Avaluate the line integral $$\int_C(x-y^2)dx $$ over the curve $C:$ $y=x-\frac{x^2}{4}$, from $(0,0)\to(4,0)$.
In my textbook, the expression for a line integral over one of the derivatives is:
$$\int_Cf(x,y)dx=\int_a^b f(x(t);y(t))x'(t)dt$$ so it seems fair that if I take $x=t$, I'd change $y$ for the expression of the parable and the integral would just be:
$$\int_{x_1}^{x_2}(x-(x-\frac{x^2}{4})^2)dx$$
However, there's also another expression: $$\int_C f(x,y)dx=\int_{x_1}^{x_2}f(x, y(x))\sqrt{1+(\frac{dy}{dx})^2}dx$$
Which yields a completely different result, so one of them must be wrong.
I'm just very confused here. Which one do I use and what's the difference between the two? Where do each of them come from?
You can parametrize the curve $ C $ as
$$x=2t$$ $$y=2t-t^2$$
the result is $$2\int_{t=0}^{t=2}(2t-(2t-t^2)^2)dt$$
Your last integral is given by $$\int_{x_1}^{x_2}f(x,y)dl$$ which is a circulation along the curve and where $$dl=\sqrt{1+(\frac{dy}{dx})^2}dx$$