T(n) S(n) 0.217206 0.214602 0.215253 0.214602 0.214765 0.214602 32 0.214643 0.21

Question

T(n) S(n) 0.217206 0.214602 0.215253 0.214602 0.214765 0.214602 32 0.214643 0.214602 0.214612 0.214602 3. On the spreadsheet, we can also tabulate the absolute errors la where r is the exact value of the integral and cis the approximation. We combine these values in the above table to get T(n) Absolute Error in T(n) Absolute Error in S(n) 4 0.217206 0.214602 2.60 x 10 3.78 x 10 8 0.215253 0.214602 6.51 x 10 5.91 x 10 10 16 0.214765 0.214602 x 10 T 1.63 9.24 x 10 12 32 0.214643 0.214602 407X 10-3 1.44 x 10 m 64 0.214612 0.214602 1.02 x 10

Explanation / Answer

5)

a)

for T_n factor = 2.60 * 10^(-3) / ( 6.51 * 10^(-4) ) = 3.999 = 4 (approx )

For S_n   factor = 3.78 * 10^(-8) / ( 5.91 * 10^(-10 ) = 63.95 = 64 ( approx )

6)

1)

matlab code to calculate simpson and trepezoidal method is shown below

%%%% simpson %%%

clc;
clear all;
close all;
format short
a=0;
b=pi;
I_actual=pi*log(1+sqrt(3)/2);
for n=5:100;

x=a:(b-a)/n:b;
for j=1:length(x)
y(j)=log(2+cos(x(j)));
end

sum=0;
for k=1:length(x)
if (k==1||k==length(x))
sum=sum+y(k);
else if (rem(k,2)==1)
sum=sum+2*y(k);
else
sum=sum+4*y(k);
end
end

end
I=(b-a)/n*1/3*sum;
if (abs(I_actual-I)<=10^(-6))
break;
end

end

fprintf('Integration by simpsons method =%d ', I);
fprintf('Integration by Actual method =%d ', I_actual);
fprintf('N to get value within given tolerence =%i ', n);

output :

Integration by simpsons method =1.959760e+000
Integration by Actual method =1.959759e+000
N to get value within given tolerence =10   

2) for b part just change the function value of a and b and atual value

%%%%

clc;
clear all;
close all;
format short
a=0;
b=2;
I_actual=log(32)-2;
for n=5:100;

x=a:(b-a)/n:b;
for j=1:length(x)
y(j)=(x(j)^2+1)/(x(j)+2);
end

sum=0;
for k=1:length(x)
if (k==1||k==length(x))
sum=sum+y(k);
else if (rem(k,2)==1)
sum=sum+2*y(k);
else
sum=sum+4*y(k);
end
end

end
I=(b-a)/n*1/3*sum;
if (abs(I_actual-I)<=10^(-6))
break;
end

end

fprintf('Integration by simpsons method =%d ', I);
fprintf('Integration by Actual method =%d ', I_actual);
fprintf('N to get value within given tolerence =%i ', n);

output :

Integration by simpsons method =1.465737e+000
Integration by Actual method =1.465736e+000
N to get value within given tolerence =20

7)

for I3

clc;
clear all;
close all;
format short
a=0;
b=pi/2;
n=10;
x=a:(b-a)/n:b;
for j=1:length(x)
y(j)=(x(j)^2+1)/(x(j)+2);
end

sum=0;
for k=1:length(x)
if (k==1||k==length(x))
sum=sum+y(k);
else if (rem(k,2)==1)
sum=sum+2*y(k);
else
sum=sum+4*y(k);
end
end

end
I=(b-a)/n*1/3*sum;

fprintf('Integration by simpsons methodusing N=10 =%d ', I);

output : =

Integration by simpsons methodusing N=10 =4.571427e-001
>>

for I4

clc;
clear all;
close all;
format short
a=0.00001;
b=1;
n=10;
x=a:(b-a)/n:b;
for j=1:length(x)
y(j)=x(j)^2*log(x(j))^3;
end

sum=0;
for k=1:length(x)
if (k==1||k==length(x))
sum=sum+y(k);
else if (rem(k,2)==1)
sum=sum+2*y(k);
else
sum=sum+4*y(k);
end
end

end
I=(b-a)/n*1/3*sum;

fprintf('Integration by simpsons methodusing N=10 =%d ', I);

output :

Integration by simpsons method using N=10 =-7.440744e-002

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