Are regular/Riemann integrals a special case of a line integral?

Question

It would seem that this is the case since Riemann integration is integrating $f(x,y)$ along a trajectory in $\mathbb{R}^2$-- a segment of, or the entirety of, the x-axis. It can be represented as integrating $f(x,0)$ with respect to $x$, and the curve can be parameterized as $$C=\left\{\begin{matrix} x=t\\ y=0 \end{matrix}\right.$$ for some $t\in [t_0,t_1]$. Does this make sense? Thank you!

Answer 1

Actually, you're integrating $f(x)$, not $f(x,y)$. You don't need to set the curve inside $\Bbb R^2$ (or $\Bbb R^3$). Line integrals make sense on "curves" in $\Bbb R$, the simplest case being the directed line segment $[a,b]$. You can indeed think of $\displaystyle\int_a^b f(x)\,dx$ as the work done by the force field $\vec F = f(x)\vec i$ moving a test object along the interval. (Indeed, the spring compression problems and such that appear in standard Calc II classes are examples of this.)

Answer 2

Classical definition, taking for example Tom M. Apostol, Calculus, Volume II_ Multi-Variable Calculus and Linear Algebra, define line integral for piecewise smooth path $\alpha$ defined in $n$ space on interval $[a,b]$ with $C$ as graph of $\alpha$ and vector field $F$, defined on $C$, by equation $$\int\limits_{C} F d \alpha = \int\limits_{a}^{b}F(\alpha(t))\cdot \alpha^{'}(t)dt$$ whenever integral on right hand exists.

Now, if we take, $\alpha :[t_0,t_1] \to \mathbb{R}^2$, defined as $\alpha(t)=(t,0)$, $F=(f,0)$ as field defined on graph of $\alpha$, then we will have $$\int\limits_{C} F d \alpha =\int\limits_{a}^{b}f(t)dt$$ which, as I guess, you wanted to express.