If $f:\mathbb{R}\rightarrow\mathbb{R}$ is an integrable function.
then $\int_a^b f(x)dx$ can be considered as the area between the graph and the x-axis.
But what if $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$?
Let $\gamma:[0,1]\rightarrow \mathbb{R}^n$ be a smooth curve.
what is the geometrical meaning of $\int_\gamma f\cdot dl$? (or in case $n=m$, $= \int_0^1 f(\gamma(t))\gamma '(t)\cdot dt$?)
Thanks :)
(for simplicity, you can assume $m,n$ are small numbers... i.e $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ )
There is another common interpretation of integration that you might be familiar with, which is an interpretation of a continuous sum. In case you aren't familiar with this, I'll explain it in terms of the area interpretation.
If you imagine drawing the curve $y=f(x)$, the area under the curve can be roughly interpreted as the sum of all the infinitesimal-width vertical 'strips' which connect the point $(x,0)$ to the point $(x,f(x))$. Of course, this doesn't make much sense, because there's no such thing as a rectangle with 'infinitely small width'. One way of turning this intuitive picture into a sensible idea is the Riemann sums, which are roughly of the form:
$$\sum_{i=1}^{n}f(x_{i}^{*})(x_{i+1}-x_{i})$$ where $x_{0},\ldots,x_{n}$ are points that partition your interval $[a,b]$ into the intervals $[x_{i-1},x_{i}]$, and $x_{i}^{*}$ is some point in that interval. But this can also be viewed as a weighted sum of the values $f$ takes in the interval $[a,b]$. If we take a suitable limit of the partition which makes the intervals $[x_{i-1},x_{i}]$ uniformly small, we get your area interpretation of the integral.
So, the line integral $$\int_{\gamma}f(x)dx:=\int_{a}^{b}f(\gamma(t))|\gamma'(t)|dt$$ can be interpreted as a continuous sum of vector contributions of $f$, as we travel along small portions of the curve $\gamma$. The $|\gamma'(t)|$ factor, reminiscent of the substitution rule for 1-D integrals, can be thought of as a scale factor relating how 'fast' we go along the curve $\gamma$ - this ensures that if we reparametrise the curve so that we go faster or slower in places, we don't change the value of the integral.
For a physical example, let's suppose $\gamma$ is a uniformly charged wire, and you want to calculate the electric field at a point $x$ due the wire. Well, there is a charge contribution from each point $\gamma(t)$, which is given approximately by $\lambda |\gamma'(t)|dt|$, where we have written $\lambda$ for the charge per unit length of the wire. The electric field due to this point-like charge is simply $$E(\gamma(t))=\int_{a}^{b}\frac{x-\gamma(t)}{4\pi\epsilon_{0}|x-\gamma(t)|^{3}}\lambda|\gamma'(t)|dt = \int_{\gamma}\frac{x-x'}{4\pi\epsilon_{0}|x-x'|^{3}}\lambda dx'$$
I hope that's a satisfying example - if you haven't seen much electromagnetism then it probably won't make much sense, but electric fields are the simplest example I can think of of a meaningful vector-valued line integral. If you don't insist on the integrand being vector valued, then the work done on a point particle is probably a simpler and more intuitive example.