Step 1

Given \(\displaystyle{P}={y}+{e}^{{x}}{\quad\text{and}\quad}{Q}={6}{x}+{\cos{{y}}}\)

Green's theorem is

\(\displaystyle\int_{{C}}{P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}=\int\int_{{C}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle=\int\int_{{C}}{\left({6}-{1}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={5}\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

Since C is closed triangle with vertices (0,0) , (0,2) & (2,2) which is oriented counter clockwise .

Since \(\displaystyle\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\) is the area of that triangle.

Step 2

Then

\(\displaystyle\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}=\frac{{1}}{{2}}\cdot{2}\cdot{2}={2}\)

Hence

\(\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}={5}\cdot\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

=5*2=10

Therefore (2) option is correct.

Given \(\displaystyle{P}={y}+{e}^{{x}}{\quad\text{and}\quad}{Q}={6}{x}+{\cos{{y}}}\)

Green's theorem is

\(\displaystyle\int_{{C}}{P}{\left.{d}{x}\right.}+{Q}{\left.{d}{y}\right.}=\int\int_{{C}}{\left(\frac{{\partial{Q}}}{{\partial{x}}}-\frac{{\partial{P}}}{{\partial{y}}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle=\int\int_{{C}}{\left({6}-{1}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={5}\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

Since C is closed triangle with vertices (0,0) , (0,2) & (2,2) which is oriented counter clockwise .

Since \(\displaystyle\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\) is the area of that triangle.

Step 2

Then

\(\displaystyle\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}=\frac{{1}}{{2}}\cdot{2}\cdot{2}={2}\)

Hence

\(\displaystyle\int_{{C}}{\left({y}+{e}^{{x}}\right)}{\left.{d}{x}\right.}+{\left({6}{x}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}={5}\cdot\int\int_{{C}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

=5*2=10

Therefore (2) option is correct.