Calculating work done when the lower bound of integral is greater than the upper bound

Question

In this video, Dr. Peter Dourmashkin explained friction as an example of a force by which the work done is not path independent. In $2$$:$$50$ min of the video, when we're coming back, he said, $d\vec{s}$ will still be $dx\hat{i}$. But if so, then $dx$ is negative now since $d\vec{s}$ is in the direction of $-\hat{i}$ now. In other words, we want $d\vec{s}$ to be $dx\hat{i}$ instead of $-dx\hat{i}$ because $dx$ is the infinitely small negative number in our integral $\int_{x_{a}}^{x_{f}} \mu_{k}mg\hat{i} \cdot dx\hat{i}$ since $x_{a}$ $>$ $x_{f}$. Am I correct?

Answer 1

This example is a good one to illustrate the difference between the magnitude of a vector and a component of a vector.

The work is defined as $\displaystyle \int_{\rm start}^{\rm finish} \vec F \cdot d\vec s$.

In this example $\mu_{\rm k},\,m$ and $g$ are all positive quantities and so $\mu_{\rm k}mg$ is the magnitude of the frictional force as that product is always positive.
When the object moves in the $\hat i$ direction the frictional force is in the $(-\hat i)$ direction so $\vec F = \mu_{\rm k}mg(-\hat i)= -\mu_{\rm k}mg\hat i$. When the object moves in the $(-\hat i)$ direction the frictional force is in the $\hat i$ direction so $\vec F = \mu_{\rm k}mg\hat i$.

In the video the lecturer treats the displacement $d\vec s = dx \hat i$ in a different way.

In this case $dx$ is the component of the displacement in the $\hat i$ direction and as such can be either positive or negative.
The sign of the component $dx$ is entirely dependent on the limits of integration.
This can be shown by just considering the displacement from $x_{\rm f}$ to $x_{\rm a}$, $\displaystyle \int _ {x_{\rm f}} ^{x_{\rm a}}dx \hat i = (x_{\rm a} - x_{\rm f})\hat i$ and then from $x_{\rm a}$ to $x_{\rm f}$, $\displaystyle \int _ {x_{\rm a}} ^{x_{\rm f}}dx \hat i = (x_{\rm f} - x_{\rm a})\hat i$.
As there is an underlying assumption that $x_{\rm a}$ and $x_{\rm f}$ are both positive, the first displacement is in the $\hat i$ direction, and the second displacement is in the $(-\hat i)$ direction, exactly as one might expect.