Thanks. But then how to implement condition like 0<x1<x2<7? I would be
happy to know that.
On 8 January 2013 18:41, David Winsemius <dwinsem...@comcast.net> wrote:
> Please reply on list.
>
>
> On Jan 8, 2013, at 10:27 AM, Naser Jamil wrote:
>
> Hi David,
>> x[2] is the second variable, x2. It comes from the condition 0<x1<x2<7.
>>
>
> No, it doesn't come from those conditions. It is being grabbed from some
> "x"-named object that exists in your workspace.
>
> If your limits were 7 in both dimensions, then the code should be:
>
> adaptIntegrate(f, lowerLimit = c(0, 0), upperLimit = c(7,7))
> #----
> $integral
> [1] 228.6667
>
> (At this point I was trusting R's calculus abilities more than yours. I
> wasn't too trusting of mine either, and so tried seeing if Wolfram Alpha
> would accept this expression:
> integrate 2/3 (x+y) over 0< x<7, 0<y<7
>
> ; which it did and calculating the decimal expansion of the exact fraction:
>
> > 686/3
> [1] 228.6667
> >
>
> --
> David.
>
>
>
>
>> Thanks.
>>
>> On 8 January 2013 18:11, David Winsemius <dwinsem...@comcast.net> wrote:
>>
>> On Jan 8, 2013, at 9:43 AM, Naser Jamil wrote:
>>
>> Hi R-users.
>>
>> I'm having difficulty with an integration in R via
>> the package "cubature". I'm putting it with a simple example here. I wish
>> to integrate a function like:
>> f(x1,x2)=2/3*(x1+x2) in the interval 0<x1<x2<7. To be sure I tried it
>> by hand and got 114.33, but the following R code is giving me 102.6667.
>>
>> ------------------------------**------------------------------**-------
>> library(cubature)
>> f<-function(x) { 2/3 * (x[1] + x[2] ) }
>> adaptIntegrate(f, lowerLimit = c(0, 0), upperLimit = c(x[2],7))
>>
>>
>> What is x[2]? On my machine it was 0.0761, so I obviously got a
>> different answer.
>>
>> --
>> David Winsemius, MD
>> Alameda, CA, USA
>>
>>
>>
> David Winsemius, MD
> Alameda, CA, USA
>
>
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