316 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 6.5 TRAPEZOIDAL RULE Putting n =1 in equation (2) and taking the curve y = f(x) through (x 0 , y 0 ) and (x 0 , y 0 ) as a polynomial of degree one so that differences of order higher than one vanish, we get ()() 0 0 00 010 01 1 () 2 22 2 xh x hh fxdx hy y y y y y y +  =+∆= +−=+    ∫ Similarly, for the next sub interval () 00 ,2, xhx h ++ we get () 00 00 2 12 1 (1) ( ) ( ) ( ) 22 xnh xh nn xh x n h hh f x dx y y f x dx y y + + − ++− =+ = + ∫∫ Adding the above integrals, we get () 0 0 012 1 () 2( ) 2 xnh nn x h f x dx y y y y y + − =+++++  ∫ which is known as Trapezoidal rule. 6.6 SIMPSON’S ONE-THIRD RULE Putting n = 2 in equation (2 ) and taking the curve through 00 11 (,),(,) xy xy and 22 (,) xy as a polynomial of degree two so that differences of order higher than two vanish, we get 0 0 2 2 00 0 1 () 2 6 xh x fxdx hy y y +  =+∆+∆   ∫ () 010210 012 2 66 (2 ) (4 ) 63 hh yyyyyy yyy  =+−+−−=++  Similarly, () 0 0 4 234 2 ( ) 4 , 3 xh xh h fxdx y y y + + =++ ∫ 0 0 21 (2) () ( 4 ) 3 xnh nnn xnh h fxdx y y y + −− +− =++ ∫ Adding the above integrals, we get ()( ) 0 0 0131242 () 4 2( ) 3 xnh nn n x h fxdx y y y y y y y y + −− = + + + ++ + + ++  ∫ which is known as Simpson’s one-third rule. Note: Using the formula, the given interval of integration must be divided into an even number of sub- intervals. NUMERICAL DIFFERENTIATION AND INTEGRATION 317 6.7 SIMPSON’S THREE-EIGHT RULE Putting n = 3 in equation (2) and taking the curve through (x 0 , y 0 ), (x 1 , y 1 ), (x 2 , y 2 ) and (x 3 , y 3 ) as a polynomial of degree three so that differences of order higher than three vanish, we get 0 0 3 23 00 0 0 33 1 () 3 24 8 xh x fxdx hy y y y +  =+∆+∆+∆   ∫ ()( ) 0 10 2 10 3 2 10 3 812 6 2 (33 ) 8 h y yy yyy yyyy  =+−+−++−+−  () 0123 3 33 8 h yyyy =+++ Similarly, () 0 0 6 3456 3 3 ( ) 3 3 , 8 xh xh h fxdx y y y y + + =+++ ∫ 0 0 6 321 (3) 3 () ( 3 3 ) 8 xh nnnn xnh h fxdx y y y y + −−− +− =+++ ∫ Adding the above integrals, we get 0 0 0) 1245 2 1 36 3 3 ( ) [( 3( ) 2( )] 8 xnh nnnn x h fxdx y y y y y y y y y y y + −− − =+++++++++++ ∫ which is known as Simpson’s three-eighth rule. Note: Using this formula, the given interval of integration must be divided into sub-intervals whose number n is a multiple of 3. 6.8 BOOLE’S RULE Putting n = 4 in equation (2) and neglecting all differences of order higher than four, we get 0 0 4 4 23 4 00 0 0 0 0 (1)(1)(2)(1)(2)(3) () 2! 3! 4! xh x rr rr r rr r r fxdx h y r y y y y dr + −−−−−−  = +∆+ ∆+ ∆+ ∆   ∫∫ (By Newton’s forward interpolation formula) 4 4 2432 23 0 00 0 0 0 (2 3) ( 2) 3 11 43 212 24 5234! y nnn nn nnn hy y y y n   ∆ −− =+∆+ −∆+ ∆+−+−      23 4 0000 0 52 7 42 3390 hyyyy y  = +∆+∆+∆+∆   01234 2 (7 32 12 32 7 ) 45 h yyyyy=++++ Similarly, 08 04 45678 2 ( ) (7 32 12 32 7 ) 45 h h x x h fxdx y y y y y + + =++++ ∫ and so on. 318 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Adding all these integrals from x 0 to x 0 + nh, where n is a multiple of 4, we get 0 0 012345678 2 ( ) (7 32 12 32 14 32 12 32 14 ) 45 xnh x h fxdx yyyyyyyyy + = ++++++++ ∫ This is known as Boole’s rule. Note: Using Boole’s rule, the number of sub-intervals should be taken as a multiple of 4. 6.9 WEDDLE’S RULE Putting n = 6 in equation (2) and neglecting all differences of order higher than six, we get 0 0 6 6 23 00 0 0 0 (1) (1)(2) () 2! 3! xh x rr rr r fxdx h y ry y y + −−−  =+∆+ ∆+ ∆   ∫∫ 45 00 (1)(2)(3) (1)(2)(3)(4) 4! 5! rrrr rrrrr yy −−− −−−− +∆+ ∆ 6 0 ( 1)( 2)( 3)( 4)( 5) 6! rr r r r r ydr −−−−−  +∆   232 4 2323 00 0 0 11 2232 64 rrr r hry y y r r y     =+∆+−∆+−+∆          54 3 6 4 3 24 5 25 00 1311 1 3550 3212 24 5 2 3 120 6 4 3 rr r r r r ry r ry   +−+−∆+ −+−+∆     78 4 3 52 6 6 0 0 1 5 225 274 17 60 720 7 2 4 3 rr r r rry         +−+−+−∆ = 63 9 2 4 41 20 11 20 41 840 00 2 0 3 0 4 0 5 0 6 0 hy y y y y y y ++ + + + + L N M O Q P ∆∆ ∆ ∆ ∆ ∆ 2345 6 00 0 0 0 0 0 641 20 60 90 80 41 11 20 42 h yy y y y y y  = +∆+∆+∆+∆+∆+∆   ()( )( ) 010 210 3210 3 20 60 90 2 80 3 3 10 h yyy yyy yyyy  =+−+−++−+−  43210 54 3 210 41(464 )11(510105 ) yyyyy yy y yyy+ −+−++ −+−+− 65 4 3 210 ( 6 15 20 15 6 )] yy y y yyy+−+−+−+ 3 41 42 1 ≈ L N M O Q P () 0 12 34 56 3 565 10 h yyyyyyy = + ++ ++ + NUMERICAL DIFFERENTIATION AND INTEGRATION 319 Similarly, () () 0 0 12 6 7 8 9 10 11 12 6 3 56 5 10 xh xh h fxdx y y y y y y y + + =++++++ ∫ 0 0 654321 (6) 3 () ( 5 6 5 ) 10 xnh nnynnnn xnh h fxdxyyyyyyy + −−−−−− +− =++++++ ∫ Adding the above integrals, we get 0 0 0 1 2 3 4 5 6 7 8 9 10 11 12 3 () ( 5 6 5 2 5 6 5 2 ) 10 xnh x h fxdx yyyyyyyyyyy y y + = + ++ ++ + + ++ + + + + ∫ which is known as Weddle’s rule. Here n must be a multiple of 6. Example 1. Use Trapezoidal rule to evaluate 1 0 1 . 1 dx x+ ∫ Sol. Let h = 0.125 and y = f(x) = 1 1 x+ , then the values of y are given for the arguments which are obtained by dividing the interval [0,1] into eight equal parts are given below: x 0 0.125 0.250 0.375 0.5 0.625 0.750 0.875 1.0 1 1 y x = + 1.0 0.8889 0.8000 0.7273 0.6667 0.6154 0.5714 0.5333 0.5 y 0 y 1 y 2 y 3 y 4 y 5 y 6 y 7 y 8 Now by Trapezoidal rule 1 0 1234567 8 0 1 [2( )] 12 h dx y yyyyyyy y x = + ++++++ + + ∫ 0.125 [1 2(0.8889 0.800 0.7273 0.6667 0.6154 0.5714 0.5333) 0.5] 2 = + ++++++ + 0.125 [1.5 2(4.803)] 2 =+ = 0.125 2 [11.106] = 0.69413. Ans. Example 2. Evaluate 1 2 0 1 dx x+ ∫ using (i) Simpson’s 1 3 rule taking 1 4 h = (ii) Simpson’s 3 8 rule taking 1 6 h = (iii) Weddle’s rule taking 1 6 h = Hence compute an approximate value of π in each case. 320 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. (i) The values of 2 1 () 1 fx x = + at 123 0, , , ,1 444 x = are given below: () 01234 11 3 01 42 4 16 1 0.8 0.64 0.5 17 x fx yyy y y By Simpon’s 1 3 rule () 1 04 13 2 2 0 [4()2] 3 1 dx h yy yy y x =++++ + ∫ {} 116 (1 0.5) 4 .64 2(0.8) 0.78539215 12 17  =++++ =   Also, 1 11 2 0 0 tan tan 1 4 1 dx x x −− π  ===  + ∫ ∴ 0.785392156 3.1415686 4 π =⇒π≈ . Ans. (ii) The values of 2 1 () 1 fx x = + at x = 0, 12345 ,,,,,1 66666 are given below: () 0123456 12345 0 1 66666 36 9 4 9 36 1 1 37 10 5 13 61 2 x fx yyyyyyy By Simpson’s 3 8 rule [] 1 06 1245 3 2 0 3 ()3( )2 8 1 dx h yy yyyy y x =++++++ + ∫ = 3 1 6 8 1 1 2 3 36 37 9 10 9 13 36 61 2 4 5 F H I K + F H G I K J ++++ R S T U V W + F H G I K J L N M O Q P = 0.785395862 Also, 1 2 0 4 1 dx x π = + ∫ ∴ 0.785395862 4 π = ⇒ π = 3.141583. Ans. NUMERICAL DIFFERENTIATION AND INTEGRATION 321 (iii) By Weddle’s rule () 1 0123456 2 0 3 565 10 1 dx h yyyyyyy x =++++++ + ∫ 1 3 36949361 6 15 6 5 10 37 10 5 13 61 2        = + ++ ++ +         = 0.785399611 Since 1 2 0 4 1 dx x π = + ∫ ∴ 0.785399611 4 π = ⇒ π = 3.141598 Example 3. Evaluate 6 2 0 1 dx x+ ∫ by using (i) Trapezoidal Rule (ii) Simpson’s one-third rule (iii) Simpson’s three-eighth rule (iv) Weddle’s rule. Sol. Divide the interval (0, 6) into six parts each of width h = 1. The value of 2 1 () 1 fx x = + are given below: x 01 2 3 4 5 6 f(x) 1 0.5 0.2 0.1 1 17 1 26 1 37 y 0 y 1 y 2 y 3 y 4 y 5 y 6 (i) By Trapezoidal rule, 6 06 12345 2 0 [( ) 2( )] 2 1 dx h yy yyyyy x = + + ++++ + ∫ 11 11 120.50.20.1 237 1726    =++++++       = 1.410798581. Ans. (ii) By Simpson’s one-third rule, ()( ) 6 06 135 24 2 0 [4 2()] 3 1 dx h yy yyy yy x =++++++ + ∫ 11 1 1 140.50.120.2 337 26 17     =++ ++++         = 1.366173413. Ans. 322 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES (iii) By Simpson’s three-eighth rule, 6 06 1245 3 2 0 3 [( ) 3( ) 2 ] 8 1 dx h yy yyyy y x =++++++ + ∫ = 31 11 1 3 0.5 0.2 2(0.1) 837 1726    ++ ++++       = 1.357080836. Ans. (iv) By Weddle’s rule, 6 0123456 2 0 3 (5 6 5 ) 10 1 dx h yyyyyyy x =++++++ + ∫ 3111 1 5(0.5) 0.2 6(0.1) 5 10 17 26 37   =+ ++ ++ +     1.373447475.= Ans. Example 4. Using Simpson’s one-third rule, find 6 2 0 (1 ) dx x+ ∫ . (B. Tech. 2002) Sol. Divide the interval [0,6] into 6 equal parts with 60 6 h − = = 1. The values of 2 1 (1 ) y x = + at each points of sub-divisions are given by x 01 2 3 4 5 6 2 1 (1 ) y x = + 1 0.25 0.11111 0.0625 0.04 0.02778 0.02041 y 0 y 1 y 2 y 3 y 4 y 5 y 6 By Simpson’s one-third rule, we get [] 6 0 135 24 6 2 0 4( ) 2( ) 3 (1 ) dx h y yyy yy y x =++++++ + ∫ = [] 1 1 4(0.25 0.0625 0.02778) 2(0.11111 0.04) 0.02041 3 +++ + ++ = [] 1 1.02041 4(0.34028) 2(0.15111) 3 ++ = [] 11 1.02041 1.36112 0.30222 (2.68375) 33 ++ = 0.89458. Ans. Example 5. Evaluate 4 2 0 1 dx x x+ ∫ using Boole’s rule taking (i) h = 1 (ii) h = 0.5 Compare the rusults with the actual value and indicate the error in both. NUMERICAL DIFFERENTIATION AND INTEGRATION 323 Sol. (i) Dividing the given interval into 4 equal subintervals (i.e. h = 1), the table is as below: x 01234 y 1 1 2 1 5 1 10 1 17 y 0 y | y 2 y 3 y 4 Using Boole’s rule, 1 4 0234 0 2 (7 32 12 32 7 ) 45 h ydx y y y y y=++++ ∫ = 21 45 71 32 1 2 12 1 5 32 1 10 7 1 17 () ()+ F H I K + F H I K + F H I K + F H I K L N M O Q P = 1.289412 (approx.) ∴ 4 2 0 1.289412. 1 dx x = + ∫ Ans. (ii) Dividing the given interval into 8 equal subintervals (i.e. h = 0.5), the table is as below: x 0 0.5 1 1.5 2 2.5 3 3.5 4 y 1.0 0.8 0.5 4 13 0.2 4 29 0.1 4 53 1 17 y 0 y 1 y 2 y 3 y 4 y 5 y 6 y 7 y 8 Using Boole’s rule, [] 4 012345678 0 2 7321232143212327 45 h ydx y y y y y y y y y =++++++++ ∫ = () () () 14441 7(1) 32(8) 12(5) 32 7 .2 7 .2 32 12 .1 32 7 45 13 29 53 17      +++ +++ + + +           = 1.326373 ∴ dx x1 2 0 4 + z = 1.326373 But the actual value is () 0 4 4 11 2 0 tan tan (4) 1.325818 1 dx x x −− === + ∫ Error in I result = 1.325818 1.289412 100 2.746% 1.325818 −  ×=   Error in II result = 1.325818 1.326373 100 0.0419% 1.325818 −  ×=−   324 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 6. Evaluate the integral 6 3 0 1 dx x+ ∫ by using Weddle’s rule. Sol. Divide the interval [0,6] into 6 equal parts each of width 60 1 6 h − == . The value of 3 1 1 y x = + at each points of sub-divisions are given below: x 0123456 y 1.0000 0.5000 0.1111 0.0357 0.0153 0.0079 0.0046 y 0 y 1 y 2 y 3 y 4 y 5 y 6 By Weddle’s Rule, we get [] 6 0152436 3 0 3 5( ) 6 10 1 dx h yyyyyyy x =++++++ + ∫ [] 3 1.0000 5(0.5000 0.0079) 0.1111 0.0153 6(0.357) 0.0046 10 =++++++ ()() 3 1.131 5 0.5079 6 0.0357 10 =+ +   [] 3 1.131 2.5395 0.2142 10 =++ 3 (3.8847) 1.1654. 10 == Ans. Example 7. Evaluate the integral 3 1.5 0 1 x x dx e − ∫ by using Weddle’s rule. Sol. Dividing the interval [0, 1, 5] into 6 equal parts of each of width 1.5 0 0.25 6 h − == and the values of 3 1 x x y e = − at each points of sub-interval are given by x 0 0.25 0.50 0.75 1.00 1.25 1.50 y 0 0.0549 0.1927 0.3777 0.5820 0.7843 0.9694 y 0 y 1 y 2 y 3 y 4 y 5 y 6 Now by Weddle’s rule, we get [] 3 1.5 0152436 0 3 5( ) 6 10 1 x xh dx y y y y y y y e =++++++ − ∫ [] 3(0.25 0 5(0.0549 0.7843) 0.1927 0.5820 0.9694 6(0.3777) 10 = + + ++++ = 0.075[1.7441 + 5(0.8392) +6(0.3777)] NUMERICAL DIFFERENTIATION AND INTEGRATION 325 = 0.075[1.7441 + 4.196 + 2.2662] = 0.075(8.2063) = 0.6155. Ans. Example 8. Evaluate the integral 5.2 4 log , xdx ∫ using Weddle’s rule. Sol. Divide the interval [4, 5.2] into 6 equal sub-interval of each width 5.2 4 0.2 6 − == and values of y = log x are given below: x 4.0 4.2 4.4 4.6 4.8 5.0 5.2 y 1.3862 1.4350 1.4816 1.5261 1.5686 1.6094 1.6486 y 0 y 1 y 2 y 3 y 4 y 5 y 6 By Weddle’s Rule, we get 5.2 0152436 4 3 log [ 5( ) 6 ] 10 h xdx y y y y y y y= + + +++ + ∫ 3(0.2) [1.3862 5(1.4350 1.6094) 1.4816 1.5686 6(1.5261) 1.6486] 10 =++++++ 0.6 [1.3862 5(3.0444) 1.4816 1.5686 6(1.526) 1.6486] 10 = + +++ + 0.6 [6.085 5(3.0444) 6(1.5261)] 10 =++ 0.6 [6.085 15.222 9.1566] 10 =++ == 0.6 (30.4636) 1.8278 10 Ans. Example 9. A river is 80 m wide. The depth y of the river at a distance ‘x’ from one bank is given by the following table: x 01020304050607080 y 0 4 7 9 12 15 14 8 3 Find the approximate area of cross section of the river using (i) Boole’s rule. (ii) Simpson’s one-third rule. Sol. The required area of the cross-section of the river. 80 0 ydx= ∫ (1) Here no. of sub intervals is 8 (i) Boole’s rule, [] = ++++++++ ∫ 80 012345678 0 2 7321232143212327 45 h ydx yyyyyyyyy . 316 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES 6.5 TRAPEZOIDAL RULE Putting n =1 in equation (2) and taking the curve y = f(x) through (x 0 , y 0 ) and (x 0 , y 0 ) as a polynomial of. approximate value of π in each case. 320 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Sol. (i) The values of 2 1 () 1 fx x = + at 123 0, , , ,1 444 x = are given below: () 01 234 11 3 01 42. 0.0419% 1.325818 −  ×=−   324 COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES Example 6. Evaluate the integral 6 3 0 1 dx x+ ∫ by using Weddle’s rule. Sol. Divide the interval [0,6] into 6 equal parts each of width