4D integral using integral and integral3
显示 更早的评论
I tried 4D integral using integral and integral3 like the following link: https://jp.mathworks.com/help/matlab/ref/integral3.html
however, it does not work when a function used become complecated.
Here I show a simplified code I used.
f = @(a, b, c, d) a+b+c+d;
I = @(a, b, c) integral(@(d) f(a, b, c, d), 0, 1);
f2=@(a, b, c) I(a,b,c)+1;
I2=integral3(f2, 0, 1, 0, 1, 0, 1);
It showed an error like "The array size does not fit this operation." (translated from Japanese).
But I did not figure out which array it said.
My question is what is the acutual issue and how to solve it to achieve the 4D integral.
采纳的回答
f = @(a, b, c, d) a+b+c+d;
I = @(a, b, c) integral(@(d) f(a, b, c, d), 0, 1,'ArrayValued',1);
f2=@(a, b, c) I(a,b,c)+1;
I2=integral3(f2, 0, 1, 0, 1, 0, 1)
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Numerical Integration and Differential Equations 的更多信息
Translated by 
