I've to find the line integral
$\int_{x=a}^{0} \vec{F} \cdot \vec{d l}$
evaluated over a line in the $x-y$ plane given as $y=1$ where $\vec{F}=\hat{\imath}$.
My solution:
Since we are moving from $x=a$ to $x=0$ ,our length element should be $\vec{d l}=-d x \hat{i}$. The line integral is then :$\int_{a}^{0} \vec{F} \cdot d l=\int_{a}^{0} \hat{\imath} \cdot(-d x \hat{i})=-\int_{a}^{0} d x=a$. Which is incorrect.
What is incorrect in my approach?
The point is that when you write $\vec{dl} = -dx\hat{i}$, you are already setting the orientation of the path, but the orientation is already included in your order of integration from $a$ to $0$, which says you are going backwards. So, you should use one of the following: $$\int_{0}^{a}\hat{i}\cdot(-dx\hat{i}) \quad \mbox{or} \quad \int_{a}^{0}\hat{i}\cdot dx\hat{i}$$
Note that there is another approach which is more rigorous and prevents thes sort of mistakes. You have $\vec{F}(\vec{r}) = (1,0)$ and the parametrized curve is $\gamma(t) = (t, 1)$, where $t$ goes from $a$ to $0$. Now, $\gamma'(t) = (1,0)$ and $F(\gamma(t))=(1,0)$, so: $$\int_{C}\vec{F}(\vec{r})\cdot d\vec{l} = \int_{a}^{0}\vec{F}(\gamma(t))\cdot\gamma'(t)dt = \int_{a}^{0}(1,0)\cdot (1,0)dt = \int_{a}^{0}dt = -a$$