mid-point method Integration

Find numerically , to a 1/10000 accuracy, the values of the following definite integral:

0 to inf 1/(x^2+1) dx

use the mid-point method.

not show how to answer this i went about integrating it. My knowledge of the midpoint rule is limited.

the width of the sub intervals would be 1/10000 but how would you go about dividing it by infinity. I did the integration:

if true
  % code
syms x
a1= int(1/(x^2+1),x,0,inf)
end

[John D'Errico]

Why do you think you need an "if true" in there??????

[Youssef Khmou]

the instruction if true appeared because the poster clicked on "Code" button

Answer 1

[Youssef Khmou]

Theoretically that integral equals pi/2, here is version, try to adjust it :

% MidPoint test integration
clear;
f=inline('1./((x.^2)+1)');
N=20000;
dx=1/1e+2;
F=0;
x1=0;
for t=1:N
  xi=(dx/2)+x1;
  F=F+dx*f(xi);
  x1=x1+dx;
end

% For verification try :

 quad(f,0,1e+18)

[Tony]

thank you for your time question. N should equal 10000 no? and dx=1/100 right? just don't understand why dx is 1/100.

trying to understand since i don't fairly understand the midpoint rule. it should be something like ((b-a)/subintervals) * f(x1)+f(x2)...f(xn). I don't see how it fits in that general form

[Youssef Khmou]

dx was taken arbitrarily, try with dx=1e-4 . that code is based on the formula in this file, try to look at this file first

http://www.math.ohiou.edu/courses/math3600/lecture22.pdf

Answer 2

[Akshay satpute]

tell me program of integration of {x^2 (sinh(X)+cosh(x))dx} between limit 0 to 1 in 100 parts