Vector Line Integral Direction of Limits

[Master1022]

Homework Statement

Evaluate the line integral for the vector field $v = (2x - y) \hat i + (-yz^2) \hat j + -(y^2 z) \hat k$ from (1,1) to (0,1)

Relevant Equations

Integration

Hi,

I apologise as I know I have made similar posts to this in the past and I thought I finally understood it. However, this solution seems to disagree on a technicality. I know the answer ends up as 0, but I still want to understand this from a conceptual point.

Question: Evaluate the line integral for the vector field $v = (2x - y) \hat i + (-yz^2) \hat j + -(y^2 z) \hat k$ from (1,1) to (0,1) when we are in the xy-plane (i.e. z = 0).

Approach:
Given that we are in the xy-plane where $z = 0$ and at $y = 1$, the vector field becomes:
$$ v = (2x - 1) \hat i $$

Then when I set up the integral, $d \vec r = -dx \hat i$ and thus:
$$ \int_0^1 \vec v \cdot d \vec r = \int_0^1 (2x - 1) \hat i \cdot -dx \hat i = - \int_0^1 (2x - 1) dx = 0 $$

However, the answer writes the integral as:
$$ - \int_1^0 (2x - 1) dx = 0 $$

I know that the answers are the same, but if the integral wasn't 0, then the answers would be different. I thought the convention was to define $d \vec r$ in the direction of the path and the limits in terms of the increasing parameter.

I am sure this has been answered in similar posts but am unable to find them. Any guidance would be greatly appreciated.

Thanks.

 

[BvU]

Seems to me the book answer is wrong ...
(@haruspex ?)

 

[haruspex]

Seems to me the book answer is wrong ...
(@haruspex ?)

I agree.
(Odd to write the range with endpoints expressed as only 2D.)

 

[Master1022]

Thank you very much @BvU and @haruspex for clarifying!