Nested Numerical Integral in Matlab
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Hello, I am fairly new in using Matlab and was wondering if a nested numerical integral was possible. I have seen a number of other questions here where the outer variable of integration appears in the limits of the inner integral but the function being integrated over just depends on one variable. So I was wondering how or if it's possible to do, say:
z = integral( e^(-integral(f(x,y),x,0,1)),y,0,1)
3 Comments
Babak
on 12 Dec 2012
If not, what is your f(x,y) function? Can you find g(.),h(.) such that f(x,y)=g(x)*h(y)?
Accepted Answer
Teja Muppirala
on 12 Dec 2012
Rather than trying to do it all in one expression, it's much simpler if you break it up into two parts.
Step 1. Make the inner part a separate function and save it to a file.
function F = innerF(y)
F11 = integral(@(x) exp(x+y) ,0,1);
F21 = integral(@(x) exp(x-y) ,0,1);
F12 = integral(@(x) sin(x+y) ,0,1);
F22 = integral(@(x) cos(x-y) ,0,1);
F = det([F11 F12; F21 F22]);
Step 2. From the command line, call INTEGRAL to do the outer integral
integral(@innerF, 0, 1, 'ArrayValued', true)
2 Comments
Sundar Aditya
on 28 Jan 2017
Hi,
I was wondering how the syntax would change if the limits of x were a function of y. Specifically, the expression I need to evaluate is of the form integral( e^( -integral( f(x,y),x,0,g(y) ) ),y,0,1). This is what I tried:
integral(@(x) exp(-integral(@(x,y) f,0,@(y) y)),0,1)
where f is the function handle for f(x,y). I get the following error message:
Function 'subsindex' is not defined for values of class 'function_handle'.
I'm unable to figure out what I'm doing wrong. Any help will be greatly appreciated.
More Answers (2)
Richard
on 18 Jul 2017
Seems to me that this is what you are looking for. I assumed a random expression for f function since you did not specify it.
f =@(x,y) x+y;
integral(@(y) exp(-integral(@(x) f(x,y),0,y)),0,1,'ArrayValued',true)
1 Comment
Roger Stafford
on 12 Dec 2012
The functions 'dbsquad' and 'quad2d' are designed to numerically solve just your kind of problem. The former uses the the kind of fixed integration limits that you have described and the latter allows variable limits. Be sure to read their descriptions carefully so you can define the integrand function properly.
Roger Stafford
0 Comments
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