Calculate the expectation value of #x^3# for a particle in the state n=2 in a square well potential?

Well, after several integration by parts...

#<< x^3 >> = ((4pi^2 - 3)L^3)/(16pi^2)#

DISCLAIMER: REALLY LONG ANSWER!

I assume you mean the one with

#V(x) = { (0, 0 < x < L),(oo, "otherwise"):}#

For this, I assume you already have the known wave function solution as:

#psi_n(x) = sqrt(2/L) sin((npix)/L)#

In that case, for #n = 2# we have #psi_2(x) = sqrt(2/L) sin((2pix)/L)#.

The #x^3# expectation value for a normalized wave function (which this is) is:

#<< x^3 >> = << psi_2 | x^3 | psi_2 >>#

#= 2/L int_(0)^(L) x^3sin^2((2pix)/L)dx#

Use the identity that #1/2 (1 - cos(2u)) = sin^2u# so that:

#= 1/L int_(0)^(L) x^3(1 - cos((4pix)/L))dx#

#= 1/L int_(0)^(L) x^3dx - 1/L int_(0)^(L) x^3cos((4pix)/L)dx#

Unfortunately there is not much we can do except integration by parts...

Let:

#u = x^3#
#dv = cos((4pix)/L)dx#
#v = L/(4pi) sin((4pix)/L)#
#du = 3x^2dx#

First we get:

#1/L int_(0)^(L) x^3cos((4pix)/L)dx#

#= 1/L [L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)int_(0)^( L)x^2sin((4pix)/L)dx]#

Repeat again. Let:

#u = x^2#
#dv = sin((4pix)/L)dx#
#v = -L/(4pi)cos((4pix)/L)#
#du = 2xdx#

Then we get:

#= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L)/(4pi) int_(0)^(L)xcos((4pix)/L)dx]}#

And we have one more integration by parts... Let:

#u = x#
#dv = cos((4pix)/L)dx#
#v = L/(4pi)sin((4pix)/L)#
#du = dx#

Then we get:

#= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L)/(4pi) {L/(4pi)xsin((4pix)/L) - L/(4pi)int_(0)^(L)sin((4pix)/L)dx}]}#

Evaluating the innermost integral,

#= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L)/(4pi) {L/(4pi)xsin((4pix)/L) - L/(4pi)(-L/(4pi)cos((4pix)/L))}]}#

And now we simplify it, before evaluating. Factor in the #L/(4pi)# on the inside:

#= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L)/(4pi) {L/(4pi)xsin((4pix)/L) + L^2/(16pi^2)cos((4pix)/L)}]}#

Factor in the #(2L)/(4pi)#:

#= 1/L {L/(4pi) x^3sin((4pix)/L) - (3L)/(4pi)[-L/(4pi)x^2cos((4pix)/L) + (2L^2)/(16pi^2)xsin((4pix)/L) + (2L^3)/(64pi^3)cos((4pix)/L)]}#

Factor in the #-(3L)/(4pi)#:

#= 1/L {L/(4pi) x^3sin((4pix)/L) + (3L^2)/(16pi^2)x^2cos((4pix)/L) - (6L^3)/(64pi^3)xsin((4pix)/L) - (6L^4)/(256pi^4)cos((4pix)/L)}#

Lastly, factor in the #1/L#:

#= 1/(4pi) x^3sin((4pix)/L) + (3L)/(16pi^2)x^2cos((4pix)/L) - (6L^2)/(64pi^3)xsin((4pix)/L) - (6L^3)/(256pi^4)cos((4pix)/L)#

And now we evaluate this from #0# to #L#... Lots of stuff goes to #0#.

#= [cancel(1/(4pi) L^3sin(4pi))^(0) + (3L)/(16pi^2)L^2cancel(cos(4pi))^(1) - cancel((6L^2)/(64pi^3)Lsin(4pi))^(0) - (6L^3)/(256pi^4)cancel(cos(4pi))^(1)] - [cancel(1/(4pi) 0^3sin((4pi cdot0)/L))^(0) + cancel((3L)/(16pi^2)0^2cos((4picdot0)/L))^(0) - cancel((6L^2)/(64pi^3)cdot0cdot sin((4picdot0)/L))^(0) - (6L^3)/(256pi^4)cancel(cos((4picdot0)/L))^(1)]#

#= [(3L)/(16pi^2)L^2 - cancel((6L^3)/(256pi^4))] - [- cancel((6L^3)/(256pi^4))]#

#= (3L^3)/(16pi^2)#

Now to finish this off:

#color(blue)(<< x^3 >>) = << psi_2 | x^3 | psi_2 >>#

#= 1/L int_(0)^(L) x^3dx - (3L^3)/(16pi^2)#

#= 1/L L^4/4 - (3L^3)/(16pi^2)#

#= L^3/4 - (3L^3)/(16pi^2)#

#= color(blue)(((4pi^2 - 3)L^3)/(16pi^2))#