8260 3767.9978 901.6740 8 6697.3663 The mixed (combined) Simpson 1/3 and 3/8 rules give I I1 I 2 4364.1197 6697.3663 11061 Comparing the truncated error of Simpson 1/3 rule 5 b a Et f (18) 2880 With Simpson 3/8 rule (See Equation 12), it seems to offer slightly more accurate answer than the former. However, the cost associated with Simpson 3/8 rule (using 3rd order polynomial function) is significantly higher than the one associated with Simpson 1/3 rule (using 2nd order polynomial function). The number of multiple segments that can be used in the conjunction with Simpson 1/3 rule is 2, 4, 6, 8, … (any even numbers). h I 1 f x0 4 f x1 f x2 f x2 4 f x3 f x4 ..... f xn 2 4 f xn 1 f xn 3 n 1 n 2 h f x0 4 f xi 2 f xi f xn (19) 3 i 1, 3,... i 2 , 4 , 6... However, Simpson 3/8 rule can be used with the number of segments equal to 3,6,9,12,.. (can be certain odd or even numbers that are multiples of 3). If the user wishes to use, say 7 segments, then the mixed Simpson 1/3 rule (for the first 4 segments), and Simpson 3/8 rule (for the last 3 segments) would be appropriate. Computer Algorithm for Mixed Simpson 1/3 and 3/8 Rule for Integration Based on the earlier discussion on (single and multiple segments) Simpson 1/3 and 3/8 rules, the following “pseudo” step-by-step mixed Simpson rules can be given as Step 1 User inputs information, such as f (x ) = integrand 07.08.10 Chapter 07.08 n1 = number of segments in conjunction with Simpson 1/3 rule (a multiple of 2 (any even numbers) n2 = number of segments in conjunction with Simpson 3/8 rule (a multiple of 3) Step 2 Compute n n1 n2 ba h n x0 a x1 a 1h x 2 a 2h . . xi a ih . . x n a nh b Step 3 Compute result from multiple-segment Simpson 1/3 rule (See Equation 19) n1 1 n1 2 h I1 f x0 4 f xi 2 f xi f xn1 3 i 1, 3,... i 2, 4, 6... Step 4 Compute result from multiple segment Simpson 3/8 rule (See Equation 17) n2 2 n2 1 n2 3 3h I 2 f x0 3 f xi 3 f xi 2 f xi f xn2 8 i 1, 4, 7... i 2, 5,8... i 3, 6, 9,... Step 5 I I1 I 2 and print out the final approximated answer for I . SIMPSON’S 3/8 RULE FOR INTEGRATION Topic Simpson 3/8 Rule for Integration Summary Textbook Chapter of Simpson’s 3/8 Rule for Integration Major General Engineering Authors Duc Nguyen Date May 25, 2017 Web Site http://numericalmethods.eng.usf.edu (19, repeated) (17, repeated) (20)