The expression for the mean value of an observable A in the normalized state |Ψ> is <A> = <Ψ|A|Ψ>. 
If |Ψ> is not normalized then < A> = <Ψ|A|Ψ>/<Ψ|Ψ>.
<x> = <Ψ|X|Ψ> = ∫ d³r <Ψ|r><r|X|Ψ>  = ∫ d³r Ψ*(r) x Ψ(r).
<p_x> = <Ψ|P_x|Ψ> = ∫ d³r <Ψ|r><r|P_x|Ψ>  = ∫ d³r Ψ*(r) (ħ/i)∂Ψ(r)/∂x.

The root mean square deviation ΔA characterizes the dispersion of the measurement around <A>.
ΔA = (<(A - <A>)²>)^1/2.

ΔA = (<(A - <A>)²>)^1/2 = (<Ψ|(A - <A>)²|Ψ>)^1/2
=  (<Ψ|A²|Ψ> - 2<A><Ψ|A|Ψ> + <A>²<Ψ|Ψ>)^1/2
=  (<A²> - 2<A><A> + <A>²)^1/2
= (<A²> - <A>²)^1/2.
ΔA is a measure of the spread that one should expect in the result of a measurement of the observable A.

Problem:

Assume the wave function of a particle is Ψ(x) = N exp(ip₀x/ħ)/(x² + a²)^1/2.
Here a and p₀ are real constants and N is a normalization constant.
(a)  Find N so that ψ(x) is normalized.
(b)  If the position of the particle is measured, what is the probability of finding the particle between  -a/√3  and  +a/√3?
(c)  Calculate the mean value of the momentum of the particle. 

• Concepts:
  The postulates of Quantum Mechanics, the mean value
• Reasoning:
  When a physical quantity described by the operator A is measured on a system in a normalized state |ψ>, the probability of measuring the eigenvalue a_α is given by dP(a_α) = |<u_α|ψ>|²dα, where |u_α> is the eigenvector corresponding to the eigenvalue a_α; we assume a_α is a non-degenerate continuous eigenvalue of A.
  The expression for the mean value of an observable A in the normalized state |ψ> is <A> = <ψ|A|ψ>. 
• Details of the calculation:
  (a) <Ψ|Ψ> = 1.  ∫_-∞^+∞Ψ*(x)Ψ(x) dx = 1.
  N²∫_-∞^+∞exp(-ip₀x/ħ)/(x² + a²)^1/2 exp(ip₀x/ħ)/(x² + a²)^1/2
  = N²∫_-∞^+∞dx/(x² + a²)  = 1.
  ∫_-∞^+∞dx/(x² + a²)  = (1/a)tan^-1(x/a)|_-∞^+∞  = (1/a)[π/2 - (-π/2)] = π/a.
  N² = a/π,  N = (a/π)^1/2.

  (b)  Let P denote the probability of finding the particle between -a/√3 and +a/√3.
  Let y = a/√3.
  P = ∫_-y^+y |Ψ(x)|² dx = (a/π)∫_-y^+ydx/(x² + a²)  = (1/π)tan^-1(x/a)|_-y^+y

  = (1/π)[tan^-1(1/√3) - tan^-1(-1/√3)]
  = (1/π)[π/6 + π/6] = 1/3.

  (c)  Let <P> denote the mean value of the momentum of the particle. 
  <P> = ∫_-∞^+∞Ψ*(x) ((ħ/i)∂/∂x) Ψ(x) dx
  = (a/π)∫_-∞^+∞exp(-ip₀x/ħ)/(x² + a²)^1/2 (ħ/i)[(ip₀/ħ)exp(ip₀x/ħ)/(x² + a²)^1/2 - x exp(-ip₀x/ħ)/(x² + a²)^3/2]
  = (a/π)p₀∫_-∞^+∞dx/(x² + a²)  - (a/π)(ħ/i)∫_-∞^+∞ xdx/(x² + a²)² = p₀.
                        (π/a)                                 0, odd function

The generalized uncertainty relation
Let A and B be two observables (Hermitian operators).  In any state of the system
ΔA ΔB ≥ ½|<i[A,B]>|.
In particular, for any two Hermitian operators Q and P for which [Q,P] = iħ,
ΔQ ΔP ≥ ½|<iiħ>| = ħ/2.

Proof:
Let |Ψ> be any state vector and let A₁ = A - <A>I and B₁ = B - <B>I.
Let |Φ> = A₁|Ψ> + ixB₁|Ψ> with x an arbitrary real number. 
Then <Φ| = <Ψ|A₁ - ix<Ψ|B₁.
<Φ|Φ> = <Ψ|A₁²|Ψ> + x²<Ψ|B₁²|Ψ> - ix<Ψ|B₁A₁|Ψ> + ix<Ψ|A₁B₁|Ψ>
= <A₁²> + x²<B₁²> + x<Ψ|i[A₁,B₁]Ψ> = (ΔA)² + x²(ΔB)² + x<i[A,B]>
for all x.  But <Φ|Φ> ≥ 0.  Therefore  = (ΔA)² + x²(ΔB)² + x<i[A,B]> ≥ 0 for all x.
For this to be true we either need that the quadratic equation y = ax² + bx + c either has no zeros, or is equal to zero for only one value of x.  (The equation describes a parabola.  The vertex of the parabola can either touch the x-axis or must lie above the x-axis.)  To find a root we set x = (b ± (b² - 4ac)^1/2)/(2a).  If b² < 4ac the equation has no roots.  If b² = 4ac or x = -b/2a  the equation is zero.  We therefore need b² < 4ac .
Therefore |<i[A,B]>|² ≤ 4(ΔA)²(ΔB)².