Question

find line integral
\[ \int_{C}(x^2y^3-sqrt x)dy\] C is the arc of the curve y =\[\ sqrt x\] from (1,1) to (4,2)
Please, help

Answer 1

I havent done line integrals but you can try wolfram alpha, sorry I couldnt help

Answer 2

\[y^2+2e^{-xy}=6\\\frac{d}{dx}\left[y^2+2e^{-xy}\right]=\frac{d}{dx}[6]\\2y\frac{dy}{dx}-2\left(y+x\frac{dy}{dx}\right)e^{-xy}=0\\2\frac{dy}{dx}\left(y-xe^{-xy}\right)=2ye^{-xy}\\\frac{dy}{dx}=\frac{ye^{-xy}}{y-xe^{-xy}}=\frac{y}{ye^{xy}-x}\\\quad\implies\frac{dy}{dx}=\frac2{2e^0-0}=1\\\frac{d^2y}{dx^2}=\frac{ye^{xy}-x-y(ye^{xy}(y+x\frac{dy}{dx})+\frac{dy}{dx}e^{xy}-1)}{(ye^{xy}-x)^2}\\\quad\implies\frac{d^2y}{dx^2}=\frac{2e^0-0-2(2e^0(2+0)+e^0-1)}{(2e^0-0)^2}=\frac{2-2(4)}{4}=\frac{2(-3)}{4}=-\frac32\]