Resolvent kernel of Fredholm integral equation.

For the linear integral equation
$y(x)=x+\int_{0}^{1/2} y(t) dt$.
Find Resolvent kernel $R(x,t,1)$.

I tried to find resolvent kernel of Volterra integral equation by taking kernel as 1.Then I got $R(x,t,1)=e^{(x-t)}$.But I don't know how to find resolvent kernel of nonhomogeneous Fredholm integral equation of second kind.please guide me.Thanks a lot.

$$k(x,t)=k_1(x,t)=1$$, $$k_2(x,t)=\int_{0}^{1/2} k(x,z)k_1(z,t) dz=\frac {1}{2}$$
$$k_3(x,t)=\int_{0}^{1/2} k(x,z)k_2(z,t) dz=\frac {1}{2^2}$$...... $$k_n(x,t)=\int_{0}^{1/2} k(x,z)k_{n-1}(z,t) dz=\frac {1}{2^{n-1}}$$
Hence $$R(x,t,\lambda)=\sum_{i=0}^\infty \lambda^i k_{i+1}(x,t)$$
$$R(x,t,1)=\frac{1}{2}+\frac {1}{2^2}+\frac {1}{2^3}+....=\frac{1}{1-(1/2)}=2$$