Calculate integral with an external formula
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Orongo
am 11 Okt. 2017
Kommentiert: Orongo
am 15 Okt. 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??
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David Goodmanson
am 11 Okt. 2017
Bearbeitet: David Goodmanson
am 11 Okt. 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 Kommentare
David Goodmanson
am 12 Okt. 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.
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