Step 1

Using Green theorem the integral becomes:

\(\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{\left({1}-{x}{y}^{{3}}\right)}{\left.{d}{y}\right.}=\int\int_{{S}}{\left(\frac{\partial}{{\partial{x}}}{\left({1}-{x}{y}^{{3}}\right)}-\frac{\partial}{{\partial{y}}}{x}{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

where S is the surface enclosed by C.

Step 2

Thus, solve the integral noting that \(\displaystyle{x}\le{y}\le{x}+{3}{\quad\text{and}\quad}−{1}\le{x}\le{1}\):

\(\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{\left({1}-{x}{y}^{{3}}\right)}{\left.{d}{y}\right.}=\int\int_{{S}}{\left(\frac{\partial}{{\partial{x}}}{\left({1}-{x}{y}^{{3}}\right)}-\frac{\partial}{{\partial{y}}}{x}{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle=\int\int_{{S}}{\left(-{y}^{{3}}-{2}{x}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{{x}=-{1}}}^{{1}}}{\int_{{{y}={x}}}^{{{x}+{3}}}}{\left(-{y}^{{3}}-{2}{x}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{{x}=-{1}}}^{{1}}}{\left(-\frac{{{\left({x}+{3}\right)}^{{4}}-{x}^{{4}}}}{{4}}-{x}{\left({\left({x}+{3}\right)}^{{2}}-{x}^{{2}}\right)}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle={\int_{{{x}=-{1}}}^{{1}}}\frac{{-{12}{x}^{{3}}-{78}{x}^{{2}}-{144}{x}-{81}}}{{4}}{\left.{d}{x}\right.}\)

Step 3

Neglecting all odd powers of x:

\(\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{\left({1}-{x}{y}^{{3}}\right)}{\left.{d}{y}\right.}={\int_{{{x}=-{1}}}^{{1}}}\frac{{-{78}{x}^{{2}}-{81}}}{{4}}{\left.{d}{x}\right.}\)

\(\displaystyle=\frac{{1}}{{4}}{\int_{{{x}=-{1}}}^{{1}}}-{78}{x}^{{2}}-{81}{\left.{d}{x}\right.}\)

\(\displaystyle=\frac{{1}}{{2}}{\left(-\frac{{78}}{{3}}-{81}\right)}\)

\(\displaystyle=-\frac{{107}}{{2}}\)

Step 4

Thus, the integral is \(\displaystyle-\frac{{107}}{{2}}=-{53.5}\).

Using Green theorem the integral becomes:

\(\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{\left({1}-{x}{y}^{{3}}\right)}{\left.{d}{y}\right.}=\int\int_{{S}}{\left(\frac{\partial}{{\partial{x}}}{\left({1}-{x}{y}^{{3}}\right)}-\frac{\partial}{{\partial{y}}}{x}{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

where S is the surface enclosed by C.

Step 2

Thus, solve the integral noting that \(\displaystyle{x}\le{y}\le{x}+{3}{\quad\text{and}\quad}−{1}\le{x}\le{1}\):

\(\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{\left({1}-{x}{y}^{{3}}\right)}{\left.{d}{y}\right.}=\int\int_{{S}}{\left(\frac{\partial}{{\partial{x}}}{\left({1}-{x}{y}^{{3}}\right)}-\frac{\partial}{{\partial{y}}}{x}{y}^{{2}}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle=\int\int_{{S}}{\left(-{y}^{{3}}-{2}{x}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{{x}=-{1}}}^{{1}}}{\int_{{{y}={x}}}^{{{x}+{3}}}}{\left(-{y}^{{3}}-{2}{x}{y}\right)}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={\int_{{{x}=-{1}}}^{{1}}}{\left(-\frac{{{\left({x}+{3}\right)}^{{4}}-{x}^{{4}}}}{{4}}-{x}{\left({\left({x}+{3}\right)}^{{2}}-{x}^{{2}}\right)}\right)}{\left.{d}{x}\right.}\)

\(\displaystyle={\int_{{{x}=-{1}}}^{{1}}}\frac{{-{12}{x}^{{3}}-{78}{x}^{{2}}-{144}{x}-{81}}}{{4}}{\left.{d}{x}\right.}\)

Step 3

Neglecting all odd powers of x:

\(\displaystyle\int_{{C}}{x}{y}^{{2}}{\left.{d}{x}\right.}+{\left({1}-{x}{y}^{{3}}\right)}{\left.{d}{y}\right.}={\int_{{{x}=-{1}}}^{{1}}}\frac{{-{78}{x}^{{2}}-{81}}}{{4}}{\left.{d}{x}\right.}\)

\(\displaystyle=\frac{{1}}{{4}}{\int_{{{x}=-{1}}}^{{1}}}-{78}{x}^{{2}}-{81}{\left.{d}{x}\right.}\)

\(\displaystyle=\frac{{1}}{{2}}{\left(-\frac{{78}}{{3}}-{81}\right)}\)

\(\displaystyle=-\frac{{107}}{{2}}\)

Step 4

Thus, the integral is \(\displaystyle-\frac{{107}}{{2}}=-{53.5}\).