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I know how the line integral is calculated, but I want to know what the meaning of line integral with respect to x is, I mean, intuitively.
The integral is $\int_C f(x,y) dx$ or it can be $\int_C f(x,y) dy$. ($C$ is a curve here)
Thanks.

Sida
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  • Finds the area under the curve, innit... Take a bit of the line, multiply its length by the value of the function, then sum the little bits. – Lost1 Jan 05 '14 at 14:00
  • @Lost1 Thanks, but isn't it the line integral with respect to arc length? What I want, is the meaning of line integral with respect to x. – Sida Jan 05 '14 at 14:05
  • Do you $\int_C x , ds$ where $C$ is some curve in the pane? – Ulrik Jan 05 '14 at 14:06
  • @Svinepels No, I mean ∫f(x,y)dx over C. I want to understand the meaning of this integral. – Sida Jan 05 '14 at 14:10
  • Oh i misunderstood you. There are people more qualified to answer this question than me but i thought it is the same... This is why you'd benefit from writing down the integral in the question. – Lost1 Jan 05 '14 at 14:21
  • @Lost1 Yes, you are rigth. I'm going to edit my question. – Sida Jan 05 '14 at 14:27
  • Wikipedia has a nice illustration for the meaning of the line integral: http://en.wikipedia.org/wiki/File:Line_integral_of_scalar_field.gif – Mark McClure Jan 05 '14 at 16:10
  • take a look at this http://math.stackexchange.com/questions/1080027/interpretation-of-a-line-integral-with-respect-to-x-or-y – KFkf Apr 23 '15 at 07:26

2 Answers2

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You can think of it as representing work done moving along the curve C with a force of strength $f$ pointing in the $x$-direction. (So if $x$ is vertical, then f might represent gravity, and the integral would be work done against gravity).

Mathematically, it essentially integrates the dot product of the unit tangent vector of C with a vector (f,0,0), thus measuring how much the curve lines up with or against a vector field of strength f in the x direction.

Brian Rushton
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The line integral can still be understood in terms of an area under a curve.

When $f>0$, the integral $$\int\limits_Cf(x,y)ds$$ calculates the area of the "curtain" beneath a curve, where $C$ is the base of the curtain, and at any point $(x,y)$, $f(x,y)$ is the height. Imagine the line C weaving through the $x,y$ plane, and a curtain extending off of this line to a height $f(x,y)$.

inversquare
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