Why are neutron absorption cross sections high at low incident energy? For example, U-235 fission cross section looks like this:

(source: science20.com) 
As I understand it, the resonances peaks correspond to discrete quantum states of the excited compound nucleus. As you go higher, the density of states is too high to be resolved and you get that continuum. 
But at thermal energies (left part of the graphic), I don't really understand what's going on, since the available states should be low. Consequently. I expect the cross section to be low too. 
Is it a tail of a resonance peak corresponding to low energy states? Is the 1/v behavior dominating the decline of that resonance peak?
I'm expanding the question a little, since I'm not satisfied with the answers. Here is what I believe should be happening (the example is done with the absorption of a neutron by Indium-115):

Left is before absorption, right is after. The orange level is not a level in the compound nucleus, so absorption would be diminished.
This also happens with Uranium-238, so the question is not about fission only.

 A: This is because U-235 is fissile, that is you only have to deliver the neutron to the nucleus for the magic to happen. Unlike U-238 where just delivering it doesn't do the job, there you also have to impart the nucleus with the neutrons kinetic energy.
Once we know this, it becomes clear that for low energy neutrons, their de Broglie wavelength is very big. So the cross-section is effectively determined by the quantum size of the neutron, rather than any other dynamics, so roughly $\sigma \approx \pi \lambda_{dB}^2\approx \frac{1}{E}$
A: The energy Eigenstates of the final nucleus (after the neutron has been captured) form a complete set. That means that any wave function can be written as a superposition of these states; in particular we can express the incoming neutrons wave function in terms of these states,
$$ \psi_{in}(x,t) = \sum_{n=1}^\infty a_n(t) \psi_n(x) e^{i E_n t}.$$
Lets say the neutron is captured at time $T$. The coefficients $a_n(T)$ correspond to the probability amplitudes that, when measured, the neutron will be in the bound state with energy $E_n$.
The incoming kinetic energy of the neutron is going to be an average of the $E_n$,
$$ E_{Kinetic} =  \sum_n \vert a_n(T) \vert^2 E_n,$$
if this incoming energy isn't exactly equal to one of the energy levels there is no problem; this just means that there is  a finite probability of landing in higher and lower energy levels. This phenomenon is sometimes described as the uncertainty in the energy.

There is a slight problem here because the odds are that the neutron is prepared in some state of definite momentum like $e^{ipz}$ which doesn't actually live in our hilbert space. A way around this is to approximate the incoming wave function as,
$$ \psi_{in} = A e^{ipz} e^{-r^2/L^2}, $$
and to take the limit as $L\rightarrow \infty$ at the end (this is motivated by the theory of generalized functions ) . 
