Does a Line Integral that sums to a vector exist ? And is it useful? We know that we can integrate a vector field on a curve to get a scalar sum: this is how the line integral is defined. Is there a use for another sort of integral whose sum is a vector ? I can't seem to find that defined anywhere.
 A: Integrals are (limits of) sums of many small things. If the small things are vectors, then the result is a vector too.
The simplest example is probably the following: The sum of many small displacements is a displacement. More specifically, let a particle follow the path $\vec p(t)$, where $t$ is time. Then we have the velocity vector $\vec v(t) =\vec p'(t)$.
The net displacement of this particle from $t$ to $t+\Delta t$ is roughly $\vec v(t)\cdot\Delta t$. For small $\Delta t$, this is a small vector. This can be turned into an integral.
If you integrate the velocity vector with respect to time from $t=t_0$ to $t=t_1$, the end result is the net displacement $\vec p(t_1)-\vec p(t_0)$:
$$
\int_{t_0}^{t_1}\vec v(t)\,dt=\vec p(t_1)-\vec p(t_0)
$$
A: When you refer to "the line integral," you're talking of course about the work line integral $\displaystyle\int_C \vec F\cdot d\vec r$. So this amounts to integrating a scalar function. But in physics you'll also encounter a line integral like $\displaystyle\int_C \vec F\times d\vec r$ (e.g., Biot-Savart), and so you'll be integrating a vector function along the length of the curve.
A: If you have a (scalor-valued) function $f:[a,b]\to\mathbb{R}$ with "nice" properties, you can define the Riemann integral:
$\int_a^b f(x)dx$, the result of which is a scalar.
If you have a vector-valued function $f:[a,b]\to\mathbb{R}^n$, where each component  of $f=(f_1,\cdots,f_n)$ is a scalar (integrable) function, you have
$$
\int_a^b f(x)dx:=(\int_a^b f_1(x)dx,\cdots,\int_a^b f_n(x)dx) \tag{1}
$$
the result of which is a vector.
More generally, if you have a vector-valued function $f:X\to\mathbb{R}^n$ where $(X,\mu)$ is some measure space,  then you have the analog of (1) as
$$
\int_Xfd\mu:=(\int_Xf_1d\mu,\cdots,\int_Xf_nd\mu)
$$
Even more generally, if you replace $\mathbb{R}^n$ with some Banach space, you have the notion of Bochner integrals.
