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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: 1,477
Registered: 11/8/10
Re: Integration with variable limits
Posted: Jul 31, 2013 3:09 AM
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"Torsten" wrote in message <ktacg3$fkj$1@newscl01ah.mathworks.com>...
> "Sanaa" wrote in message <kt9bf4$56h$1@newscl01ah.mathworks.com>...
> > "Torsten" wrote in message <kst5mn$h75$1@newscl01ah.mathworks.com>...
> > > "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.

> >
> > Sorry for not posing my question correctly. The function I want to integrate is
> > y(s)*(1-y(s)) ds from (n*0.25) to t, and t/in(n*0.25, (n+1)*0.25).
> > Do you have any idea on how to format my code above to solve my problem?
> > Thanks a lot in advance

>
> int_{n*0.25}^{t} (y(s)*(1-y(s)) ds = (1/2*t^2 - 1/3*t^3) - (1/2*(n*0.25)^2 - 1/3*(n*0.25)^3)
> for t/in(n*0.25, (n+1)*0.25).
> Is it that what you asked for ?
>
> Best wishes
> Torsten.


No sorry, this is wrong - I integrated s*(1-s).
For integration, you will have to know the explicit form of y as a function of s,
e.g. y(s)=sin(s) or something like that.

Best wishes
Torsten.



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