Borrar filtros
Borrar filtros

Calculate integral with an external formula

2 visualizaciones (últimos 30 días)
Orongo
Orongo el 11 de Oct. de 2017
Comentada: Orongo el 15 de Oct. de 2017
I calculating an integral over a function that is defined in seperate m-file. My program looks like:
% main.m
mu_x=@(t) lx(t);
l_x = exp(-integral(@(t) mu_x(t),0,98));
% lx.m
function res=lx(x)
a=0.006782872;
b=5.44781E-08;
c=0.137849468;
if x>97
res = a+b*exp(c*97)+(x-97)*0.001;
else
res=a+b*exp(c*x);
end
I expect lx(x)=lx(98) => res=a+b*exp(c*97)+(x-97)*0.001; to be calculated but instead res=a+b*exp(c*x) is done. I want to keep this if-function in lx.m, is there anyway I can do this? Also, what is going wrong??

Respuesta aceptada

David Goodmanson
David Goodmanson el 11 de Oct. de 2017
Editada: David Goodmanson el 11 de Oct. de 2017
Hi Lenovo,
Your 'if' check is not working correctly, because the res function takes vector input for x. One way to do the task is to create an index that shows whether or not x > 97:
x = 0:.01:120;
plot(x,lx(x)) % demo
function res=lx(x)
a=0.006782872;
b=5.44781E-08;
c=0.137849468;
res = zeros(size(x));
ind = x>97;
res(ind) = a+b*exp(c*97)+(x(ind)-97)*0.001;
res(~ind) =a+b*exp(c*x(~ind));
end
  3 comentarios
David Goodmanson
David Goodmanson el 12 de Oct. de 2017
I just plugged your expression
0.0067828+5.447812E-08*EXP(0.137849*97)+(98-97)*0.001
verbatim into Matlab, changed EXP to exp and got 0.0427, just like the demo plot in the posted answer gives and Excel gives. So I do not know where the .385 might be coming from.
Orongo
Orongo el 15 de Oct. de 2017
Ok. Thanks. I still getting the problem and will close this discussion and open a new one because I think the problem is rippling.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Programming en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by