Question 2.2: A particle in the infinite square well has the initial wave ...

First we need to determine A, by normalizing  Ψ(x,0):

1=\int_{0}^{a}\left|\psi (x,0)\right|^{2}dx=\left|A\right| ^{2}\int_{0}^{a} x^{2}(a-x)^{2}dx=\left|A\right|^ {2} \frac {a^{5}}{30}

so

A=\sqrt{\frac{30}{a^{5}}}

The nth coefficient is (Equation 2.40)

c_{n}=\sqrt{\frac{2}{a} }\int_{0}^{a}{\sin{ \left (\frac{n\pi}{a} x\right) }}\psi (x,0) dx                        (2.40)

c_{n}=\sqrt{\frac{2}{a} }\int_{0}^{a}{\sin{ \left (\frac{n\pi}{a} x\right) }} \sqrt{\frac{30}{a^{5} }} x(a-x)dx

=\frac{2\sqrt{15} }{a^{3}} \left[a\int_{0}^{a}{x\sin\left(\frac{n\pi}{a} x\right) }-\int_{0}^{a}{x^{2}\sin\left(\frac{n\pi}{a} x\right)}dx \right]

=\frac{2\sqrt{15} }{a^{3}}\left\{a\left [\left (\frac{a}{n\pi} \right)^{2}\sin\left(\frac{n\pi}{a} x\right) dx-\frac{ax}{n\pi}\cos\left(\frac{n\pi}{a} x\right) \right]|_0^a -\left[2\left(\frac{a}{n\pi} \right) ^{2} x\sin \left(\frac{n\pi}{a} x\right)-\frac{(n\pi x/a)^{2}-2}{(n \pi /a)^{3}}\cos\left( \frac{n \pi}{a} x\right) \right] \mid ^{a}_{0} \right\}

=\frac{2\sqrt{15} }{a^{3}}\left[-\frac{a^{3}}{n\pi}\cos(n\pi ) +a^{3}\frac{(n\pi)^{2}-2}{(n\pi)^{3}}\cos \left(n\pi \right)+a^3 \frac{2}{(n \pi )^3}cos(0) \right]=\frac{4\sqrt{15} }{(n\pi)^{3}} [\cos(0)-\cos(n\pi)]

=\begin{cases}0,&n  even\\ 8\sqrt{15}/ (n\pi)^{3} , &n  odd.\end{cases}

Thus (Equation 2.39):

\psi (x,t)= \sum\limits_{n=1}^{\infty }{c_n\sqrt{\frac{2}{a}} \sin \left(\frac{n\pi}{a}x\right)e^{-i(n^{2}\pi^{2} \hbar t/2ma^{2})t} }                      (2.39)

\psi (x,t)=\sqrt{\frac{30}{a} } \left(\frac{2}{\pi} \right) ^{3}\sum\limits_{n=1,3,5}{\frac{1}{n^{3}} \sin \left(\frac{n\pi}{a}x\right)e^{-in^{2}\pi^{2} \hbar t/2ma^{2}} } .