How do I numerically integrate a multivariable function wrt only one variable?

조회 수: 5 (최근 30일)
Hi guys,
I have a problem, since I want to numerically intergrate a multivariable function wrt only one variable. However, I only succeed by doing this with the following commands:
if true
clear
syms r s t
p0 = .75;
p1 = 3;
y0 = 720;
b = [5 -14 0.08];
end
if true
function demand = qn(x,b)
z1=b(1)+b(2)*x(1)+b(3)*x(2);
demand = z1;
end
if true
w(r,s) = int(qn([t, y0],b),t,r,s);
cs = w(p0,p1);
end
I know this is not a numerical integration. This returns me a symbolic variable (which I do not want), but this was the only way that I was able to retrieve some value. To clarify, I want to numerically integrate the function qn with respect to x(1) from p0 to p1. Could somebody help me with this please? Many thanks in advance! Cheers!

채택된 답변

Matt J
Matt J 2018년 7월 9일
편집: Matt J 2018년 7월 9일
p0 = .75;
p1 = 3;
b = [5 -14 0.08];
x2=...
result = integral( @(x1) qn([x1,x2],b) , p0, p1)
  댓글 수: 4
Matt J
Matt J 2018년 7월 9일
In this case, it would be much easier if you wrote qn() with scalar arguments,
function demand = qn(t,y,b)
z1=b(1)+b(2)*t+b(3)*y;
demand = z1;
end
then this would work
result = integral( @(t) qn(t,y0,b) , p0, p1)
Martijn Mouw
Martijn Mouw 2018년 7월 9일
Thanks this works! I know that indeed with scalar arguments it would be easier, but it has to stay the way it is for other commands work (I did not include them in the post)

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

제품


릴리스

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by