Question

1. Use the Trapezoidal Rule to numerically integrate the following polynomial from a = 0.5 to b = 1.5 f(x) = 0.2 + 25x – 200x2 +675x3 – 900x4 + 400x5 Use three different number of segments and show that higher number of segments give lesser relative error considering the exact value of the integral, which is 49.3667. 2. Write a MATLAB or OCTAVE code to solve the above problem numerically and verify your result. Copy/paste the code and answers in the answer script.

Answer 1

anes f(x) = 0.2 +25%-2003 +675x3_900 x4 + 90b azos to b=1.5 Exact value = 49.3567 Formule for trapezoidal rule: I I= (f(x) + f(x) +24%)+F(X) + -) S ha boa is at i*) is no. of segments mai h=1-5-0.5 - 1 - Ic 5 [f(x=0.9) + f(x=1-5) *[0.2725(0.5)-260(65+67563-91006094 +4070 55€ + 0.2+25(1-5)-200 (1.52 +675(15)3-900(1-5)+ 400(1-5) -{3.325 + 347.075) I =175.2 Error s E-Exactuale) xico Exact value I nco .|75.2- 04.36674 woo = 254,895 % 49.3667

for nez halsos, os Is toros 566 -0.5) - 3.325 : q= 1.0 F(x,-1.0 - 0.2 = 1-5 → Flix=1.5) = 347.075 .I- {f(x) +f(x) + 2 (F(X.) -0.5 (3.325 + 347.075 + 2x0.2] I: Essor 87.5 = I- Exact value) ooo 1 Exact value 87.7 .49.3667 o = 77.65 49.367 Tor 123 ha 15-as - 0.333333 To=0.5 = f(x0.5) - 3.325 21=0.833333 = 4*40.833333) = -0.5072976 42 = 1.166666 = f(x= ) = 26.2157444 43=ts = f(x3 = +-5) = 347.075 I- (F(Xo) +FCx3) + 2 (PCXd tf(x)] 22043.325+347.075+21-0.5072976 26.2154464)] s son227 66.9694 Error = I- Exact vakia | xloo , Exact vake!

166.9694-49-36677 xloo 49.3667 = 35.6SY Relative emer is decreasing segments are increasing as the no. of

MatLab program:

clear
clc
% function
f=@(x) 0.2+25*x-200*x^2+675*x^3-...
        900*x^4+400*x^5;
% limits
a=0.5;
b=1.5;
A=49.3667;
fprintf('\nSolution by analytical method');
fprintf(' is %f',A);
% no. of segments
nn=[1 2 3];
for k=1:length(nn)
    n=nn(k);
    h=(b-a)/n;
    for i=1:n+1
        x(i)=a+(i-1)*h;
        y(i)=f(x(i));
    end
    I=0;
%formula I=(h/2)*(y0+yn+2(y1+y2+.....yn-1)
I=(h/2)*(2*sum(y(2:end-1))+y(1)+y(n+1));
fprintf('\n\nn=%d\n',n);
fprintf('\nSolution I=%f',I);
Er=abs((A-I)/A)*100;
fprintf('\nTrue percent error is %f%%',Er);
end

Screenshot:

1 clear 2 clc 3% function 4 f=@(x) 0.2+25*x-200*x^2+675*x^3-... 5 900*x^4+400*x^5; 6% limits 7 a=0.5; 8 b=1.5; 9 A=49.3667; 10 fprintf('\nSolution by analytical method'); 11 fprintf(' is %f',A); 12 % no. of segments 13 nn=[1 2 3]; 14 for k=1:length(nn) n=nn(k); 16 h=(b-a)/n; for i=1:n+1 18 x(i)=a+(i-1)*h; 19 y(i)=f(x(i)); 20 end I=0; 22 %formula I=(h/2) * (yo+yn+2(y1+y2+.....yn-1) 23 I=(h/2)*(2*sum(y(2:end -1))+y(1)+y(n+1)); 24 fprintf('\n\nn=%d\n',n); 25 fprintf('\nSolution I=%f',1); 26 Er=abs ((A-I)/A) *100; 27 fprintf('\nTrue percent error is %f%%', Er); 28 end 15 17 21

Save the above program and execute it.

Result:

Solution by analytical method is 49.366700 n=1 Solution I=175.200000 True percent error is 254.895101% n=2 Solution I=87.700000 True percent error is 77.650116% n=3 Solution I=66.969547 True percent error is 35.657330%

I hope this will help you.