Since you didn't mention anything about the Hamiltonian, I assume it's just a particle in a box. The wavefunction vanishes at the boundaries.
You are obviously familiar with the eigenfunctions of the Hamiltonian! Is $\psi(x)$ such an eigenfunction? If not, can it be expressed in terms of eigenfunctions? How?
For the last part, think how probabilities relate to the coefficients in the expansion wrt. a certain basis.
[Update:] According to your comment you only know the energies, eg. the eigenvalues of the Hamiltonian $E_n = \frac{\hbar^2}{2m}\frac{\pi^2}{L^2}n^2$ And you know - at least you say so yourself - how to calculate the expectation-value of the enery. Divide this number by $E_1$ then!
The wavefunction $\psi(x)$ does not carry the quantum number $n$, because it is not an eigenstate. It's rather a superposition of different eigenstates (different $n$).
[/Update]
If you need further hints, please ask, but I advice you to try it for yourself.
And when you're done, post your own answer!
...it makes no sense to calculate any property of the particle with the "initial" wave-function, since this is simply the incorrect wave-function for the new well?
The wavefunction can't be "incorrect for the well". Your wavefunction is just an initial condition for time-dependent Schrödinger equation. Here's how it would evolve if you solve the time-dependent equation (I ignore normalization here):
We see that since the actual particle's wave-function is not defined when $x>L$...
The wavefunction is defined for all $x\in\mathbb R$. It's just zero outside of $(0,L)$ because the potential is infinite there.
the second term will be zero
This remains true, however, because of what I said above.
the wave-functino is normalized for $0<x<L$
Actually, again, the wavefunction is normalized period. It's defined for the whole real line, and zero outside the well, so when you normalized using the integral over the well, it's the same as if you integrated over $\mathbb R$. If it were not this way, your "normalization for some domain" would not make any sense, i.e. it'd not be a normalization at all.
Your further calculation looks OK to me.
I think maybe my difficulty with "visualzing" the problem is that I do not fully understand the expression for $c^2_n$, and how this gives the probability for measuring the energy level $n$.
That this gives the probability for measuring energy level $n$ is known as Born rule. You find projection of your actual wave function onto the eigenstate of energy, namely on state $n$. By Born rule, its magnitude squared is the probability of measuring the system to appear in that eigenstate.
The fact that you used only the unchanged original wavefunction for calculations, even though it almost immediately drastically changes as time goes on, is that despite its change in shape, its coefficients $c_n$ actually only change their phase like
$$c_n\propto\exp\left(-\frac i\hbar E_nt\right),$$
but remain the same in magnitude — because the potential is independent of time. Thus you can measure the energy after some time passes, and will still get the same result.
Best Answer
Hints: It is easy to do the integral yourself without the aid of Mathematica.
For instance use the product-to-sum-formula for the two sines.
Then integrate by part to get rid of the $x$ power.
You will need the following primitive integrals (aka. antiderivatives or indefinite integrals): $$ \int \!dx ~\cos(bx)~=~ \left\{\begin{array}{ccc} \frac{\sin(bx)}{b} &\text{for}& b\neq 0, \\ x&\text{for}& b= 0,\end{array} \right. $$ and $$ \left\{\begin{array}{ccc}\int \!dx ~\frac{\sin(bx)}{b} &=& \frac{1-\cos(bx)}{b^2} &\text{for}& b\neq 0, \\ \int \!dx~x &=& \frac{x^2}{2}&\text{for}& b= 0.\end{array} \right. $$ Here the various integration constants have been chosen such that the $b=0$ case can be viewed as the limit $b\to 0$ of the $b\neq0$ case.
Then the above comment of Jerry Schirmer clearly applies: You can either recover the limit $b\to 0$ from the $b\neq0$ case, or treat the $b=0$ case separately.