I am reading a book on PDEs, and I am near the beginning where the author is talking about the heat equation and, specifically, solving the non-homogenous equation $u_t={\alpha}^2u_{xx}+f(x,t).$ The author talks about "breaking up" the heat source/sink $f(x,t)$. This is what the author says, continued from previous parts of the problem:
Hence our decomposition of the heat source has the form
$$f(x,t)=f_1(t)\sin(\pi x)+f_2(t)\sin(2\pi x)+\cdots+f_n(t)\sin(n\pi x)$$
Finally, to find the functions $f_n(t)$, we merely multiply each side of this equation by $\sin(m\pi x)$ and integrate from zero to one (with respect to $x$); hence, we have
$$\int_0^1f(x,t)\sin(m\pi x)\mathrm dx = \sum_{n=1}^{\infty}f_n(t) \int_0^1\sin(m\pi x)\sin(n\pi x)\mathrm dx $$
$$=\frac 12 f_m(t)$$
I understand that the sine functions are mutually orthogonal, so the integral sums to $\frac 12$, but what I do not understand is 1. why the final result is $f_m(t)$ (you sum all of the $f_n(t)$'s and you get $f_m(t)$?), and also 2. why the author is able to interchange sum and integral when integrating from $0$ to $1$ and stick the $f_n(t)$ outside the integral but still inside the sum.
$$ \int_0^1 f(x, t) \sin (m \pi x)dx = \int_0^1 [\sum_{k = 0}^n f_k(t) \sin (k \pi x)] \sin (m \pi x)dx = \int_0^1 \sum_{k = 0}^n f_k(t) \sin (k \pi x) \sin (m \pi x)dx = \sum_{k = 0}^n \int_0^1 f_k(t) \sin (k \pi x) \sin (m \pi x)dx = \sum_{k = 0}^n f_k(t) \int_0^1 \sin (k \pi x) \sin (m \pi x)dx = \sum_{k=1}^n f_k(t) \frac{1}{2} \delta_k(m) = \frac{1}{2}f_m(t) $$ where $\delta_k(m)$ is the Kronecker delta function.