Show that the form $\omega = 2xy^3dx + 3x^2y^2dy$ is closed and calculate $\int_{c}\omega$, where $c$ is given by $y=x^2$, $(0,0)$ to $(x,y)$.
My attempt.
$\begin{eqnarray*} d\omega &=& d(2xy^3)\wedge dx + d(3x^2y^2)\wedge dy\\ &=& (2y^3dx + 6xy^2dy)\wedge dx + (6xy^2dx + 6xy^2dy)dy\\ & = & -6xy^2dx\wedge dy + 6xy^2dx\wedge dy\\ &=& 0. \end{eqnarray*}$
Now, by definition
$$\int_{c}\omega = \int_{a}^{b}\left(\sum_{i}a_{i}(t)\frac{dx_{i}}{dt}\right)dt$$
where $\omega = \sum_{i}a_{i}dx_{i}$. Then
$$\sum_{i}a_{i}(t)\frac{dx_{i}}{dt} = 2x(t)y(t)^3\frac{dx}{dt} + 3x(t)^2y(t)^2\frac{dy}{dt}.$$
We can write $c: [0,x] \to \{(x,x^2)\}$ given by $c(t) = (t,t^2)$, by taking $x(t) = t$ and $y(t) = t^2$. So, we have
$\begin{eqnarray*} \int_{c}\omega &=& \int_{0}^{x}\left(2x(t)y(t)^3x'(t) + 3x(t)^2y(t)^2y'(t)\right)dt\\ &=& \int_{0}^{x}2t^{7}dt + \int_{0}^{x}6t^{7}dt\\ & = & \frac{t^{8}}{4}\Bigg|_{0}^{x} + \frac{3t^{8}}{4}\Bigg|_{0}^{x}\\ & = & t^{8}\Bigg|_{0}^{x}\\ &=& x^{8} \end{eqnarray*}$
Is correct? This is my first problem about Line Integrals on $1$-forms.
The differential form is even exact: $$ \omega = 2xy^3\,dx + 3x^2y^2\,dy = \frac{\partial(x^2y^3)}{\partial x} dx + \frac{\partial(x^2y^3)}{\partial y} dy = d(x^2y^3). $$ From this follows that $\omega$ is closed, since $d^2\omega = 0$ for any differential form $\omega.$
Therefore we can calculate the integral as $$ \int_c \omega = \int_{(0,0)}^{(x,x^2)} d(x^2y^3) = x^2y^3 {\Bigg|}_{(0,0)}^{(x,x^2)} = x^2(x^2)^3 - 0^2 \cdot 0^3 = x^8. $$
Thus, your result is correct.