On Jan 8, 2013, at 10:51 AM, Naser Jamil wrote:
Thanks. But then how to implement condition like 0<x1<x2<7? I would
be happy to know that.
Multiply the function by the conditional expression:
> f<-function(x) { 2/3 * (x[1] + x[2] )*(x[1] < x[2]) }
> adaptIntegrate(f, lowerLimit = c(0, 0), upperLimit = c(7,7))
$integral
[1] 114.3333
--
David.
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
David Winsemius, MD
Alameda, CA, USA
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