I was studying line integral for some pdfs and came up with the following question: $ \int _ {\gamma} (x-2y ^ {2}) dy, \gamma $ is the arc of the parabola $ y = x ^ {2} $ from $(-2,4)$ to $(1,1).$
I used the parameterization $\gamma (t)=(t,t^2)$ , $\gamma {}'(t)=(1,2t)$ e $\left \|\gamma'(t) \right \|=\sqrt{1+4t^2}$, but when i do $\int_{-2}^{1}(t-2t^4)\sqrt{1+4t^2}dt$ It does not work.
According to the PDF the answer is 48.
Best Answer
Unless you copied it wrong, the integral is with respect to $y$. This means that it is a line integral of a vector field. Or, if you want, $dy=y'(t)\,dt=2t\,dt$. So your integral is $$ \int_{-2}^1 (t-2t^4)\,2t\,dt=48. $$