Hey I'm doing a line integral question, and am struggling to find ds in terms of dx, I have a suspicion I'm just being stupid but I've been trying to do this for 20 minutes.
$$\int_{C}^{} F(x,y) ds$$ where $F(x,y) = 1$, over the curve $$y = \frac{1}2 x^2 - \frac{1}4 lnx$$ from $\left(1,\frac{1}2\right)$ to $(2, \left(2-\frac{1}4ln2)\right) $
I know I need to express it in terms of $x$, and that $$ds=\sqrt{1^2+\left(\frac{dy}{dx}\right)^2}dx$$ Worked out dy/dx as $$\frac{dy}{dx}=x - \frac{1}{4x}$$
Substituting into $ds$
$$ds= \sqrt{1+\left(x-\frac{1}{4x}\right)^2} dx$$
I can't seem to solve this
Notice that $$dS = \sqrt{1+\left(x-\frac{1}{4x}\right)^2} = \sqrt{x^2+\frac{1}{16x^2}+\frac{1}{2}} = \sqrt{\left(x+\frac{1}{4x}\right)^2}$$