I have this problem:$$\int_C f(x,y)ds$$where $f(x,y)=2x$, and $C-> y=2x^2\in (-1,2) \cup(1,2)$
I have parameterized $x=t$ and $y=2t^2$. After transforming $ds$ to $dt$ and after simplifying, I am left with $$\frac{1}{16}\int_? u^\frac{1}{2}du$$
At this point, I do not know what bounds (C) to set. My inkling is to use t=-1 to 1 but am unsure.Any advice?
Taking it back one step, you must have had $$ I = 2\int_{-1}^1t\sqrt{1+16t^2} dt $$ which is zero because the integrand is odd and the limits of integration are symmetric about $t=0$
It looks like you made the substitution $u=1+16t^2$
but that relation is not invertible on the interval $(-1,1)$
the integral would need to be split into separate integrals from $-1$ to $0$ and from $0$ to $1$ .
If you did this carefully you would get $I=0$