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Subj: math solve the following questions ( handwritten) solution only neatly and clean it must be complete read instructions. thankyou sample solution below Topic: Numerical

Subj: math

solve the following questions ( handwritten) solution only neatly and clean it must be complete read instructions. thankyou

sample solution below

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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

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