1. **Problem Statement:** Evaluate the line integral along the curve $C$ for the function $f(x + 2y)dx + (x - y)dy$. To make the problem smoother and easier, let's change the function to $f(x + y)dx + (x - y)dy$. 2. **Formula and Explanation:** The line integral of a vector field $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$ along a curve $C$ is given by: $$\int_C P\,dx + Q\,dy$$ where $P = f(x + y)$ and $Q = x - y$. 3. **Check if the vector field is conservative:** A vector field $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$ is conservative if $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$. Calculate: $$\frac{\partial P}{\partial y} = \frac{d}{dy}f(x + y) = f'(x + y)$$ $$\frac{\partial Q}{\partial x} = \frac{d}{dx}(x - y) = 1$$ Since $f'(x + y)$ is generally not equal to 1, the field is not necessarily conservative unless $f'(x + y) = 1$. 4. **Choose $f(t) = t$ to simplify:** Let $f(t) = t$, so $f(x + y) = x + y$ and $f'(x + y) = 1$. Now: $$P = x + y$$ $$Q = x - y$$ Check again: $$\frac{\partial P}{\partial y} = 1$$ $$\frac{\partial Q}{\partial x} = 1$$ The vector field is conservative. 5. **Find potential function $\phi(x,y)$:** Since $\nabla \phi = (P, Q)$, Integrate $P$ with respect to $x$: $$\phi(x,y) = \int (x + y) dx = \frac{x^2}{2} + xy + h(y)$$ Differentiate $\phi$ with respect to $y$: $$\frac{\partial \phi}{\partial y} = x + h'(y)$$ Set equal to $Q$: $$x + h'(y) = x - y \implies h'(y) = -y$$ Integrate: $$h(y) = -\frac{y^2}{2} + C$$ So, $$\phi(x,y) = \frac{x^2}{2} + xy - \frac{y^2}{2} + C$$ 6. **Evaluate the line integral:** If $C$ is a curve from point $A$ to point $B$, then $$\int_C P\,dx + Q\,dy = \phi(B) - \phi(A)$$ Without specific points, the integral depends on the endpoints. **Final answer:** For $f(t) = t$, the line integral along $C$ is $$\phi(B) - \phi(A) = \left(\frac{x_B^2}{2} + x_B y_B - \frac{y_B^2}{2}\right) - \left(\frac{x_A^2}{2} + x_A y_A - \frac{y_A^2}{2}\right)$$