Let K : L 2 ( Y , μ ) → L 2 ( X , ν ) K:{L_2}(Y,\mu ) \to {L_2}(X,\nu ) be continuous and linear and assume ( K f ) ( x ) = ∫ Y k ( x , y ) f ( y ) d μ ( y ) (Kf)(x) = \smallint _Y^{} {k(x,y)f(y)\,d\mu (y)} . Define k x {k_x} by k x ( y ) = k ( x , y ) {k_x}(y) = k(x,y) . Assume K has the property that (a) k x ∈ L 2 ( Y , μ ) {k_x} \in {L_2}(Y,\mu ) for all x ∈ X x \in X and (b) if K f = 0 ν Kf = 0\;\nu -a.e., then ( K f ) ( x ) = 0 (Kf)(x) = 0 for all x ∈ X x \in X . For example, if X = Y = [ 0 , 1 ] X = Y = [0,1] , μ = ν \mu = \nu is Lebesgue measure and if k ( x , y ) k(x,y) satisfies a Lipschitz condition in x, then K has the above property. Assume K satisfies this property and f 0 {f_0} is a minimum L 2 {L_2} norm solution of the first-kind integral equation ( K f ) ( x ) = g ( x ) (Kf)(x) = g(x) for all x ∈ X x \in X . It is shown that f 0 {f_0} is the L 2 {L_2} -norm limit of linear combinations of the k x i {k_{{x_i}}} ’s. It is then shown how to choose constants c 1 , … , c n {c_1}, \ldots ,{c_n} to minimize ‖ f 0 − ∑ j = 1 n c j k x j ‖ \left \| {{f_0} - \sum \nolimits _{j = 1}^n {{c_j}{k_{{x_j}}}} } \right \| without knowing what f 0 {f_0} is. This paper also contains results on how to choose the k x j {k_{{x_j}}} ’s as well as numerical examples illustrating the theory.