Facing problem while solving this.

    p1 = @(x) (1./F(x)) .* ((0.5 .* hbar(x)) - Q);
    P = @(x) integral(@(x) p1(x),0,x,'ArrayValued', true);

     xi = linspace(0,1) ; 
    P1 = zeros(size(xi)) ; 
    for j = 1:length(xi)  
        P1(j) = P(xi(j));
    end
    

Using the abpve approach I am getting datasets of P1 but not able to figure out how to approach in the following method given below

    for i = 1:100
        p1(i) =(((0.5*hbar(i))-Q)/F(i));
%         P = @(x) integral(@(x) p1(x),0,x,'ArrayValued', true);  %%%% how
%         to write this line here?
    end

Answer 1

[Walter Roberson]

    p1{1} = @(x) (1./F(x)) .* ((0.5 .* hbar(x)) - Q);
    P{1} = @(x) integral(p1{1}, 0, x, 'ArrayValued', true);

    xi = linspace(0,1) ; 
    P1 = zeros(size(xi)) ; 
    for j = 1:length(xi)  
        P1(j) = P{1}(xi(j));
    end
    
    for i = 1:100
      p1{i} = @(x) (((0.5*hbar(i))-Q)/F(i));
      P{i} = @(x) integral(p1{i}, 0, x, 'ArrayValued', true);
    end
 

Note that this would overwrite the original p1{1} and P{1}

Also note that your p1{i} does not involve x, so the body would effectively be constant, and the result of the integral would be x times the constant minus 0 times the constant, which is predictable ahead of time.

[AVINASH SAHU]

Appreciating your help. But I want to say that I am getting the results (i.e. datasets of P1) using the following approach

    p1 = @(x) (1./F(x)) .* ((0.5 .* hbar(x)) - Q);
    P = @(x) integral(@(x) p1(x),0,x,'ArrayValued', true);

     xi = linspace(0,1) ; 
    P1 = zeros(size(xi)) ; 
    for j = 1:length(xi)  
        P1(j) = P(xi(j));
    end
    

and I want to get the same results using the below approach but don't know how to proceed

       
        p1(i) =(((0.5*hbar(i))-Q)/F(i));   % same as above p1
%        P = ?????

%         P = @(x) integral(@(x) p1(x),0,x,'ArrayValued', true);  %%%% how
%         to write this line here in the form of indices and the obtain the
%         datasets of uppercase "P1"
       
       xi = linspace(0,1) ; 
       P1 = zeros(size(xi)) ; 
       for j = 1:length(xi)  
            P1(j) = P(xi(j));
       end

Thank you very much for your help. Hope I am able to clarify my problem.

[Walter Roberson]

You cannot do that.

Your original code defines p1 as an anonymous function handle. Your new version defines p1(i) as a numeric constant, not as a function handle. It is not useful to integrate a numeric constant.