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!
.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Programming dans Help Center et File Exchange

Produits

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by