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

Mathematics Asked by Pendronator on December 1, 2021

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 $tin [t_0,t_1]$. Does this make sense? Thank you!

2 Answers

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 $$intlimits_{C} F d alpha = intlimits_{a}^{b}F(alpha(t))cdot alpha^{'}(t)dt$$ 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 $$intlimits_{C} F d alpha =intlimits_{a}^{b}f(t)dt$$ which, as I guess, you wanted to express.

Answered by zkutch on December 1, 2021

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 $displaystyleint_a^b f(x),dx$ 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.)

Answered by Ted Shifrin on December 1, 2021

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