그린 정리

Direct evaluation will lead to three integrals, one for each side of the triangle. The integral from $(0,0)$ and $(1,-2)$ and the integral from $(1,2)$ to $(0,0)$ will each contain six (nonconstant) terms. This does not look very efficient. So let's try Green's theorem. Denote by $T$, the triangle

$$T = \big\{ \ (x,y)\ \big|\ 0\le x\le 1,\ -2x\le y\le 2x\ \big\} {-50pt}[40pt][40pt]{  }$$

It has boundary $\partial T=C$, oriented counterclockwise as desired. So, by Green's theorem,

$$\begin{aligned} \int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} &= \int_{\partial T} \big\{\big(\tfrac{3}{2}y^2 + e^{-y} +\sin x\big)\mathrm{d}x +\big(\tfrac{1}{2}x^2+x-x e^{-y}\big)\mathrm{d}y\big\} \\ &= \iint _T \big\{ \tfrac{\partial}{\partial x}\big(\tfrac{1}{2}x^2+x-x e^{-y}\big) -\tfrac{\partial}{\partial y}\big(\tfrac{3}{2}y^2 + e^{-y} +\sin x\big) \big\}\mathrm{d}x\mathrm{d}y \\ &= \iint _T \big\{ \big(x + 1-e^{-y}\big) - \big(3y - e^{-y} \big) \big\}\mathrm{d}x\mathrm{d}y \\ &= \iint _T \big\{ x -3y + 1 \big\}\mathrm{d}x\mathrm{d}y \end{aligned}$$

Now

$$\begin{aligned} \iint _T \mathrm{d}x\mathrm{d}y &=\text{Area}(T) = \frac{1}{2}(4)(1) = 2 \\ \iint _T y\ \mathrm{d}x\mathrm{d}y &= 0 \qquad\text{since $y$ is odd under $y\rightarrow-y$} \\ \iint _T x\ \mathrm{d}x\mathrm{d}y &=\int_0^1\mathrm{d}x\int_{-2x}^{2x} \mathrm{d}y\ x =\int_0^1 4x^2\ \mathrm{d}x =\frac{4}{3} \end{aligned}$$

So

$$\int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} = \frac{4}{3}-3\times 0 +2 =\frac{10}{3}$$