negative $dx$ in line integral

Question

In the third line of this solution, the author added a negative sign in front of $dx$. Does anyone know why that is?

My solution has $\int^0_1x\sqrt{1+(\frac{dy}{dx})^2}dx$ instead of $\int^0_1x\sqrt{1+(\frac{dy}{dx})^2}(-dx)$.

Answer 1

Note that the differential in the integral $$\int_{\cal C} x\>ds\tag{1}$$ is $ds$, and not a $dx$ or $dy$, as occurring in connection with vector fields ${\bf F}$. This scalar line element $ds$ is an "unsigned" quantity, like a volume element. When we compute $(1)$ we compute, e.g., the total heat content of the wire ${\cal C}$ when the temperature at the point $(x,y)$ is $\>=x$. It follows that you may parametrize ${\cal C}_3$ to your liking, e.g., in the form $${\cal C}_3:\quad x\mapsto\bigl(x,1-x^2\bigr)\qquad(0\leq x\leq1)\ .$$ You then obtain $$\int_{{\cal C}_3}x\>ds= \int_0^1x\>\sqrt{1+\left({\partial y\over\partial x}\right)^2}\>dx=\int_0^1x\sqrt{1+4x^2}\>dx=\ldots$$