linear transformation matrix under the line integral

Question

Is there a general methodology/approach for evaluating an integral of this form?

$$ \int_C {\bf Ax} \cdot \mathrm{d}{\bf x} $$

Assume ${\bf x} \in \mathbb{R}^{n \times 1}$ and ${\bf A} \in \mathbb{R}^{n \times n}$. Or is more information needed about ${\bf A}$? I'm not necessarily looking for someone giving me an elaborate explanation but a pointer in the right direction would be nice.

EDIT: When I say "more information", I don't mean knowing the exact value of ${\bf A}$, but rather some qualitative information. For example, ${\bf A}$ is positive-definite or symmetric or maybe non-invertible.

Answer 1

If $C$ is sufficiently smooth, then we can parameterize the curve $C$ and write

\begin{align} \int_C (\bar A\cdot x)\cdot dx&=\sum_{i}^n\,\sum_{j}^n\,A_{ij}\int_{t_1}^{t_{2}} x_j(t)\,x'_i(t)dt\\\\ &=\sum_{i}^n\,\sum_{j}^n\,A_{ij}\left([x_i(t_2)x_j(t_2)-x_i(t_1)x_j(t_1)]-\int_{t_1}^{t_{2}} x_i(t)\,x'_j(t)dt\right) \end{align}

where we used integration by parts to obtain the last expression. If $\bar A$ is symmetric, then by definition $A_{ij}=A_{ji}$ and

$$\sum_{i}^n\,\sum_{j}^n\,A_{ij}\int_{t_1}^{t_{2}} x_j(t)\,x'_i(t)dt=\sum_{i}^n\,\sum_{j}^n\,A_{ij}\left([x_i(t_2)x_j(t_2)-x_i(t_1)x_j(t_1)]-\int_{t_1}^{t_{2}} x_j(t)\,x'_i(t)dt\right)$$

which reveals that

$$\sum_{i}^n\,\sum_{j}^n\,A_{ij}\int_{t_1}^{t_{2}} x_j(t)\,x'_i(t)dt=\frac12 \sum_{i}^n\,\sum_{j}^n\,A_{ij}[x_i(t_2)x_j(t_2)-x_i(t_1)x_j(t_1)]$$

as reported earlier by Kobe.

Answer 2

There is a method of evaluation when $\mathbf{A}$ is symmetric. If $\mathbf{A}$ is symmetric, then $\mathbf{Ax}$ is the gradient of the scalar field $F(\mathbf{x}) = \frac{1}{2}\sum\limits_{i, j = 1}^n a_{ij}x_i x_j$. So if $C$ is a space curve starting at $\mathbf{p}$ and ending at $\mathbf{q}$, then $\int_C \mathbf{Ax}\cdot d\mathbf{x} = F(\mathbf{q}) - F(\mathbf{p})$.