I was studying for multivariable calculus and I came across the line integral section. Visually, I perfectly understand why $\int_{t_1}^{t_2} f(x(t),y(t))s'(t)\,dt$ computes the area under $f(x,y)$ along the curve C whose point $(x,y)$ given $t$ is $x = x(t)$, $y = y(t)$.
However, can anyone explain why we couldn't just do $\int_{t_1}^{t_2} f(x(t),y(t))\,dt$ to compute this area? Graphically, what is the quantity $\int_{t_1}^{t_2} f(x(t),y(t))\,dt$?
Consider a simple example like: $C$ is a circle of radius 2 and $f(x,y)=1$.
In this case $\int_C f(x,y)\,ds=4\pi$ since it computes the area of under the portion of $f(x,y)=1$ which lies above the circle (this is a cylinder of radius 2 and height 1).
Parameterize $C$ by $(2\cos(t),2\sin(t))$ with $0 \leq t \leq 2\pi$. The if you neglect $s'(t)$ (the $ds$ part) of the line integral, you get: $\int_0^{2\pi} f(x(t),y(t))\,dt = \int_0^{2\pi} 1\,dt = 2\pi$...the WRONG ANSWER!
So why is $ds$ necessary?
Well, area is width times height. Consider a point on your curve: $(x(t),y(t))$. Then $f(x(t),y(t))$ gives the height above the curve. If we want to compute a little strip of area, we need a "bit of width". This should be a little chunk of arc length (i.e. $ds$).
Take a little change in $t$, say $\Delta t$. Then for such a small change in $t$ our curve stays approximately linear, so we can approximate it with its tangent line there. The length of the corresponding piece of tangent line is $\sqrt{x'(t)^2+y'(t)^2}\,\Delta t$. So our little strip of our region has area approximately $f(x(t),y(t))\sqrt{x'(t)^2+y'(t)^2}\,\Delta t$.
Adding up these strips and taking limits ($\Delta t \to 0$) we end up with the line integral.
Now what is the integral without the "$s'(t)$" part? Not really anything. This integral depends on the way you parameterize $C$, so it doesn't have a reasonable geometric interpretation. :(