Vfrdtyky

Consider I have a simple formula for the work along some path (in 1 dimension):

$$W~=~int_{x_0}^{x_1}vec{F}cdot dvec{x}.$$

If I now move from left to right ($x_1 > x_0$) along the axis (assuming the X-axis points right) it is obvious that (using unit vectors along the axis):

$vec{F}=Fvec{i}$

$vec{x}=xvec{i} => dvec{x}=dxvec{i}$

$vec{F}dvec{x} = (Fvec{i})(dxvec{i})=Fdx$

Integration gives (assuming constant force along path for simplicity):

$int_{x_0}^{x_1}vec{F}dvec{x} = int_{x_0}^{x_1}Fdx = F int_{x_0}^{x_1}dx = F(x_1-x_0) > 0$

Consider now I move from right to left ($x_1 < x_0$) but the force is also reversed, then:

$vec{F}=-Fvec{i}$

$vec{x}=-xvec{i} => dvec{x}=-dxvec{i}$

$vec{F}dvec{x} = (-Fvec{i})(-dxvec{i})=Fdx$

$int_{x_0}^{x_1}vec{F}dvec{x} = int_{x_0}^{x_1}Fdx = F int_{x_0}^{x_1}dx = F(x_1-x_0) < 0$

Obviously, there is some terrible mistake here in my reasoning... If we move from one point to another along the force direction the work must be positive.

Your mistake is in thinking that $dx$ doesn't account for the direction of motion. More explicitly, $$dboldsymbol{x} neq -dx hat{boldsymbol{i}}$$ in the second scenario. Instead, $$dboldsymbol{x} = dx hat{boldsymbol{i}}$$ as in the first case.

This answer amplifies @PiKindOfGuy's answer and @Dale's comment.

I think there is a confusion in the use and/or interpretation of
$dvec x$

• as an infinitesimal element of a directed path
  and


• as an increment of the x-coordinate.
  
  

Let's use $dvec s$ for the infinitesimal element of a path.

Let P and Q refer to start and end of the path.

So, $$W=int_P^Q vec Fcdot dvec s$$

Before I begin, I want to emphasize that the Force $vec F$ is generally independent of the path (and its direction) of the displacement. For instance, the gravitational force always points down... but the path could be arbitrary.

In your example, when you turn, you are changing the direction of the force and the displacement... let's say forward from P to Q then backward to R, where
$x_P ,x_R < x_Q$.

Along a horizontal displacement and constant magnitude force
...with no mention of any directions...
begin{align}
W_{AB}
&=int_A^B vec Fcdot dvec s\
&=int_A^B (F_x hat i + F_y hat j) cdot (dx hat i)\
&=int_{x_A}^{x_B} F_x dx\
&=F_x (x_B-x_A)\
end{align}

So, for the forward force $F_x=F$ along the forward path (displacement $(x_Q-x_P)>0$), we have
$$W_{PQ}= (F )(x_Q-x_P)>0.$$

For the backward force $F_x=-F$ along the backward path (displacement $(x_R-x_Q)<0$), we have
$$W_{QR}= (-F)(x_R-x_Q)>0.$$

Note that the lower limit and upper limit refer to the start and end of the oriented path... the limits control the orientation of the path.

Note the lower limit and the upper limits are not necessarily the smaller and larger of x-coordinates.