May I ask everyone to help me solve this problem

mu = [0, 0];                      % 均值向量（行向量）
cov_2d = [1 0.5; 0.5 1];          % 协方差矩阵

% 定义二维正态分布函数
fun = @(x, y) mvnpdf([x, y], mu, cov_2d);  % 直接传递 mu 作为行向量

% 使用 integral2 计算二维积分
p = integral2(fun, -2, 3, 3, 4);  % x从-2到3，y从3到4
错误使用 mvnpdf (第 67 行)
X 和 MU 的列数必须相同。

出错 jifen>@(x,y)mvnpdf([x,y],mu,cov_2d) (第 6 行)
fun = @(x, y) mvnpdf([x, y], mu, cov_2d);  % 直接传递 mu 作为行向量

出错 integral2Calc>tensor (第 240 行)
Z = FUN(X,Y);  NFE = NFE + 1;

出错 integral2Calc>integral2t (第 55 行)
[Qsub,esub,FIRSTFUNEVAL,NFE] = tensor(thetaL,thetaR,phiB,phiT,[],[], ...
    
出错 integral2Calc (第 9 行)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);

出错 integral2 (第 105 行)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);

出错 jifen (第 9 行)
p = integral2(fun, -2, 3, 3, 4);  % x从-2到3，y从3到4

Answer 1

[Torsten]

I'm not sure, but if you want to integrate the normal distribution, use "mvncdf". integral2 can be very inaccurate when integrating probability density functions (not in the case given):

format long
mu = [0, 0];                     
cov_2d = [1 0.5; 0.5 1];
p = mvncdf([-2,3],[3 4],mu,cov_2d)

p = 
   0.001241214814729

fun = @(z) mvnpdf(z, mu, cov_2d);  
p = integral2(@(x,y)arrayfun(@(X,Y)fun([X,Y]),x,y),-2,3,3,4)

p = 
   0.001241214813916

%or - most probably faster -
p = integral2(@(x,y)reshape(fun([x(:),y(:)]),size(x)),-2,3,3,4)

p = 
   0.001241214813916