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Subj: math

Topic: Numerical Integration ATT O + sinx - 1 X = ex 0 1+ 1cosx Used a ) 4 and 6 segment Trapezoidal Rule b) 6 segment Simpson's /3 Rule () l segment Simpson's / RuleDATE Assignment : 5.2 4 log x dx 1.) Trapezoidal n : 6 Error : he 5.2 - 4 = 0.2 Error = 0.0105 %% I = 2 (0.2) [4(4) + 2:(4.2) + 20 ( 4.4 ) + 2p ( 4.6 ) + 214.8 ) + 24( 5) + 85. 2) ] I = (0.2) (0.6070594913+2(0:023 24 92904) + 2(0. 0434526765 ) + 2 (0.60 275783 17 ) + 2 (0.081 2412374 ) + 2(0.698 9700043) + 0. 71 60033436) I = 0.7437405416 => actual value I = 0. 79 38 24 0433 => true value 2) 1/3 Rule 1 = 6 h = 0 . 2 J = = ( 0.2 ) [ F ( 4 ) + 41 ( 4.2 ) + 2*( 4.4) + 4F ( 4.6 ) + 2f( 48 ) + 4( 5 ) + F(5.2) ] 1 = = (0.2) 60.4020 599 913 + 4 (0.62324 92904)+ 2(0. 143 43 26745) + 4 (0. 642757 8317 ) + 2 (0. 681 241 2374 ) + 4 (0.6989700043 ) + 0.7160633436] I = 0. 79382 39 779 => actual value I = 0. 7938 24 0433 =) True value Error : 0.00 000 823 % or : 8. 23 X 10 %DATE 3 ) 3 / 8 Rule 1= 6 h : 0. 2 J = = (0.2 ) f ( 4 ) + 3 f ( 4.2) + 3 f ( 4.4) + 2 f14.6) + 3F ( 4.8) + 3+(5) + +15.2) J : 2 (0 .2 ) 10 . 0 0205 9 9 9 1 3 + 3 ( 0 . 423 24 9 2 90 4 ) + 3 ( 0 . 64 3 4 5 207 45 ) + 2 ( 0.462757 8317) 1 3 ( 0 . 68 1 2 4 1 2 3 7 4 ) + 3 ( 0 . 6 9 8 9 7 0 0 0 43 ) + 0.71 600 33436 I = 0. 7938 238908 => actual value Error : 0. 00001 8455 I = 0. 793824 0433 => true value = 1. 8455 x 10-5 4) 2 18 Rule n = 9 h = 5.2- 4 2 15 - 8 ( 15 ) + ( 4 ) + of ( 4 2 ) + 3f ( /8 ) + 2f ( ? 3 ) + of ( be ) + 3+ ( 1 4 ) + 2f ( 24 ) + of ( is ) +of ( 75 ) + + ( 5.2)] J: B ( 2) 0.6020 59 9 913 + 3 ( 0 . 61630 04304 ) + 3 ( 0.430088 7149 ) + 2 ( 0. 04 34 5 24 705 ) + 3 ( 0 . 45 4 4 17 6 5 37 ) + 3 ( 0 . 0 6 9 0 0 0781 ) + 2 (0. 6812412374 ) + 3 ( 0 . 6 9 3 1 4 0 4 0 0 7 ) + 3 (0 . 70 4 7 2 2 3 3 32 ) + 0. 71 400 3 3436 I : 0. 7938 24 0142 = actual Error = 3.4039 x 10" I = 0 . 39 38 24 0 433 = 7 truevalue = 0. 0000 034039\f( 1) Solution of x = f / IT 141 / Sinn After compairing x = [ bfin, du we get f ( n ) = sin x " a = 0 6 = TT Here the value of function for) = It corn at b= TJ does not exist . Therefore the value of X = [ 's sinn ) ax does not exist by any given it corn methods . 5 .2 (2 ) Solution of X = / logn dn . Actual Value of X = 1.0270 4741 After compairing it from x = [ f ( my) do we gett a = y b = s . 2 9 f ( n ) = logix . Ja Now we solve above integral by given method. ( a ) Trapezoidal Rule , Trapezoidal Rule for findy is given as [ fend x = = [ f ( 1 0 ) + 2 { f ( m ) + f ( x , ) + . . . + f ( x n . > y + f ( 20 ) where , h= ath all: Trapezoidal Rule with n = 4 . For ny we get the trapezoidal Rule from eq" + 5. 2 7 , log n ax = = [ log ( no ) + 2 4 log( m) + log ( 12 ) +log ( 7> > > + log (x4))S ( C) Simpson's 13 10) Rule for n= 6 ( mod Simpson's Bla) Rule for net in given and + 2 f ( 2 3 ) ] Ja Now after simplifying as we do in previous weget loon do = 1. 0 2 7 04 7071 Estimated Value logs da - 1 . 8 27 0 4 741 Actual Value Exor = E. V - A. V A. V( b) Simpson's ( 13) ra Rule ? Simpson's ( /3 ) " full for ting du is given as [ fox ) da = = [ { f ( x o ) + f ( and j + 4 . f ( = > + f ( *3 ) + - - - +finn- 1 3+ For n = 6 7 Joy - 3[ log no + log xc ) + 4 / leg m + log * 3 + log ns ) + 2 ( log * 2 + egg zy ) Now we do same process for hi up, my - - - 26 ap in previous solution we have done . 5. 2 7 / logn dx = 32 [ log 4 + log s . 2 ) + 4 ( log 4.2 + log. 4.6 + log 5 ) + 4 2 ( dog 4 . 4 + log 4 . 82 ] = 3 2 / ( 1 3 0 6 2 9 4 3 6 1 + 1 6 4 0 65 0 6 26 ) + 4 ( 1 4 3 50 0 4 5 25 + 1.526056 303 + 1 . 6 0 9 4 3 7 9 1 2 ) + 2 ( 1 4 8 1 6 0 4 5 4 1 + 1 . 56 8 61 59180 52 Jy Rogna x = 1. 027047250 - Estimate value Error = 1 1 . 027 047 250 - 182 704 741/ X 100 = 0 00.00.08./0 1. 827 04 74 )ali For n= 6 Trapezoidal Bule . from egh DD weget ( b ) If ( x ) d x = = [ f ( x 1 0 ) + 2 d f ( 70 + f ( x 2 ) + f ( 13 ) + finy ) + fins ) ] + find Sin f fonda = bo [ log no + 2 { log * + log * s + log ng + log my + log ns * + logic * Ja Now calculate the value of h, no, 24 212, 213, my, Mr, nc wegel Fo h = b - A _ 5 . 2 - 4 - 12 12 = 012 no = a = y H = ath = 4 + 0 . 2 = 4. 2 R F = at sh = 4+ 1 = 5 26 2 = 9 +2 5 = 4+0.4 = 14.4 212 76 = at 6 h = 4+1 12 = 52 R g = Q+3h = 4+ 0'6= 4.6 214 = 9+4 R= 4+0.0=4.0 After putting this value in eq" mmy we get- " log max - % lagy + 20 log 4 2 + log 4 4 + 109 4 . 6 + 109 4 . 0 + logs ] tlog s. 2] = Diz 1: 30 6 2 9 43 6 1 + 2 / 1: 4 3 5 0 09 4 525 + 1. 48 16 04 541 + 1 52 605 6323 + 1 5 6 0 6 1 5 910 + 1 . 6 0 9 4 3 7912 ) + 1648 658626 - 1 0276 55139 log x dx = 1 . 0 2 76 5 5 139 / Estimated value 2 H Jy logn da = 1 . 02704741- Actual Value E . V. - A . V) XIDD - O.01 90 A. VNow find the value of h, no, Mi, M2, ng, & My for ney . A h = b - a _ s. 2ch _ 1: 2 = 0 . 3 7 * = atoh = y 262 = 91+3 h = 4+ 0. 9 = 4.9 Mi= ath = 4+ 0.3 = 4.3 My = a+ 4 h = 4+1.2 =5- 2 2 2 = at 2h = 4 + 0.6 = 4.6 Now but these values in egh ED!). we get y " logkan = 2 2 [ log ( 4 ) + 2 1 log ( 4 : 3 ) + log ( 4 . " ) + log ( 1 9 / } + log ( sued* = 0 . 15 / 1 30 6 2 9 4 3 6 1 + 2 0 / 1 0 4 5 0 6 1 5 0 23 + 1 . 52 60 5 6 303 + 1- 50 92 35 205/ +1 64065 0 626 / 2 501 = 0'15 / 1 . 30 6 2 94 3 61 + 9. 147 013062+164 86586261 = 1 . 0 2 7 41 4 9 07 71 log n d x = 1 027 41 4 9 07 / Estimated Value Jy Error = /Estimated Value - Actual valecel Actual Value = (1 02 7 4 1 4 907 - 1 02 704741/ X100 = 0102 1 0 1 . 02 7 04 7 4 1 Error = 0.02 70 Y.A