Question

How to calculate this?`ointxdx`

Answer 1

0

Explanation:

`xdx = 1/2 d(x^2)`

Thus, its integral over any curve is the value of `1/2x^2` at the end minus its value at the end. Since for a closed curve, the beginning and the end are the same points, the two values cancel

`oint xdx = 0`

This is true in general whenever the differential to be integrated over a closed loop is an exact differential.