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Topic: Integration with variable limits
Replies: 17   Last Post: Aug 20, 2013 10:20 AM

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Torsten

Posts: 1,450
Registered: 11/8/10
Re: Integration with variable limits
Posted: Jul 26, 2013 2:41 AM
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"Sanaa" wrote in message <kss3h5$50n$1@newscl01ah.mathworks.com>...
> "Torsten" wrote in message <ksrcel$2k$1@newscl01ah.mathworks.com>...
> > "Sanaa" wrote in message <ksrbia$r0l$1@newscl01ah.mathworks.com>...
> > > Hi all,
> > > How to integrate a function within a loop and the limits of integration are variables?
> > > For instance, y(i)*(1-y(i)) is the function I wish to integrate from 0.25*i to t where
> > > t \in (0.25*i, 0.25*(i+1)).
> > > Any help is really appreciated.
> > > Sanaa Moussa

> >
> > I think y(i)*(1-y(i)) is just a real number, not a function, isn't it ?
> >
> > Best wishes
> > Torsten.

>
> Thanks for replying. You are right, but I mean the function is y*(1-y) which I want to put it into a loop
> My code is
> itermax=300;min=itermax-9;% That is we plot from 291:300 i.e. 10 values of x only.
> r=0.25;
> for rho=0:0.001:4
> %x0=0.1;
> x(1)=0.1;
> for i=1:itermax-1
> %t=linspace(i*r,(i+1)*r,itermax)
> y(i+1)=x(i);
> x(i+1)=x(i)+ rho*int(y(i)*(1-y(i)),i*r,(i+1)*r);
> end
> %fix(y)
> plot(rho*ones(10),x(min:itermax),'b.','linewidth',0.1)
> hold on
> end
> fsize=15;
> xlabel('\rho','FontSize',fsize)
> ylabel('\itx','FontSize',fsize)
> %title('r=0.25, \alpha=1','FontSize',fsize)
> hold off
> % print(gcf, '-djpeg', '-zbuffer', 'bif.png');
>
> I get the error
> Function 'int' is not defined for values of class 'double'.
> what does this mean and how to fix it please?
> Many thanks in advance.


It means what I said before:
y(i)*(1-y(i))
is a scalar value and not a function.
So "int" does not know how to handle this because it expects a function, not a scalar as its first input argument.

Best wishes
Torsten.



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