In dealing with a question about equilibrium values for the heat equation $$\frac{\partial u}{\partial t}=\frac k r \frac{\partial}{\partial r}\left(\frac{\partial u}{\partial r}\right)$$
after a number of computations, we find that the condition for equilibrium implies
$$0=\frac{\mathrm{d}}{\mathrm{d}t}\int^b_ar\,u(r,t)\,\mathrm{d}r=\int^b_ar\,\frac{\partial u}{\partial t}\,\mathrm{d}r=k\int^b_a\,\frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r}\right)\mathrm{d}r$$
Which makes sense. However, the calculation then does:
$$\dots=b\frac{\partial u}{\partial r}\big|_{r=b}-a\frac{\partial u}{\partial r}\big|_{r=a}$$
How does this occur? Perhaps I'm rusty on my integration, but wouldn't both the partial derivative and $r$ in the integrand be evaluated? And by that, wouldn't integration by parts be used? It seems that $r$ wasn't even touched, since its integral is $\frac {r^2} 2$...
Basically, could someone walk me through how to perform the integration
$$\int^b_a\,\frac{\partial }{\partial r}\left(r\frac{\partial u}{\partial r}\right)\mathrm{d}r$$