The correct answer is $\boxed{\left( {{\text{y}}_2^2 – {\text{y}}_1^2} \right) + \left( {{\text{x}}_2^2 – {\text{x}}_1^2} \right)}$.
The line integral $\int\limits_{{{\text{P}}_1}}^{{{\text{P}}_2}} {\left( {{\text{ydx}} + {\text{xdy}}} \right)} $ is the integral of the vector field $\left( {{\text{y}}, {\text{x}}} \right)$ along the curve $C$ from $P_1$ to $P_2$. In this case, the vector field is $\left( {{\text{y}}, {\text{x}}} \right)$ and the curve $C$ is the semicircle $P_1, P_2$ shown in the figure.
To evaluate the line integral, we can use the formula $$\int\limits_{{{\text{P}}_1}}^{{{\text{P}}_2}} {\left( {{\text{ydx}} + {\text{xdy}}} \right)} = \left( {{\text{y}}_2^2 – {\text{y}}_1^2} \right) + \left( {{\text{x}}_2^2 – {\text{x}}_1^2} \right)$$
Substituting the values of $P_1$ and $P_2$, we get $$\int\limits_{{{\text{P}}_1}}^{{{\text{P}}_2}} {\left( {{\text{ydx}} + {\text{xdy}}} \right)} = \left( {{\text{y}}_2^2 – {\text{y}}_1^2} \right) + \left( {{\text{x}}_2^2 – {\text{x}}_1^2} \right)$$
Therefore, the line integral $\int\limits_{{{\text{P}}_1}}^{{{\text{P}}_2}}
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