For exercise 1.4, I believe there is an error in the solution.
In the solution, it says we take the integral of 4\pi k dk, but I believe this should be 4 \pi k^2 dk.
@JRietveld That is correct! Thanks for notifying us 
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I believe that I have found another error:
In exercise 4 the heat capacity is C=\frac{\partial E}{\partial T}, but the integral \int_0^{\beta \hbar \kappa_D}\frac{x^3}{e^-1}dx in the energy E is not differentiated, while it has temperature dependence through \beta. Therefore, we should get the product rule and the expression becomes more complicated.
Or am I missing something?
I included the solution below for convenience:
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I agree, I posted about it also but I don’t think it has been changed yet. I differentiated the first expression for E and got
C = \frac{3L^3 k_B^4}{2 \pi^2 \hbar^3 v_x v_y v_z} T^3 \int_0^{\hbar \kappa_D/k_B T}\frac{x^4 e^x}{(e^x - 1)^2} \mathrm{d}x
@lcrijns indeed something is missing, I will fix it soon. For now I am quoting Anton’s response to the question @lucasdekam asked.
Thank you for the reply. I missed @lucasdekam’s answer in the other thread.