Integrating x and y components of speed separately to get distance traveled?

Question

Let's say I have a particle moving in the xy-plane, with its position at any $t$ being $(x(t),y(t))$. In order to calculate distance traveled, I have been taught to integrate $\sqrt{(x'(t))^2+(y'(t))^2} dt$. However, I'm wondering why I can't just integrate $|x'(t)|dt$ to get the distance traveled in the x-direction, and integrate $|y'(t)|dt$ to get the distance in the y-direction, and then add the two? In order to find displacement, we can drop the absolute value and it seems to work just fine.

Answer 1

If you move in a circle, with $$ \left\{ \begin{align} x(t)&=R\cos t,\\ y(t)&=R\sin t, \end{align} \right. \qquad t\in[0,2\pi], $$ you get $$ \int_0^{2\pi}\sqrt{x'^2+y'^2}dt=2\pi R $$ while $$ \int_0^{2\pi}|x'|dt=\int_0^\pi R\sin tdt+\int_\pi^{2\pi}(-R\sin t)dt=2R+2R=4R $$ so you get the distance traveled along $x$, a diameter in the go and another diameter in return.