There is a line integral over a straight line as follows (the problem here: https://tutorial.math.lamar.edu/Solutions/CalcIII/LineIntegralsPtI/Prob1.aspx):
Line Integral: $$\int 3x^2 - 2y\ ds$$ Equation of the line: $$2y=7x-9$$
In the solution, The author suggests that due to the reverse direction of line, parametrization is necessary but I don't know why. In fact, I can't figure out why my solution (below) leads to a wrong result (negative the correct one), despite the fact that I'm defining the integration interval($x$) from 3 to 1 so it is supposed to be consistent with the direction. $$\int 3x^2 - 2y\ ds \rightarrow \int_{3}^{1} 3x^2 - 2y \sqrt{1+\left (\frac{7}{2} \right )^2} dx $$
Edit: once I simplified the main function, but now it is as in the problem.
You are traversing the line in the opposite direction than the "natural one" (usually $x$ increases), this is why the result should be negative.
If you want your answer to be correct, you need to do $\int_3^1$, not $\int_1^3$, since you should start from the point where $x$ is 3, and finish where $x$ is 1.
Edit: The line segment is the set of points $(x,\frac{7x-9}2)$ from $x=3$ to $x=1$. Thus $ds=-\sqrt{1+\big(\frac72\big)^2}\,dx$, and we have $$\int_\mathcal C 3x^2-2y\,ds = \int_3^1\Big(3x^2-2\Big(\frac{7x-9}{2}\Big)\Big)\Big(-\sqrt{1+\Big(\frac72\Big)^2}\,dx\Big) = 8\sqrt{53}.$$ Notice that $ds$ has a minus sign because the orientation is reversed. If this is confusing, you can just pick a parametrisation where the directions are aligned with the integral. The strangeness here occurs because we are going backwards. (Intuitively, it should make sense that $\Delta s \approx - c\,\Delta x$.)