Vectorization and plotting user-defined function

I am new to MATLAB and am having trouble understanding how to plot userdefined functions.

I have defined a function f via

fun=@(t,r)(t.^2-t.^4).*exp(r.*t.^2);
fun1=@(t,r)exp(r.*t.^2);
f=@(r)integral(@(t)fun(t,r),0,1)./integral(@(t)fun1(t,r),0,1);

Now if I specify say

r=linspace(-10,10);

and type

x=f(r);

I get errors such as

Matrix dimensions must agree.
Error in @(t,r)(t.^2-t.^4).*exp(r.*t.^2)

whereas

x=sin(r);

is accepted. I seem to have been careful to use .* etc. What is the difference between my f and sin, and how do I plot a parametrized graph (f(r),g(r))?

Answer 1

[Adam Danz]

To evaluate the integrals at each value in r,

x = arrayfun(f,r);

[John Ball]

Many thanks Adam for the superfast response. But if I do

>> x=arrayfun(T,r);

>> y=arrayfun(g,r);

>> fplot(T,g, [-100 100])

I get a plot but still get vectorization errors

Warning: Function behaves unexpectedly on array inputs. To improve performance,
properly vectorize your function to return an output with the same size and shape as
the input arguments. 

So it seems I am not understanding something.

[Adam Danz]

I don't get that error when I apply this solution to the function in you question (I doubt you get that error with those functions, either). So, something is different between T and g compared to f.

What line is producing the error? Could you provide a new minimal working example that reproduces the error?

Answer 2

[Steven Lord]

By default integral calls the integrand function with a vector of values at which the integrand should be evaluated. There's no guarantee that this vector will have a size compatible with the size of your r vector.

If you tell integral that your function is array-valued by passing the name-value pair 'ArrayValued', true then integral will call your integrand with a scalar and expect an array as output. Try running the "Vector-Valued Function" example on the documentation page for the integral function with and without the name-value pair and compare the results.

[John Ball]

Thanks very much for this. Combining your and Adam's answer works fine, and I think I even understand why. I accepted his answer, but I would like to have accepted both, but don't see how. In any case, this being my first use of this forum I found the quick responses so helpful.

[Adam Danz]

@John, voting (thumbs up icon) is another way to show support/appreciation. Glad to hear that you found the forum useful!