91Ó°ÊÓ

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

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.

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

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.

$\frac{∂Q}{∂x}-\frac{∂P}{∂y}=0$

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.

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

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