You were close. I'm going to define values for the variables for demonstration purposes
hlpexpvar = 1;
univar = 2;
Now to define the three functions
funone =@(x) exp(-hlpexpvar.*x);
funtwo =@(x) 1-(x./univar(1));
You can't multiply function handles. However you can multiply the values returned by evaluating the function handles.
fun = @(x) funone(x).*funtwo(x);
Let's check by comparing the integral of fun (using funone and funtwo) and the explicitly specified function fun2.
fun2 = @(x) exp(-hlpexpvar.*x).*(1-(x./univar(1)));
answer1 = integral(fun, 0, 1)
answer2 = integral(fun2, 0, 1)
