Question

Use Simpson's 1/3 rule with n segments to approximate the integral of the following function on interval [1, 13] f(t) = 1.945 · sin (27) The exact value of the integral is Teract = 15.4821 Fill in the blank spaces in the following table. Round up your answers to 4 decimals. Relative error et is defined as I - Ieract * 100% Et = Texact n, segments I, integral Et(%) 2 8

Answer 1

lo According Simpsons rule. 1 (f(no) Como) & hf (mi) + 2 f(22) + ... + f(nn) 3 when ne2 ans 13-1 11 12 2 6. 2. 13 77>?, f(no) = 1.945. sin (0) zo, F(MI) = 1.965, sin (3) = 1.9401 F(2) = 1.945esis (3) = 0.2744 1- CFemelt ufcm.) + Feas) - 2[+Gr! -]]= z[©turl.9401 * 0-2746) ) = 278.0348 = 16.0696 Therefore I = 16.0696

Now cuben n = 8 an= 13-1 11 12 8 العالم N $ then z = 5.5, nuo 7, x=8.5 2 Mos!, 7,-2.5, 76= 10, 217 = 1105, ng = 13. E 0.7124 f(mo) = 1.945 stisco) 70, f(m) = 1.94.57 sin(a)=0.7 FIN2) = 1.9454 sin (2) = 1.3257, F(M3) = 1.9254 Sin ) = 1.7549 f(an) = 1.945 y sin ( 3 ) = 1.9401 f(ms) = 1.945 usin ($)=1.8557 F(mp) = 1.945 a sin )=1,5133 F(mm) = 1.9454 sin () - 0.9606 f(ng) = 1.945+ sin(3) = 0.2744

then (F ( ۷۰ ) + f ( w ) + 2 ( * ) + ( ( 2 ) 2f ( ۷۰ ) + ( ( wرجه =T 3 +2( (۷۵) + F (۶) ۹۴ (72] . اج = = د اما [5 3 ۰۰۰ ۱۹۸۵۱ به ۹ با ۶۱۰۶ا +1،3257 2 + 0712ما + ا ا 27 % 6064 م ا 1 05133 2+7 $1095 47 + ۰۹۹ ) 1 = = 96.961): 15 ، 335

in, segments I, integral Ę (%) 2 16.0696 3.7942 15.4835 0.009 8 for n=2. 16.0696-15.4821 & 100 3.794.7 % 18-4821 For n=8 15.4835 - 15.4821 0.0097 Ela 15.4821