The nuclear Overhauser effect I am having some trouble with understanding the NOE. The carbon NMR spectra can't be integrated, because the protons attached to the carbon atoms will make primary carbon atom relaxation faster than tertiary, so their peaks will have bigger intensity. With a bigger acquisition time, the peaks will be more proportional. In proton NMR, nearby protons will cause the target proton relaxation to be faster. When NOE is applied, those nearby protons will be irradiated and the target proton peak will have a slightly bigger intensity. Why will it have a bigger intensity?
Consider the simplest possible NOE experiment, with two dipole-coupled protons, the 
target I and nearby proton S. 

Before the experiment there are four populated spin states: $\alpha_{I}\alpha_{S}$, $\beta_{I}\alpha_{S}$, $\alpha_{I}\beta_{S}$ and $\beta_{I}\beta_{S}$, where $\alpha$ indicates $m_{i} = \frac{1}{2}$ and $\beta$ indicates $m_{i} = -\frac{1}{2}$. The relative populations of the states at thermal equilibrium (from Boltzmann distribution) are:


$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}$,  $\alpha_{I}\beta_{S}: 1$
$\beta_{I}\beta_{S}: 1 - 2\Delta $


S is irradiated so that the population difference between the $m_{S} = \pm \frac{1}{2}$ states goes to zero. This results immediately in the following populations: 


$\alpha_{I}\alpha_{S}, \beta_{I}\alpha_{S}: 1 + \Delta $
$\beta_{I}\beta_{S}, \alpha_{I}\beta_{S}: 1 - \Delta $


What happens next typically for small molecules is fast "flip-flip" cross-relaxation, restoring the $\alpha_{I}\alpha_{S}, \beta_{I}\beta_{S} $ states to their equilibrium populations. So we have the populations:


$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}: 1 + \Delta$
$\alpha_{I}\beta_{S}: 1 - \Delta $
$\beta_{I}\beta_{S}: 1 - 2\Delta $


The population difference between the pairs $\alpha_{I}\beta_{S}, \alpha_{I}\alpha_{S}$ and  $\beta_{I}\beta_{S}, \beta_{I}\alpha_{S}$ has now increased by 50%, from $2\Delta$ to $3\Delta$. This corresponds to a 50% increase in intensity of the peak. Transition between these pairs obviously corresponds to spin-lattice relaxation.