If your $\Phi_n(x,t)$ are alreay the energy eigenstates, then they are orthogonal and therefore, if "*let the normalised wave function in one dimension be*" means you have already normalized it, then $f(x)$ at $t=0$ is normalized. 

For later time you have
$$\Phi_n(x,t)=e^{-iE_nt}\Phi_n(x,0)=e^{-iHt}\Phi_n(x,0)=U(t)\Phi_n(x,0)$$
and so 
$$\Phi(x,t)=c_1\Phi_1(x,t)+c_2\Phi_1(x,t)=U(t)(c_1\Phi_1(x,0)+c_2\Phi_1(x,0))=U(t)f(x).$$
Since $U(t)$ is unitary, the state will stay normalized. You have not given us the Hamiltonian to see if these are really the eigenstates, but the point is that if they are, then the states are orthogonal and the mixing terms are zero. 

The integration (for normalization, checking orthogonality as well as finding the particle in a finite interval) is a purely mathematical problem.