Integral operator in a for loop

Hi folks! I am trying to calculate the integral of a function while it is in a loop,but MATLAB gives error. How can I change "v" each time? this is the code

f=@(y) sqrt(y)/(1+exp(y/v))
for v=-5:0.1:5
 q(i)=integral(f,0,inf);
end

[Shweta Rajoria]

If we want to calculate double integral in a loop then what will be the modification in code below. If any body know the answer please help me. Any suggestions will be appreciated. Thanks in advance.

v= 1:1:5;
L= length(v)
for i=1:L
    w=v(i)
    fun= @(x,y) w.*x.*exp(-x.*y./w);
    c(i)= @(y) integrate(@(x)fun(x,y), 0,Inf);
    fun1= @(y) y+c(x)
    c1(i)= integrate(fun1, 0,1)
end
diplay(c1())

[Walter Roberson]

Use integral instead of integrate -- just a small mistake in the name of the function.

[José Anchieta de Jesus Filho]

so, I need to create a loop that calculates integrals at each point. Say I have the following function handle

D = @(x,y) x.*(Z(x,y).^2);

from that function, I need to calculate a double integral and create the following loop

int1 = @(y) integral(@(x)D(x,y),x1,x2,'ArrayValued',true);

int2 = integral(@(y)int1,y1,y2,'ArrayValued',true);

c1 = int2./g % g is a scalar number

int11 = @(y) integral(@(x)D(x,y),x2,x3,'ArrayValued',true);

int22 = integral(@(y)int11,y1,y2,'ArrayValued',true);

c11 = int22./g % g is a scalar number

int111 = @(y) integral(@(x)D(x,y),x3,x4,'ArrayValued',true);

int222 = integral(@(y)int111,y1,y2,'ArrayValued',true);

c111 = int222./g % g is a scalar number

.

.

.

i'm trying to do like this

x = 1:0.1:5

for i = 1:length(x)

D = @(x,y) x.*(Z(x,y).^2);

int1(i) = @(y) integral(@(x)D(x,y),x(i),x(i)+0.5,'ArrayValued',true);

int2(i) = integral(@(y)int1(i),y1,y2,'ArrayValued',true);

c = int2(i)./g

end

I need to plot a graph of c versus x

[Walter Roberson]

Could you confirm that Z is a function?

It is confusing that you are using the same variable name x for your original vector that your for i is for, and also being the parameter name for integrate() which will be a vector reflecting local x conditions within the bounds. With similar concerns for y

[Walter Roberson]

It gets confusing when the same question is posted in multiple places! https://www.mathworks.com/matlabcentral/answers/509422-double-integration-with-loop

[José Anchieta de Jesus Filho]

I thought that this way it would be easier to understand. I'm sorry for this

dF = @(z1,r) 2.*pi.*v.*ro.*(Brt(z1,r).^2).*r; % v, ro they are scalar numbers
p1 = @(r) integral(@(z1) dF(z1,r),z0,z0+e,'ArrayValued',true);% e is also a scalar
F = integral(@(r) p1,ri,re,'ArrayValued',true);
c = F./v 

that's my role in response to z0. what I need is for it to be calculated with the z0 varying

z0 = 0:0.001:0.03

for each variation of z0, I will have a new p1, a new F and, consequently, a new c. I want to plot a graph of z0 versus c

Answer 1

[Andrei Bobrov]

v = -5:.1:5;
f = @(y)sqrt(y)./(1+exp(y./v));
q = integral(f,0,inf,'ArrayValued',true);

Answer 2

[Matt J]

vdata=-5:0.1:5; 
n=length(vdata);
q=zeros(1,n);

for i=1:n 

   v=vdata(i); 
   f=@(y) sqrt(y)/(1+exp(y/v));

   q(i)=integral(f,0,inf);

end