Path-independence of $\int_C (x+y^3)\,dx+(e^y+3xy^2)\,dy$

1.1. Let \(P(x, y) = x + y^3\) and \(Q(x, y) = e^y + 3xy^2\) be the components of the 1-form \(\omega = P\,dx + Q\,dy\).

1.2. We seek a \(C^1\) function \(f(x, y)\) on \(\mathbb{R}^2\) such that \(\nabla f = (P, Q)\), which requires:\[\frac{\partial f}{\partial x} = x + y^3, \qquad \frac{\partial f}{\partial y} = e^y + 3xy^2.\]

1.3. Integrating \(\frac{\partial f}{\partial x} = x + y^3\) with respect to \(x\) while holding \(y\) constant yields\[f(x, y) = \int (x + y^3)\,dx = \frac{1}{2}x^2 + xy^3 + g(y),\]where \(g(y)\) is an arbitrary differentiable function of \(y\).

1.4. Differentiating this expression with respect to \(y\):\[\frac{\partial f}{\partial y} = 3xy^2 + g'(y).\]

1.5. Setting \(\frac{\partial f}{\partial y} = Q(x, y) = e^y + 3xy^2\) gives \(3xy^2 + g'(y) = e^y + 3xy^2\), so \(g'(y) = e^y\).

1.6. Integrating with respect to \(y\) gives \(g(y) = e^y + C\). Setting \(C = 0\) yields the potential function\[f(x, y) = \frac{1}{2}x^2 + xy^3 + e^y.\]

1.7. Direct verification: \(\frac{\partial f}{\partial x} = x + y^3 = P(x, y)\) and \(\frac{\partial f}{\partial y} = 3xy^2 + e^y = Q(x, y)\), so \(\nabla f = (P, Q)\) on all of \(\mathbb{R}^2\).

2.1. Let \(C\) be any piecewise smooth curve in \(\mathbb{R}^2\) parameterized by \(\mathbf{r}(t) = (x(t), y(t))\) for \(a \le t \le b\), with initial point \(A = \mathbf{r}(a)\) and terminal point \(B = \mathbf{r}(b)\).

2.2. By the chain rule:\[\int_C (x + y^3)\,dx + (e^y + 3xy^2)\,dy = \int_a^b \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\,dt = \int_a^b \frac{d}{dt}[f(\mathbf{r}(t))]\,dt.\]

2.3. By the Fundamental Theorem of Calculus:\[\int_C (x + y^3)\,dx + (e^y + 3xy^2)\,dy = f(\mathbf{r}(b)) - f(\mathbf{r}(a)) = f(B) - f(A).\]

2.4. The value of the line integral depends solely on the endpoints \(A\) and \(B\), and is completely independent of the path \(C\) connecting them.