Compute $\int_0^a {\bf e}_i^TA {\bf t} \, {\rm d} t_i$

Question

How to compute the following integral \begin{align} \int_0^a {\bf e}_i^TA {\bf t} \, {\rm d} t_i \end{align} wher ${\bf e}_i$ is standard bases vector and $A$ is some matrix square full rank matrix.

It should be a quadratic term. However, I am not sure how to do this. For example, I am not convinced that integration and vector multiplication commute in this case.

Answer 1

$$ \int u^T x d x_i = \int \sum_k u_k x_k d x_i = \int \sum_{k\neq i} u_kx_k dx_i + \int u_i x_i d x_i = \Big(\sum_{k\neq i} u_k x_k\Big)x_i + \tfrac 12u_ix_i^2 + C $$

Now, we can write it a little bit prettier as

$$ \int u^T x d x_i = (u^T x) x_i -\tfrac 12 u_i x_i ^2 + C $$

Answer 2

Assume that $A=A^T$ and consider the following scalar function, its differential, and its subsequent integral. $$\eqalign{ \phi(x) &= \frac{1}{2}x^TAx \\ d\phi &= dx^T(Ax) \\ \int_0^a dx^T(Ax) = \int_{\phi(0)}^{\phi(a)} d\phi &= \Big[\phi(a)-\phi(0)\Big] = \frac{a^2}{2}\big({\tt 1}^TA{\tt 1}\big) \\ }$$ This might be the answer to your question, but only if $A$ is symmetric and if the repetition of the $i$-index was meant to imply the Einstein summation convention.

NB: The above uses $(x)$ instead of $(t)$ as the independent variable.

Update
Based on your comment, it sounds like you have the following problem in mind.
For $i=1$, write the vector as $x={\tt [}\;t\;u\;v\;\ldots\;{\tt ]}^T$ and write the integral as $$\eqalign{ \Phi_1 &= \int_0^a dt\; e_1^TAx \cr &= e_1^TA\;\int_0^a dt\,x \cr &= e_1^TA\;{\tt [}\;\tfrac{1}{2}a^2\quad au\quad av\;\ldots\;{\tt ]}^T \cr &= e_1^TA\Big(\tfrac{1}{2}a^2e_1 + a(I-e_1e_1^T)x\Big) \cr &= \frac{a^2}{2}\big(e_1^TAe_1\big) + a(e_1^TAx) - a(e_1^TAe_1)(e_1^Tx) \cr }$$ The calculation is the same for any other value of $i$, just replace the $e_1$ basis vectors with $e_i$.