For the infinite square-well potential, find the probability that a particle in its fourth excited state is in each third of the one-dimensional box:
(0 to L/3)
(L/3 to 2L/3)
and (2L/3 to L)
The Attempt at a Solution
So from ∫ψ^2 for the first third of the problem i got that my probability equation should be
1/3-(1/((2πn))(sin(2π/3)) - 0 where n is that excited state but I am getting a wrong answer of about .10 off , is there something wrong with my equation?