How to solve this problem of integeral?

Can anypody help with this piece of code?

The problem is with the use of the integeral function

%------------------------------------------------------------------------%
clear variables; 
close all;
clc;
%------------------------------------------------------------------------%
x=-1:0.1:1;
n=length(x);
y=zeros(1,n);
for i=1:n
     y = m_cost(x);
end
plot(y)

and here is my function:

function y =m_cost(x)
fun1=inline('0-x','x');
fun2=inline('x-4','x');

syms fun1;
syms fun2;
syms x;

if (x>=0)
y=x.^2+4*x+8*int(fun1,x,0,10);
else 
y=x.^2+4*x+8*int(fun2,x,0,10);
end

[Lazaros Christoforidis]

can you type the equation of integral u trying to calcuate?

[darova]

What about this?

[Lazaros Christoforidis]

Well I dont know what inline does, I want you to post the math equation so I can try to solve it with different method if u want

[Lazaros Christoforidis]

Is it a differential equation?

[Hamzah Faraj]

Thank you. This is really informative.

If I have this cost function, for example, is using the integeral function the right option? I want to integerat int(p) over time dt (axis x).

% Assume this is the results of the temperature function v(t)=v(i-1)+A*(time);
v = -1:0.1:5;    
% A is the heaating parameter
A = 4/3;

% Cost at each v(t) is given by the expression:
% y(v(t))=(50*int(p)+10*x+30)/(x/A);
% P = x-4   if x > 4 and x <= 5
% p = 0     if x >= 0 and x <= 4
% p = -x    if x >= -1 and x < 0

y = cost(v)



function y= cost(v)
func1 = @(v) v-4;
func2 = @(v) -v;

% Cost of each v(t) at time t is given by the expression:

% y(v(t))=(50*int(p)+10*x+30)/(x/A);
% P = x-4   if x > 4 and x <= 5
% p = 0     if x >= 0 and x <= 4
% p = -x    if x >= -1 and x < 0

if (v > 4 & v <= 5)
    y = ((50*integeral(func1,0,inf)+((v./A)*10)+30)/(v./A);
elseif ( v>=0 & v<=4)
    y = (((v./A)*10)+30)/(v.A);
else
    y = ((integeral(func2,0,inf))+((v./A)*10)+30)/(v./A);
end

end

Answer 1

[Ameer Hamza]

There is no need to use inline. You can use an anonymous function. Also, int() is used for symbolic integration. Here you can use numeric integration using integral(). The function can be vectorized so there is no need to use a for-loop

%------------------------------------------------------------------------%
clear variables; 
close all;
clc;
%------------------------------------------------------------------------%
x = -1:0.1:1;
n = length(x);
y = m_cost(x);

function y =m_cost(x)
    fun= @(x) -x;
    y = x.^2+4*x+8*integral(fun,0,10);
end

Answer 2

[Walter Roberson]

int is only for use with Symbolic functions or symbolic expressions. inline objects cannot be used with int.

You should only use inline if you have a very specific reason to use it. You should be using symbolic or anonymous functions instead.

However your

syms fun

on the next line overwrites fun the same as if you had done

fun = sym('fun') ;

So when you int that it is just a symbolic constant and the integral is the constant times the difference in time. Your function is going to return a symbolic expression involving the two unresolved values x and fun. x is unresolved because the

syms x

that you did overwrote the x parameter passed to the function.

so you return a symbolic expression that contains unresolved variables and outside the function you assign that into part of y. But y is double precision and you cannot store a symbolic expression with unresolved variables in double precision.