[R] Integration in R
David Winsemius
dwinsemius at comcast.net
Tue Jan 8 19:41:59 CET 2013
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 <dwinsemius at 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
More information about the R-help
mailing list