Calculate
$$\int_\gamma (2e^{{(y-2x)}^2}-y)dx+(-e^{({y-2x})^2}+2x)dy$$
where $\gamma$ is the section of the curve $y=x^2$ from the point $(0,0)$ to $(2,4)$.
A tip I got was to first close the region by adding a line between the two points, which is given by $y=2x$. Then Green's theorem can be used to evaluate it. Afterwards calculate the line integral of the added line with the same function and subtract it from the previous calculation. In other words the answer is given by...
$$\oint_{\gamma+\gamma_1} \textbf{F}\cdot d\textbf{r}-\int_{\gamma_1} \textbf{F}\cdot d\textbf{r}$$
where $\gamma_1$ is the added line between the two points. Using Green's theorem on the first integral
$$3\int_0^2\int_{x^2}^{2x}dydx=4$$
The second integral can be calculated easily as a regular line integral with the parameterization $\textbf{r}(x)=(x,2x)$ which gives
$$\int_0^2(2-2x)+2(-1+2x)dx=4$$ The answer is therefor $4-4=0$, which is wrong as the correct answer is supposedly $8$. My professor said that regarding curls, the region should always be on our left if we were to "walk" along the boundary. My thinking is then to reverse the bounds on the line integral because on that line we are walking from the point $(2,4)$ to $(0,0)$ instead of the opposite way which was unconsciously assumed by my calculation. Doing so gives
$$3\int_0^2\int_{x^2}^{2x}dydx-\int_2^0(2-2x)+2(-1+2x)dx=4-(-4)=8$$
Is my thinking regarding the switch of the bounds correct? Or is there a flaw within the calculation itself?