Hi all,
Why
fr2-function(x)
{ if(x1){x*x}else{1}
}
integrate(fr2,-1,2)
gives the following wrong answer:
3 with absolute error 3.3e-14
Warning message:
the condition has length 1 and only the first element will be used in: if (x
1) {
while
fr-function(x){as.numeric(x1)*(x^2-1)+1}
The function need to be vectorized. Try:
fr2.new - function(x) ifelse(x 1, x * x, 1)
integrate(fr2.new, -1, 2)
1.67 with absolute error 3.4e-05
Andy
From: Lynette Sun
Hi all,
Why
fr2-function(x)
{ if(x1){x*x}else{1}
}
integrate(fr2,-1,2)
gives the following wrong
Hi,
the function 'fr' evaluates each element of the vector
'x', independently of its length. 'fr2' does not.
Try with
fr2-function(x){
ifelse(x1, x*x, 1)
}
integrate(fr2,-1,2)
1.67 with absolute error 3.4e-05
Marco
--- Lynette Sun [EMAIL PROTECTED] wrote:
Hi all,
Why
