1. **Problem Statement:** Evaluate the line integral $$\int_C (x + 2y)\,dx + (x - y)\,dy$$ along the curve $C$. To make the problem smoother and easier, let's change the curve $C$ to be the line segment from $(0,0)$ to $(1,1)$. 2. **Formula and Explanation:** The line integral of a vector field $\mathbf{F} = P\mathbf{i} + Q\mathbf{j}$ along a curve $C$ parameterized by $\mathbf{r}(t) = (x(t), y(t))$, $a \leq t \leq b$, is given by: $$\int_C P\,dx + Q\,dy = \int_a^b \left[P(x(t), y(t)) \frac{dx}{dt} + Q(x(t), y(t)) \frac{dy}{dt}\right] dt$$ 3. **Parameterization of the Curve:** For the line segment from $(0,0)$ to $(1,1)$, a simple parameterization is: $$x(t) = t, \quad y(t) = t, \quad 0 \leq t \leq 1$$ 4. **Compute Derivatives:** $$\frac{dx}{dt} = 1, \quad \frac{dy}{dt} = 1$$ 5. **Substitute into the Integral:** $$P = x + 2y = t + 2t = 3t$$ $$Q = x - y = t - t = 0$$ So the integral becomes: $$\int_0^1 \left[3t \cdot 1 + 0 \cdot 1\right] dt = \int_0^1 3t \, dt$$ 6. **Evaluate the Integral:** $$\int_0^1 3t \, dt = 3 \int_0^1 t \, dt = 3 \left[ \frac{t^2}{2} \right]_0^1 = 3 \cdot \frac{1}{2} = \frac{3}{2}$$ **Final Answer:** $$\boxed{\frac{3}{2}}$$