How do I exclude values in linspace?

Hello! I have here a code that integrates a function using the Simpson's 1/3 Rule. Supposedly, my output is the step size (h), x values in a vector, and the final value of integration (y). My question is (1) how do I display the x-values in such a way that a and b is excluded? And (2) how do I display y without using inline?

For (1): The vector should be sized at [1 17] but my vector output right now is at [1 19].

For (2): The variable y must be of data type double and I understand that I'm using a function_handle. But for this, we aren't allowed to use inline. Is there a way I can do this without using inline?

Here is what I've made so far:

function [h,x,y] = integral(a,b,n)
    x0 = a:n:b;
    y = @(x) (log(abs(2*x/(1-2*x)+1)));
    
    % step size
    h = (b - a)/n;
   
    % summation of odd and even terms
    odd = 0;
    even = 0;
    
    % for odd terms
    for i = 1:2:n-1
        x0 = a + (i*h);
        odd = odd + y(x0);
    end
    
    % for even terms
    for i = 2:2:n-2
        x0 = a + (i*h);
        even = even + y(x0);
    end
   
    % simpson 1/3 rule
    s = (h/3) * (y(a) + y(b) + 4*odd + 2*even);
    
    % values of x in vector
    x = linspace(a,b,n);
end

To call the function above, I made this:

a = 1+10*rand();
b = a+30*rand();
n = 1+round(20*rand(),0);

[h,x,y] = integral(a,b,n)

And the output I get is:

h =

    0.2468


x =

    9.3878    9.6501    9.9123   10.1746   10.4368   10.6991   10.9614   11.2236   11.4859   11.7481   12.0104   12.2727   12.5349   12.7972   13.0594   13.3217   13.5840


y =

  function_handle with value:

    @(x)(log(abs(2*x/(1-2*x)+1)))

[Luca Ferro]

for the point 1), after the linpace declaration cut the extremes:

x = linspace(a,b,n);

x=x(2:end-1);

Answer 1

[Davide Masiello]

Hi @lea.c, I have several comments about your code.

1) MatLab already implements a function called integral, so I would recommend renaming your function to something like my_integral.

2) At the moment, y seems to be the function you want to integrate rather than the final value of the integral.

3) The for loops are unnecessary.

4) It is also not necessary to define x after you have already defined x0.

5) Use dot notation for vector operation! i.e. log(abs(2*x./(1-2*x)+1))

I'd rewrite your code as below.

a = 1;
b = 10;
n = 10;
[h,x,y] = my_integral(a,b,n)

h = 0.9000

x = 1×9

1.9000 2.8000 3.7000 4.6000 5.5000 6.4000 7.3000 8.2000 9.1000

y = -16.1617

Let's now double check with the MatLab built-in function.

integral(@(x) log(abs(2*x./(1-2*x)+1)),a,b)

ans = -18.9722

The values are close but not quite the same. If we increase n and try your code again we get

n = 100;
[h,x,y] = my_integral(a,b,n)

h = 0.0900

x = 1×99

1.0900 1.1800 1.2700 1.3600 1.4500 1.5400 1.6300 1.7200 1.8100 1.9000 1.9900 2.0800 2.1700 2.2600 2.3500 2.4400 2.5300 2.6200 2.7100 2.8000 2.8900 2.9800 3.0700 3.1600 3.2500 3.3400 3.4300 3.5200 3.6100 3.7000

y = -18.7052

Which is much closer, as it should be expected. Keep increasing n and you'll get just the right answer.

function [h,x,y] = my_integral(a,b,n)
    h   = (b - a)/n;                                            % step size
    x   = a:h:b;                                                % Integration points
    f   = @(x) log(abs(2*x./(1-2*x)+1));                       % Function to integrate
    y   = (h/3)*(f(a)+f(b)+4*sum(f(x(3:2:n-1)))+2*sum(f(x(2:2:n-1))));        % Simpson's 1/3 rule   
    x   = x(2:end-1);
end

[lea.c]

Hi Davide, thank you for your help! I mixed what you sent into my original code since this is a schoolwork/exercise and we are given the starter code: function [h,x,y] = integral(a,b,n). I should've also added that we are doing it on a MATLAB Grader. Here is what I ended up with:

function [h,x,y] = integral(a,b,n)
    % step size
    h = (b - a)/n;
    
    % integration points
    x = a:h:b;
     
    % function to integrate
    f = @(x) log(abs(2*x./(1-2*x)+1));
    
    % summation of odd and even terms
    x_odd = 0;
    x_even = 0;
    
    for i = 1:2:n-1
        x = a + (i*h);
        x_odd = x_odd + f(x);
    end
    
    for i = 2:2:n-2
        x = a + (i*h);
        x_even = x_even + f(x);
    end
    
    % simpson's 1/3 rule
    y   = (h/3) * (f(a) + f(b) + 4*x_odd + 2*x_even);
    
    % x-value vector
    x = x(2:end-1);
end

And I used the a,b values you have provided, with n = 150, which gives me the correct step-size.

Now my problems are:

• The x-value now is a 1x0 empty double row vector, and
• I did obtain the value y = -18.9722 but it says that the value of the integral is wrong. Honestly, I'm not really sure what to do.

I did use what you sent above, exactly as it is, other than changing the values for n similar to the code above.

• It gives me the same step size which is h = 0.0600, but it shows that it's wrong.
• The x-values are wrong too, and it says: Use [x_odd, xeven] to combine the odd and even values of x.
• The y-value becomes y = -18.7946

[Davide Masiello]

That is because you change the value of x in the for loops, which is not needed.

Furthermore, I think you sould not include the values of the function calculated at a and b in the summation of the odd and even elements.

Try with

function [h,x,y] = integral(a,b,n)
    % step size
    h = (b - a)/n;
    
    % integration points
    x = a:h:b;
     
    % function to integrate
    f = @(x) log(abs(2*x./(1-2*x)+1));
    
    % summation of odd and even terms
    x_odd = 0;
    x_even = 0;
    
    for i = 3:2:length(x)-1
        xi = x(i);
        x_odd = x_odd + f(xi);
    end
    
    for i = 2:2:length(x)-1
        xi = x(i);
        x_even = x_even + f(xi);
    end
    
    % simpson's 1/3 rule
    y   = (h/3) * (f(a) + f(b) + 4*x_odd + 2*x_even);
    
    % x-value vector
    x = x(2:end-1);
end

[lea.c]

Thank you for your help, Davide! I have found a way to make it work. This is the working code that I have made:

function [h,x,y,a,b,n] = integral(a,b,n)
    % step size
    h = (b - a)/n;
    
    % x-values
    xi = 0;
    xf = 0;
    
    % vectors/array
    x = a:h:b;
    x(end) = [];
    x(1) = [];
    
    % for odd and even terms
    x_odd = x(1:2:end);
    x_even = x(2:2:length(x)-1);
    x = [x_odd x_even];
    
    % summation of odd terms
    for i = 1:length(x_odd)
        syi = f(x_odd(i));
        xi = xi + syi;
    end
    
    % summation of even terms
    for i = 1:length(x_even)
        syf = f(x_even(i));
        xf = xf + syf;
    end
    
    % integral and simpson's 1/3 rule
    sum1 = f(x_odd);
    sum2 = f(x_even);
    y = (h/3)*(f(a) + f(b) + 4*sum1 + 2*sum2);

    function y=f(x)
        y=log(abs(2*x/(1-2*x)+1));
    end
end

[Davide Masiello]

Hi @lea.c, my pleasure, glad you found a suitable solution.

If you think my answer helped, please consider accepting it for future reference to other users.