Let us look at a problem we can solve exactly.

A particle of mass m moves in the potential energy function

U(x) = ∞,  x < 0,  x > a,
U(x) = 0,  0 < x < a.

The energy eigenfunctions are Φ_n(x) = (2/a)^½sin(nπx/a)
with eigenvalues E_n = n²π²ħ²/(2ma²), n = 1, 2, 3, ... .
k = 2π/λ = nπ/a, a = nλ/2.

∫₀^a k dx = (2π/λ) ∫₀^a dx = (2π/λ)(nλ/2) = nπ for the eigenfunctions Φ_n(x).
∫₀^a (2/λ)dx = n,  n counts the number of half-wavelength or the number of anti-nodes of the wave function in the well.

If U(x) = finite when x < 0 or  x > a, then the wave function does not go to zero at the finite edge, but extends into the classically forbidden region. 
Therefore ∫₀^a (2/λ)dx = n - ε, where ε depends on the exact nature of the edge(s).