Effacer les filtres
Effacer les filtres

Integral with specific range (i.e t = 0:0.1:1)

8 vues (au cours des 30 derniers jours)
Fakhri Alifin
Fakhri Alifin le 23 Sep 2020
Commenté : Star Strider le 2 Oct 2020
Hello everybody,
I'm a new user of matlab and still a beginner of this software. I'm using matlab R2015a by the way.
I've made a function called intensityc (t) with this code:
%NHPP on critical component (N(t))
function lc = intensityc(t)
B = 2;
ac = 0.4;
lc = (B/ac)*((t/ac).^(B-1));
end
then I was trying to call the function to another function Lamdac (t) with this code:
%big Lamda
function Lc = Lamdac(t)
Lc = integral(@intensityc, 0 ,t);
end
I have a trouble to find the value of Lamdac(t) because the "t" is ranging from 0 up to 1. I have used 'Arrayvalued' but the code still error.
Anyone can help me? I will appreciate all suggestions given to me. Thank you.

Réponse acceptée

Star Strider
Star Strider le 23 Sep 2020
Do the integration in a loop for each value of ‘t’:
function lc = intensityc(t)
B = 2;
ac = 0.4;
lc = (B/ac)*((t/ac).^(B-1));
end
function Lc = Lamdac(t)
for k = 1:numel(t)
Lc(k) = integral(@intensityc, 0 ,t(k));
end
end
Calling it:
t = linspace(0, 1, 10);
Out = Lamdac(t).'
produces:
Out =
0
0.077160493827160
0.308641975308642
0.694444444444444
1.234567901234568
1.929012345679012
2.777777777777777
3.780864197530863
4.938271604938270
6.249999999999999
(Another option is to use arrayfun, however it is much slower and less efficient than an explicit loop.)
  4 commentaires
Fakhri Alifin
Fakhri Alifin le 2 Oct 2020
Thanks for your help, it already work!
Star Strider
Star Strider le 2 Oct 2020
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.

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