# How to do a double integral on a function with 3 variables, so the answer is dependent on the 3 variable

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Trond Oesten on 13 Mar 2015
Commented: Mike Hosea on 16 Mar 2015
Hi,
I have a function with 3 variables. I want to do a double integral on my function so it's dependent on the last variable. The variables I want to integrate is u and k so my function is dependent on x. I tryed using integral2, but that didn't work. Is it an easy way I can do this?
Thanks!
My script:
clc; clear all; close all;
C = [23.875 15.75281; 15.75281 93.9842];
mu_U = 1788.2058;
mu_K = 70.8489;
sigma_U = sqrt(C(1,1));
sigma_k = sqrt(C(2,2));
ubot = mu_U-4*sigma_U;
utop = mu_U+4*sigma_U;
kbot = mu_K-4*sigma_k;
ktop = mu_K+4*sigma_k;
mu = [mu_U;mu_K];
fun = @(x,u,k) wblpdf(x,u,k)*mvnpdf([u;k],mu,C);

Mike Hosea on 16 Mar 2015
Edited: Mike Hosea on 16 Mar 2015
Not sure what your problem with INTEGRAL2 was, but you're using matrix multiplication in your definition of FUN, which probably isn't right--I doubt you want to multiply a matrix of WBLPDF values by a matrix of MVNPDF values, and you're stacking u and k as different rows when you input to MVNPDF when I think what you should be doing is passing them as different columns. Assuming you meant to integrate over u and k first and to have a function of x coming out, this might be the sort of thing you want:
C = [23.875 15.75281; 15.75281 93.9842];
mu_U = 1788.2058;
mu_K = 70.8489;
sigma_U = sqrt(C(1,1));
sigma_k = sqrt(C(2,2));
ubot = mu_U-4*sigma_U;
utop = mu_U+4*sigma_U;
kbot = mu_K-4*sigma_k;
ktop = mu_K+4*sigma_k;
mu = [mu_U;mu_K];
fun = @(x,u,k) wblpdf(x,u,k).*reshape(mvnpdf([u(:),k(:)],mu(:).',C),size(u));
qfun_scalar = @(x)integral2(@(u,k)fun(x,u,k),ubot,utop,kbot,ktop,'AbsTol',1e-10,'RelTol',1e-6);
qfun = @(x)arrayfun(qfun_scalar,x);
I just threw the tolerances in there in case you wanted to play with them.
Now you can use qfun to evaluate for x values (scalar or arrays), or you can integrate over ranges of x, or whatever.
>> qfun(1750)
ans =
0.006956904221022
>> integral(qfun,1600,1800)
ans =
0.789977339847138
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Mike Hosea on 16 Mar 2015
Iqfun_scalar = @(a,b)integral(qfun,a,b);
As the name implies, that function is only set up for scalar a and scalar b. Now to vectorize it, I'll add a little trick using ONES to make it do scalar expansion.
Iqfun = @(a,b)arrayfun(Iqfun_scalar,a.*ones(size(b)),b.*ones(size(a)));
>> Iqfun([1600,1700],1800)
ans =
0.789977339847138 0.758915659832745
Of course you're doing a triple integral there, a couple of them, so it isn't blazingly fast.