I can't find solution of sinusoidal function's by numerical methods

I tried to write a code for a rectangle method(integral). Actually i saw same problem with other numerical methods too.

My problem is occur when i write a function like f = sin(x), sin(x^2) or x + sin(x^2) etc. (and i wrote a=0 and b = pi)

This code solves equations which doesn't have a trigonometric parameters.

How can i edit my code?

f =@(x) x^2;
a = 1;
b = 4;
n = 2;

h = (b-a)/n;

x(1) = f(1);

for j = 1:(n)
    x(j+1) = x(1) + j*h;
    c(j) = (x(j+1) + x(j)) / 2;
end

for i = 1:n
    s(i) = h * f(c(i));
end
fx = sum(s);
fprintf('\n f(x) = %f',fx);

Answer 1

[Jon]

I assume that you are trying to approximate the integral of f(x) between x = a and x = b.

You are initializing x incorrectly. You have

x(1) = f(1);

should be

x(1) = a;

Also you have n set to be equal to 2, but you will need many more points than this to get an accurate answer. Try with for example n = 100

Actually you can replace all of your code with just a few lines taking advantage of MATAB's ability to deal directly with vectors.

f =@(x) sin(x.^2);
a = 0;
b = pi;
n = 100;
h = (b-a)/n;
x = linspace(a,b,n);
c = (x(2:end)+x(1:end-1))/2;
fx = sum(h*f(c))
% check your result with MATLAB's built in integration routinne
fquad = integral(f,a,b)

If for some reason you really want to break it down and see each step in a loop it would be clearer and more efficient to just march through in one loop:

f =@(x) sin(x.^2);
a = 0;
b = pi;
n = 100;
h = (b-a)/n;
x1 = a;
fx = 0;
for j = 1:n
    x2 = x1 + h;
    c = (x1 + x2 )/2;
    fx = fx + f(c)*h;
    x1 = x2;
end

Finally, I would note that since your points are regularly spaced, there is no need to find the center of each interval, you can calculate it directly. So you could just use:

c = a+h/2:h:b

[Ahmet Keçeci]

f = @(x) ((exp(-x^2)) * sin(x^2+1));
a = 0;
b = pi/2;
n = 100;

h = (b-a)/n;

x(1) = f(a);

...

i know solution is 0.7484696904 for n = 100

but i found 0.113822 when i use this code. (That code finds true value for x^2 and like other non-trigonometric functions.)

[Jon]

I think your comment got out of synch with my answer. Please try the above suggestions. Most importantly initalize x(1) = a not x(1) = f(a). Note if a = 0 and f(a) = 0 then it won't matter which way you assign it. So your current code should still work for functions like sin(x) and sin(x)^2 as long as you start at a = 0, but still it makes no sense to have x(1) = f(a). For the function in your comment f = @(x) ((exp(-x^2)) * sin(x^2+1)) it will definitely make a difference as f(0) = 0.8415