In position-space (that is, when your functions are functions of x), the function $\int|\Psi|^2$ gives the probability of finding the particle in a given range. The expectation value of x is where you'd expect to find the particle. It is often essentially the weighted average of all the positions where the probability density, $|\Psi|^2$, is the weighting function (that's not exactly what it is, but it's a useful analogy). Similarly, you can find the expectation value for any measurable quantity. In this space, the difference between the two is that the expectation value is a number that represents the expected average position of the particle over many measurements whereas the probability is a number that gives you the probability for finding the particle within the limits of integration.
However, you can use any different basis. For example, you could choose momentum-space, $\left|\Psi\right>$ is $\Psi(p)$ (quantum physicists please don't kill me for that affront to notation). In momentum space, the integral $\int|\Psi|^2$ is now the probability of the particle having a given range of momenta. However, the expectation value of x is still the average measurement of x. What, you ask, is the point? The expectation value is a number that can be found in any basis that represents the "on-average" value of a measurement. The probability found by $\int|\Psi|^2$ is the probability that a particle will be found existing within a specified range of values for the basis you are using.
$\int_{x_1}^{x_2}|\Psi|^2dx$ is "there is #% chance that the particle will be found between $x_1$ and $x_2$"
$\left<\Psi\right|x\left|\Psi\right>$ is "the expected average position of the particle over a large number of sample measurements is at $x$=#"
$|\Psi|^2(x)$ is a function "the probability per unit length of finding the particle at this position is #%"
The probability density to find one particle at $r_1$ and the other particle at $r_2$ is the absolute square of the wave-function. The question is, so I understand, how does the wave-function look like.
If the particles a and b are distinguishable, the wave-function looks as given in the exercise.
However, if the particles are indistinguishable fermions, the wave function should be antisymmetrical at the interchange of the particles, i.e. you do antisymmetrization:
$ψ_{a,b}(r_1,r_2) = \frac{\{ψ_{a,b}(r_1,r_2) - ψ_{a,b}(r_2,r_1)\}}{\sqrt{2}}$
On the other hand, if the particles are indistinguishable bosons, you do symmetrization
$ψ_{a,b}(r_1,r_2) = \frac{\{ψ_{a,b}(r_1,r_2) + ψ_{a,b}(r_2,r_1)\}}{\sqrt{2}}$
Best Answer
All the definitions you've posted are correct, and they aren't in conflict with each other, although they are a bit imprecise. I'll try to explain in more detail what these concepts are.
Probability
I'll assume we don't need to attempt to define what probability is. :) I'll just note that it's formalized in mathematics under the aegis of measure theory.
Probability Density
Probability density is a concept that naturally arises whenever you talk about probability in connection with a continuous variable, such as position of a particle. In contrast to discrete probability (such as idealized coin-flipping or dice-rolling), we can't directly assign a probability to each individual outcome. Probabilities have to sum to one, but there are infinitely many possible positions for a particle, so any individual position would have to have probability zero. So how can we talk about probability spread over a continuous domain?
Well, we can talk about the probability of finding the particle within some region. If the region covers a nonzero length/area/volume/whatever, we can expect it to have a nonzero probability. But the exact probability will generally depend on the size/shape/position of the region in a complicated way.
Suppose we consider a very small region. To be concrete let's suppose we're talking about a 1D position and our small region is some interval $[x, x+dx]$. The interval has width $dx$. It stands to reason that a small interval will have a small probability. Moreover, if probability is "spread smoothly" in a sense that I won't bother making precise (hey, this is Physics.SE, not Math.SE!), then for small intervals the probability will be proportional to the width of the interval. If you halve the size of the interval you get half as much probability. Therefore $$dP = p(x) \, dx$$ where $dP$ is the small probability of finding the particle in this small region, and $p(x)$ is some proportionality constant. I've written it as a function of $x$ because the proportionality could vary from point to point.
Well, $p(x)$ is called the probability density, and it has the units of probability per unit length. (Or unit area, or volume, depending on how many dimensions we do this in.) It's a lot like mass density, which is mass per unit length/area/volume, or charge density is charge per unit length/area/volume, etc...
So, now you can imagine calculating the probability to find the particle in any arbitrary region, by slicing up that region into many small ones and summing up $dP$, which is $p(x) \, dx$, for all of them. Well, this is just integrating: $$P(\text{particle is in region $R$}) = \int_R p(x) \, dx$$ In particular, the particle has to be somewhere, so $p(x)$ must integrate to one: $$P(\text{particle is somewhere}) = 1 = \int_{-\infty}^\infty p(x) \, dx$$ But $p(x)$ itself isn't limited to 1. It isn't a probability but a probability per unit length/area/volume, so $p(x) > 1$ just means a region where probability is concentrated. (Much like you can drive your car at 100 km/hr without actually driving 100 km.)
The Wavefunction
All of the above was just standard probability theory, nothing to do with quantum mechanics. Now, in QM we try to predict the probability density for a particle's position (or momentum, or energy, or whatever). We could try to do this by writing an equation for how $p(x)$ changes over time, but it turns out that doesn't give us enough information; there are situations where particles start with identical $p(x)$ but do different things as time goes on.
It's found that we do get enough information to make predictions if we write an equation for a complex-valued function $\psi(x)$, and derive the probability density from it as $$p(x) = \psi^*(x) \psi(x)$$ The way the complex phase of $\psi(x)$ varies from point to point encodes additional information about the particle's momentum, which is necessary to predict its future behavior. It has units of the square root of a probability density, which is a bit weird but perfectly mathematically acceptable. This is of course the wavefunction, and the equation that determines how it varies is the Schrödinger equation.