How do we prove vector-valued functions can be integrated so easily? In applied math, you learn that you can differentiate and integrate vector-valued functions, but in intro analysis, you learn primarily the one-dimensional case.
Can anyone explain what entails the leap from one-dimensional integrals to integrals over vector-valued functions? Is this due to the algebra that vectors form?
 A: If $f\colon [a,b] \to V$ is any function, where $V$ is any finite-dimensional vector space, choose a basis $(e_1,\ldots, e_n)$ for $V$ and write $$f(t) = \sum_{i=1}^n f_i(t)e_i.$$Say that $f$ is integrable if each $f_i$ is integrable (in the one-dimensional sense), and then set $$\int_a^b f(t)\,{\rm d}t \doteq \sum_{i=1}^n \left(\int_a^bf_i(t)\,{\rm d}t\right)e_i,$$and we show that the vector $\int_a^bf(t)\,{\rm d}t\in V$ is well-defined, i.e., integrability of $f$ in terms of the components $f_i$ and the resulting vector-valued integral and independent of the choice of basis. If $(\widetilde{e}_1,\ldots,\widetilde{e}_n)$ is a second basis and we write $$\widetilde{e}_j = \sum_{i=1}^n a_{ij}e_i$$for some non-singular matrix $(a_{ij})$ and $$f(t) = \sum_{i=1}^n \widetilde{f_i}(t)\widetilde{e_i},$$then we have that $$\widetilde{f_j}(t) = \sum_{i=1}^n b_{ij}f_i(t),$$where $(b_{ij})$ is the inverse of $(a_{ij})$. At this point we already know that integrability of $f$ is basis-independent. So $$\begin{align}\sum_{j=1}^n \left(\int_a^b\widetilde{f_j}(t)\,{\rm d}t\right)\widetilde{e}_j &= \sum_{j=1}^n \left(\int_a^b \sum_{k=1}^n b_{jk}f_k(t)\,{\rm d}t\right) \sum_{i=1}^n a_{ij}e_i \\ &= \sum_{i,j,k=1}^n a_{ij}b_{jk} \left(\int_a^bf_k(t)\,{\rm d}t\right)e_i \\ &= \sum_{i,k=1}^n \delta_{ik} \left(\int_a^bf_k(t)\,{\rm d}t\right)e_i \\ &= \sum_{i=1}^n \left(\int_a^bf_i(t)\,{\rm d}t\right)e_i,\end{align}$$as required.
For infinite-dimensional spaces, the situation is more complicated. The keyword is "Bochner Integral".
