When integrating a vector expression, why is the constant of integration a vector?
Is there a certain proof that shows this is the case?
Integration constant vector
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$\begingroup$
calculus
integration
vectors
indefinite-integrals
constants
2 Answers
1
In $R^{2}$:
If r$(t) = f(t)$i + g(t)j
$\int$ r$(t)dt$ = $[\int f(t)dt ]$i + $[\int g(t)dt ]$j
Let $F'(t)=f(t)$ and $G'(t)=g(t)$, then
$\int$ r$(t)dt$ = $[F(t) + C_{1} ]$i + $[G(t) + C_{2} ]$j
$\int$ r$(t)dt$ = $[F(t)$i + $G(t)$j$]$ + $[C_{1}$i + $C_{1}$j$] = $ R$(t) + $C
where R'$(t) = r(t)$
3
We have $$\int\begin{bmatrix}f_1(x)\\ \vdots\\ f_n(x)\end{bmatrix}dx\underbrace{=}_{\text{def}}\begin{bmatrix}\int f_1(x)dx\\ \vdots\\\int f_n(x)dx\end{bmatrix}=\begin{bmatrix}F_1(x)+C_1\\ \vdots\\ F_n(x)+C_n\end{bmatrix}=\begin{bmatrix}F_1(x)\\ \vdots\\ F_n(x)\end{bmatrix}+\underbrace{\begin{bmatrix}C_1\\ \vdots\\C_n\end{bmatrix}}_{\mathbf{C}},$$ where $F_1,\ldots,F_n$ are primitives of $f_1,\ldots,f_n$ respectively.