Integrating a multivariate function w.r.t. a single variable
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Hello,
I am defining a function using
f = @(x,y) (expression in x and y)
This definition I believe it is correct as I can call f(0,0) for example and I get the numerical value.
What I need to do next is integrate f(x,y) with respect to y between a and b and call this g(x). I need then to be able to pass g(x) to fsolve in order to compute the roots. How do I do this? I tried dblquad, but it integrates w.r.t. both variables at once. quad gives me and error as I am not sure what's the correct syntax for this.
Thanks, Alex
采纳的回答
Teja Muppirala
2013-2-14
Define another function handle to be the integral over y. Like this:
% Just making some random 2d function
f = @(x,y) (x.^2-y.^2).*cos(x./(1+y.^2));
% Some limits of integration
a = 0;
b = 3;
% Define g as the integral of f(x,y) dy from a to b
g = @(x) integral(@(y) f(x,y) , a,b);
% Plot it, and find a zero
ezplot(g)
fzero(g,0)
If your version of MATLAB doesn't have the INTEGRAL function, you could use QUAD instead.
3 个评论
joseph erlich
2019-2-5
Thank you.
ARNAB PAL
2019-9-19
Dear sir,
I have a question that if the function is a matrix function like,
f=@(x,y) K_2*exp(A*(x(i)-y))*B*u;
where K_2 is a (3*6) matrix,A is a (6*6) matrix ,B is (6*3) and u is (3*1) matrix. I want to find g=@(x) integral(@(y) f(x,y),0,x(i)) but when I use this it is not giving the output as a (3*1) vector?
Finally I want to find w=u-(K_2*exp(A*x(i))*X) where X is (6*1).
Walter Roberson
2019-9-19
When you call integral(), it is required to return an array the same size as x, which will be a vector of varying sizes. However, there is the 'vectorvalued' option for integral(), and when set then the function will be passed scalars and can return multidimensional outputs as needed.
更多回答（3 个）
Walter Roberson
2013-2-13
1 个投票
You cannot do this with numeric integration.
If you have the symbolic toolbox, then expression the function symbolically and do symbolic integration with int(). Then if you need, you can use matlabFunction to turn the symbolic result into a function handle of a numeric function.
2 个评论
Adrian Martinez
2021-6-5
Can you please show an example of how to do this?
f = @(x,y) (x.^2-y.^2).*cos(x./(1+y.^2));
% Some limits of integration
a = 0;
b = 3;
syms x y
% Define g as the integral of f(x,y) dy from a to b
g(x) = int(f(x,y), y, a, b)
x0 = -100;
vpasolve(g, x0)
G = matlabFunction(g)
fsolve(G, x0)
In some cases, the int() step would be able to calculate a closed form expression, so G will not always end up with an integral() in it.
Youssef Khmou
2013-2-13
Hi, try this :
syms x y
h=exp(-x^2-y^2)
F1=int(h,x)
F2=int(h,y)
Based on F1 and F2 you make function handle :
Fx=@(x) 1/2/exp(x^2)*pi^(1/2) % truncated ERF(Y)
Y=fsolve(Fx,0.1)
2 个评论
Walter Roberson
2013-2-13
How do you get from the "F1 and F2" to the Fx ? And where do the limits of integration over y come in?
Youssef Khmou
2013-2-13
Hi,i used the undefined integration or "primitive" , for F1 you get :
F1 =
1/2/exp(y^2)*pi^(1/2)*erf(x)
you have an analytic integration w.r.t. x, so its function of y , now manually you set a function handle Fy :
Fy=@(x) 1/2/exp(y^2)*pi^(1/2)
and then solve it with "fslove" as Alexandru said he wants to use "fslove" .
Yang Zhang
2015-3-15
0 个投票
Use Symbolic Math Toolbox to your problem.
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