${\mathbf{\int }}_{\mathbf{C}}{\mathit{e}}^{\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{y}\mathit{d}\mathit{x}\mathbf{-}{\mathit{e}}^{\mathbf{x}}\mathit{s}\mathit{i}\mathit{n}\mathit{y}\mathit{d}\mathit{y}$, whereCis the broken line from$\mathit{A}\mathbf{=}\mathbf{}\mathbf{(}\mathit{l}\mathit{n}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}$to $\mathit{D}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{}$and then from $\mathit{D}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathit{B}\mathbf{=}\mathbf{(}\mathbf{-}\mathit{l}\mathit{n}\mathbf{}\mathbf{2}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{.}\mathbf{}$

The given information is ${\int }_{\mathrm{C}}{e}^{\mathrm{x}}cosydx-e^{\mathrm{x}}sinydy$.

The Green's theorem connects a line integral around a simple closed curve C to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

Consider an equation and name it as equation 1.

${\int }_{\mathrm{C}}{e}^{\mathrm{x}}cosydx-e^{\mathrm{x}}sinydy......\left(1\right)$

Compare equation (1) to Green’s Theorem.

$$\begin{gathered} P=e^{x}cosy \\ Q=-e^{x}siny \end{gathered}$$

Find the differentiation with respect to.

$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0$

This implies that the surface integral will also be zero, which further leads to the line integral along C is negative that of line integral moving from B to A.

Let the value be as follows:

$$\begin{gathered} y=0 \\ dy=0 \end{gathered}$$

Find the Integral.

$$\begin{gathered} l=-{\int }_{-\ln 2}^{\ln 2}{e}^{x}dx \\ =-\left(2-\frac{1}{2}\right) \\ =-\frac{3}{2} \end{gathered}$$

Hence, the solution to this problem is $-\frac{3}{2}.$